Quiz about IONIC-EQUILIBRIUM IN GENERAL AWARENESS

Question 1

In the third group of qualitive analysis the precipitating reagent is `NH_(4)CINH_(4)OH`. The function of `NH_(4)CI` is toA. Increase the ionization of `NH_(4)OH`B. Supress the ionization of `NH_(4)OH`C. Stabilise the hydroxides of group cationsD. Convert the ions of group third into their respective chlorides.

Accepted Answer

NA

Question 2

Which of the following solutions will have pH close to 1.0 ?A. `100 mL M10 HCl + 100 mL M10 NaOH`B. `160 mL 0.075 M H_(2)SO_(4) + 40 mL M10 NaOH`C. `55 mL M10 HCl + 45 mL M10 NaOH`D. `75 mL M5 HCl + 25 mL M5 NaOH`

Accepted Answer

NA

Question 3

The number of moles of `Ca(OH)_(2)` that must be dissolved to make 250 mL solution in water of pH = 10.65 isA. `5.6 xx 10^(-5)`B. `6.5 xx 10^(-5)`C. `4.5 xx 10^(-5)`D. `5.4 xx 10^(-5)`

Accepted Answer

NA

Question 4

Which of the following solutions will have pH close to 1.0A. 100 ml of (M10) HCl + 100 ml of (M10) NaOHB. 55 ml of (M10) HCl + 45ml of (M10) NaOHC. 10 ml of (M10) HCl + 90 ml of (M10) NaOHD. 75 ml of (M5) HCl + 25 ml of (M15) NaOH

Accepted Answer

NA

Question 5

The cyanide ion `CN^(-) N_(2)` and `N_(2)` are isoelectronic. But in contrast to `CN^(-) N_(2)` is chemically inert because ofA. Low bond energyB. Absence of bond polarityC. Unsymmetrical electron distributionD. Presence of more number of electrons in bonding orbitals

Accepted Answer

NA

Question 6

Given pH of a solution A is 3 and it is mixed with another solution B having pH 2. If both mixed then resultant pH of the solution will beA. 3.2B. 1.9C. 3.4D. 3.5

Accepted Answer

NA

Question 7

An aqueous solution of metal chloride `MCI_(2)(0.05M)` is saturated with `H_(2)S (0.1M)`. The minimum `pH` at which metal sulphide will be precipiated is `[K_(sp)MS = 5 xx 10^(-21)K_(1)(H_(2)S) = 10^(-7)K_(2)(H_(2)S) = 10^(-14)`.A. `3.25`B. `2.50`C. `1.50`D. `1.25`

Accepted Answer

NA

Question 8

During neutralisation of an acid by a base the end point refers for the completion of reaction. The detection of end point in acid -base neutralisation is usually made by an acid-base indicator. An acid-base indicator is itself a weak acid (Phenolphthlein) or a weak base (Mrthyl orange). At about `50%` ionisation which depends on the medium the anion furnished by an indicator (acid) or cation furnished by indicator (basic) imparts its characteristic colour to solution at point. For example phenolphthalein the dissociation is `underset(Colourless)(H In hArr H^(+))+underset(Pink)(In^(-)) K_(H In)= ([H^(+)][In^(-)])([H In])` favoured in presence of alkali and pink colour of phenolphalein ion is noticed as soon as the medium changes to alkaline nature. The end point of acid-base neutralisation not necessarily coincides with equivalent point but it is closer and closer to equivalence point. Also at equivalence point of acid-base neutralisation pH is not necessarliy equal to 7. Bromophenol blue is an acid indicator having dissociation constant `5.84xx10^(-5)` . The percentage of coloured ion furnished at a pH of `4.84` isA. `80%`B. `40%`C. `20%`D. `90%`

Accepted Answer

NA

Question 9

On adding sold potassium cyanide to waterA. pH will increaseB. pH will decreaseC. pH will not changeD. Electrical conducting will not change

Accepted Answer

NA

Question 10

The pH of 0.0001 N solution of KOH will beA. 4B. 6C. 10D. 12

Accepted Answer

NA

Question 11

During neutralisation of an acid by a base the end point refers for the completion of reaction. The detection of end point in acid -base neutralisation is usually made by an acid-base indicator. An acid-base indicator is itself a weak acid (Phenolphthlein) or a weak base (Mrthyl orange). At about `50%` ionisation which depends on the medium the anion furnished by an indicator (acid) or cation furnished by indicator (basic) imparts its characteristic colour to solution at point. For example phenolphthalein the dissociation is `underset(Colourless)(H In hArr H^(+))+underset(Pink)(In^(-)) K_(H In)= ([H^(+)][In^(-)])([H In])` favoured in presence of alkali and pink colour of phenolphalein ion is noticed as soon as the medium changes to alkaline nature. The end point of acid-base neutralisation not necessarily coincides with equivalent point but it is closer and closer to equivalence point. Also at equivalence point of acid-base neutralisation pH is not necessarliy equal to 7. Which of the following statements are correct ? (1) Phenolphthalein is not a good indicator for weak alkali titrations (2) Phenolphthalein does not give pink colour with weak alkalies as `NH_(4)OH` (3) Phenolphthalein is an acid indicator and imparts colour in basic medium (4) Phenolphthalein is a basic indicator and imparts colour in basic medium (5) Phenolphthalein furnishes coloured cationA. `1245`B. `1234`C. `13`D. `24`

Accepted Answer

NA

Question 12

During neutralisation of an acid by a base the end point refers for the completion of reaction. The detection of end point in acid -base neutralisation is usually made by an acid-base indicator. An acid-base indicator is itself a weak acid (Phenolphthlein) or a weak base (Mrthyl orange). At about `50%` ionisation which depends on the medium the anion furnished by an indicator (acid) or cation furnished by indicator (basic) imparts its characteristic colour to solution at point. For example phenolphthalein the dissociation is `underset(Colourless)(H In hArr H^(+))+underset(Pink)(In^(-)) K_(H In)= ([H^(+)][In^(-)])([H In])` favoured in presence of alkali and pink colour of phenolphalein ion is noticed as soon as the medium changes to alkaline nature. The end point of acid-base neutralisation not necessarily coincides with equivalent point but it is closer and closer to equivalence point. Also at equivalence point of acid-base neutralisation pH is not necessarliy equal to 7. The dissociation constant of an acid-base indicator which furnished coloured anion is `1xx10^(-5)`. The pH of solution at which indicator is `80%` in dissociated form isA. `5.2310`B. `5.6020`C. `8.3980`D. `8.4820`

Accepted Answer

NA

Question 13

During neutralisation of an acid by a base the end point refers for the completion of reaction. The detection of end point in acid -base neutralisation is usually made by an acid-base indicator. An acid-base indicator is itself a weak acid (Phenolphthlein) or a weak base (Mrthyl orange). At about `50%` ionisation which depends on the medium the anion furnished by an indicator (acid) or cation furnished by indicator (basic) imparts its characteristic colour to solution at point. For example phenolphthalein the dissociation is `underset(Colourless)(H In hArr H^(+))+underset(Pink)(In^(-)) K_(H In)= ([H^(+)][In^(-)])([H In])` favoured in presence of alkali and pink colour of phenolphalein ion is noticed as soon as the medium changes to alkaline nature. The end point of acid-base neutralisation not necessarily coincides with equivalent point but it is closer and closer to equivalence point. Also at equivalence point of acid-base neutralisation pH is not necessarliy equal to 7. Which among the following statements are correct ? (1) At equivalence point of NaOH and HCI pH = 7 (2) At equivalence point to of NaOH and `CH_(3)COOHpH gt 7` (3) At equivalence point of `NH_(4)OH` and `HCI pH lt 7` (4) an indicator shoes best results if equivalence point is within the pH range `pK_(a)` of In `+1` (5) At equivalence point of `NH_(4)OH` and formic acid `pH lt 7`A. `1234`B. `1345`C. `145`D. `1235`

Accepted Answer

NA

Question 14

During neutralisation of an acid by a base the end point refers for the completion of reaction. The detection of end point in acid -base neutralisation is usually made by an acid-base indicator. An acid-base indicator is itself a weak acid (Phenolphthlein) or a weak base (Mrthyl orange). At about `50%` ionisation which depends on the medium the anion furnished by an indicator (acid) or cation furnished by indicator (basic) imparts its characteristic colour to solution at point. For example phenolphthalein the dissociation is `underset(Colourless)(H In hArr H^(+))+underset(Pink)(In^(-)) K_(H In)= ([H^(+)][In^(-)])([H In])` favoured in presence of alkali and pink colour of phenolphalein ion is noticed as soon as the medium changes to alkaline nature. The end point of acid-base neutralisation not necessarily coincides with equivalent point but it is closer and closer to equivalence point. Also at equivalence point of acid-base neutralisation pH is not necessarliy equal to 7. The dissociation constant of an acid-base indicator which furnishes coloured cation is `1xx10^(-5)` . The pH of solution at which indicator will furnish its colour is A. 5B. 9C. 6D. 10

Accepted Answer

NA

Question 15

What `(H_(3)O^(+))` must be maintained in a saturated `H_(2)S` solution to precipitate `Pb^(2+)` but not `Zn^(2+)` from a solution in which each ion is present at a concetration of `0.01 M ? (K_(SP)`for `H_(2)S= 1.1xx10^(-22) K_(SP)`for `ZnS=1.0xx10^(-21))`

Accepted Answer

NA

Question 16

During neutralisation of an acid by a base the end point refers for the completion of reaction. The detection of end point in acid -base neutralisation is usually made by an acid-base indicator. An acid-base indicator is itself a weak acid (Phenolphthlein) or a weak base (Mrthyl orange). At about `50%` ionisation which depends on the medium the anion furnished by an indicator (acid) or cation furnished by indicator (basic) imparts its characteristic colour to solution at point. For example phenolphthalein the dissociation is `underset(Colourless)(H In hArr H^(+))+underset(Pink)(In^(-)) K_(H In)= ([H^(+)][In^(-)])([H In])` favoured in presence of alkali and pink colour of phenolphalein ion is noticed as soon as the medium changes to alkaline nature. The end point of acid-base neutralisation not necessarily coincides with equivalent point but it is closer and closer to equivalence point. Also at equivalence point of acid-base neutralisation pH is not necessarliy equal to 7. The indicator phenolphalein is a tautomeric mixture of two forms as given below  Which of the following statements are correct ? (1) The form I is referred as quinonoid form and is deeper in colour (2) The form I is referred as quinonoid form and is lighter in colour (3) The form II is more stable in alkaline solution (4) The change in pH form acidic to alkaline solution brings in the more and more conversion of I form to II form (5) The form I is more stable in acidic mediumA. `1234`B. `1345`C. `345`D. `2345`

Accepted Answer

NA

Question 17

A sample of AgCI was treated with `5.00mL` of `1.5M` `Na_(2)CO_(3)` solubility to give `Ag_(2)CO_(3)` . The remaining solution contained `0.0026g of CI^(-)` per litre. Calculate the solubility product of AgCI. `(K_(SP)for Ag_(2)CO_(3)=8.2xx10^(-12))`A. `1.1 xx 10^(-2)`B. `1.71 xx 10^(-11)`C. `1.71 xx 10^(-10)`D. `1.32 xx 10^(-9)`

Accepted Answer

NA

Question 18

An acid `HA` is `40%` dissociated in an aqueous solution. The hydronium ion concentration of its `0.2M` solution would beA. 0.08 MB. 0.4 MC. 0.2 MD. None of the above.

Accepted Answer

NA

Question 19

The hydronium ion concentration in `5xx10^(-4)` M aqueous solution of NaOH at `25^@C` is A. zeroB. `5xx10^(-4) M`C. `2xx10^(-11) M`D. `2xx10^(3) M`

Accepted Answer

NA

Question 20

Calculate the per cent error in hydronium ion concentration made by neglecting the ionisation of water in `1.0 xx 10^(-6) M NaOH`

Accepted Answer

NA

Question 21

The process in which solution containing more than one typeof ions (either cation or anion) and in which one ion undergoes almst complete precipitation followed by the second ion and similarly second ion undergoes complete precipitation followed by the third ion is known as preferential precipitation. It is the solubility of the saltnot the solubility product of the salt by means of which one can predict the preferential precipitation of the salt or ion among the lot of ions in the solution. For example in a solution containing `Cl^(-) Br^(-)` and `I^(-)` ions if `Ag^(+)` ions are added then out of three the less soluble salt is precipitated first . If the addition of `Ag^(+)` ions is continued eventually a stage is reached when the next lesser soluble salt starts precipitating along with the least soluble salt and so on. If the stiochiometry of the salt is same then the salt with the minimum `K_(sp)` ( solubility product ) will have minimum solubility and will precipitate first followed by the salt of next higher solubility product and so on. If the stiochiometry of teh salt is not same then from the solubility product data alone we cannot predict which ions will precipitate first. For example a solution containing `Cl^(-)` and `CrO_(4)^(2-)`. In order to predict which ion will precipitate first we have to calculate the amount of `Ag^(+)` ions needed to start precipitation through the solubility product data given. When `AgNO_93)` is added to the solution the minimum of the two concentrations needed to start the precipitation will be reached and thus corresponding `[Ag^(+)]` to start the precipitation increases. Its concentration eventually becomes equal to the value required for `CrO_(4)^(2-)` . At this stage practically almost the whole of `Cl^(-)` ions get precipitated . Addition of more `AgNO_(3)` causes simultaneous precipitation of both the ions together. Suppose a student of chemistry performs the following two sets of experiments in order to known the preferential precipitation of ions as well as its precipitated percentage amount. Experiment 1. He prepare a solution of anions composed of 0.1 M each of `Cl^(-)Br^(-)` and `I^(-)` . Further he adds gradually solid `AgNO_(3)` to this solution. He assumes that volume of the solution does not change after the addition of solid `AgNO_(3)(K_(sp)AgCl=1.0 xx 10^(-9)M^(2)K_(sp) AgBr=10^(-10)M^(2)` and `K_(sp) AgI=10^(-12)M^(2))` Experiment -2. He prepares a solution of cations `Cd^(2+)(0.2 M)` and `Bi^(3+) (0.3 m)` . Now he adds `S^(2-)` ions into the solution of cations in order to separate them by selective precipitation `Cd^(2+) `forms yellow precipitate of CdS and `Bi^(3+)` forms black precipitate of `Bi_(2)S_(3)` with `S^(2-)` ions respectively. `(K_(sp)Bi_(2)S_(3) = 9 xx 10^(-25) M^(5)` and `K_(sp)CdS =2 xx 10^(-20)M^(2))`.Answer the following questions ont he basis of the above write up. In experiment -2 what `%` of the metal ions get precipitated at which `S^(2-)` get saturated with another ion ?A. `10%`B. `80%`C. `20%`D. `90%`

Accepted Answer

NA

Question 22

Which of the following statement(s) is (are) correct?A. The `p^(H)` of `10^(-8)` M solution of `HCl` is `8`B. The conjugate base of `H_(2)PO_(4)^(-)` is `HPO_(4)^(2-)`C. Auto protolysis constant of water increases with temperatureD. When a solution of weak monoprotic acid is titrated against a strong base at half of neutraliastion point `p^(H) = 12p^(Ka)`

Accepted Answer

NA

Question 23

The process in which solution containing more than one typeof ions (either cation or anion) and in which one ion undergoes almst complete precipitation followed by the second ion and similarly second ion undergoes complete precipitation followed by the third ion is known as preferential precipitation. It is the solubility of the saltnot the solubility product of the salt by means of which one can predict the preferential precipitation of the salt or ion among the lot of ions in the solution. For example in a solution containing `Cl^(-) Br^(-)` and `I^(-)` ions if `Ag^(+)` ions are added then out of three the less soluble salt is precipitated first . If the addition of `Ag^(+)` ions is continued eventually a stage is reached when the next lesser soluble salt starts precipitating along with the least soluble salt and so on. If the stiochiometry of the salt is same then the salt with the minimum `K_(sp)` ( solubility product ) will have minimum solubility and will precipitate first followed by the salt of next higher solubility product and so on. If the stiochiometry of teh salt is not same then from the solubility product data alone we cannot predict which ions will precipitate first. For example a solution containing `Cl^(-)` and `CrO_(4)^(2-)`. In order to predict which ion will precipitate first we have to calculate the amount of `Ag^(+)` ions needed to start precipitation through the solubility product data given. When `AgNO_93)` is added to the solution the minimum of the two concentrations needed to start the precipitation will be reached and thus corresponding `[Ag^(+)]` to start the precipitation increases. Its concentration eventually becomes equal to the value required for `CrO_(4)^(2-)` . At this stage practically almost the whole of `Cl^(-)` ions get precipitated . Addition of more `AgNO_(3)` causes simultaneous precipitation of both the ions together. Suppose a student of chemistry performs the following two sets of experiments in order to known the preferential precipitation of ions as well as its precipitated percentage amount. Experiment 1. He prepare a solution of anions composed of 0.1 M each of `Cl^(-)Br^(-)` and `I^(-)` . Further he adds gradually solid `AgNO_(3)` to this solution. He assumes that volume of the solution does not change after the addition of solid `AgNO_(3)(K_(sp)AgCl=1.0 xx 10^(-9)M^(2)K_(sp) AgBr=10^(-10)M^(2)` and `K_(sp) AgI=10^(-12)M^(2))` Experiment -2. He prepares a solution of cations `Cd^(2+)(0.2 M)` and `Bi^(3+) (0.3 m)` . Now he adds `S^(2-)` ions into the solution of cations in order to separate them by selective precipitation `Cd^(2+) `forms yellow precipitate of CdS and `Bi^(3+)` forms black precipitate of `Bi_(2)S_(3)` with `S^(2-)` ions respectively. `(K_(sp)Bi_(2)S_(3) = 9 xx 10^(-25) M^(5)` and `K_(sp)CdS =2 xx 10^(-20)M^(2))`.Answer the following questions ont he basis of the above write up. In experiment 02 what is the maximum concentration of `S^(2-)` ion at which one of the two metal ions gets maximum precipitation ?A. `10^(-8)` MB. `2 xx 10^(-7)M`C. `2 xx 10^(-9)M`D. None

Accepted Answer

NA

Question 24

The process in which solution containing more than one typeof ions (either cation or anion) and in which one ion undergoes almst complete precipitation followed by the second ion and similarly second ion undergoes complete precipitation followed by the third ion is known as preferential precipitation. It is the solubility of the saltnot the solubility product of the salt by means of which one can predict the preferential precipitation of the salt or ion among the lot of ions in the solution. For example in a solution containing `Cl^(-) Br^(-)` and `I^(-)` ions if `Ag^(+)` ions are added then out of three the less soluble salt is precipitated first . If the addition of `Ag^(+)` ions is continued eventually a stage is reached when the next lesser soluble salt starts precipitating along with the least soluble salt and so on. If the stiochiometry of the salt is same then the salt with the minimum `K_(sp)` ( solubility product ) will have minimum solubility and will precipitate first followed by the salt of next higher solubility product and so on. If the stiochiometry of teh salt is not same then from the solubility product data alone we cannot predict which ions will precipitate first. For example a solution containing `Cl^(-)` and `CrO_(4)^(2-)`. In order to predict which ion will precipitate first we have to calculate the amount of `Ag^(+)` ions needed to start precipitation through the solubility product data given. When `AgNO_93)` is added to the solution the minimum of the two concentrations needed to start the precipitation will be reached and thus corresponding `[Ag^(+)]` to start the precipitation increases. Its concentration eventually becomes equal to the value required for `CrO_(4)^(2-)` . At this stage practically almost the whole of `Cl^(-)` ions get precipitated . Addition of more `AgNO_(3)` causes simultaneous precipitation of both the ions together. Suppose a student of chemistry performs the following two sets of experiments in order to known the preferential precipitation of ions as well as its precipitated percentage amount. Experiment 1. He prepare a solution of anions composed of 0.1 M each of `Cl^(-)Br^(-)` and `I^(-)` . Further he adds gradually solid `AgNO_(3)` to this solution. He assumes that volume of the solution does not change after the addition of solid `AgNO_(3)(K_(sp)AgCl=1.0 xx 10^(-9)M^(2)K_(sp) AgBr=10^(-10)M^(2)` and `K_(sp) AgI=10^(-12)M^(2))` Experiment -2. He prepares a solution of cations `Cd^(2+)(0.2 M)` and `Bi^(3+) (0.3 m)` . Now he adds `S^(2-)` ions into the solution of cations in order to separate them by selective precipitation `Cd^(2+) `forms yellow precipitate of CdS and `Bi^(3+)` forms black precipitate of `Bi_(2)S_(3)` with `S^(2-)` ions respectively. `(K_(sp)Bi_(2)S_(3) = 9 xx 10^(-25) M^(5)` and `K_(sp)CdS =2 xx 10^(-20)M^(2))`.Answer the following questions ont he basis of the above write up. What percent of the anions `Br^(-)` and `I^(-)` get precipitated respectively when the third ion starts precipitating in experiment -1 ?A. `90%99.9%`B. `99.9%90%`C. `80%90%`D. `90%80%`

Accepted Answer

NA

Question 25

The process in which solution containing more than one typeof ions (either cation or anion) and in which one ion undergoes almst complete precipitation followed by the second ion and similarly second ion undergoes complete precipitation followed by the third ion is known as preferential precipitation. It is the solubility of the saltnot the solubility product of the salt by means of which one can predict the preferential precipitation of the salt or ion among the lot of ions in the solution. For example in a solution containing `Cl^(-) Br^(-)` and `I^(-)` ions if `Ag^(+)` ions are added then out of three the less soluble salt is precipitated first . If the addition of `Ag^(+)` ions is continued eventually a stage is reached when the next lesser soluble salt starts precipitating along with the least soluble salt and so on. If the stiochiometry of the salt is same then the salt with the minimum `K_(sp)` ( solubility product ) will have minimum solubility and will precipitate first followed by the salt of next higher solubility product and so on. If the stiochiometry of teh salt is not same then from the solubility product data alone we cannot predict which ions will precipitate first. For example a solution containing `Cl^(-)` and `CrO_(4)^(2-)`. In order to predict which ion will precipitate first we have to calculate the amount of `Ag^(+)` ions needed to start precipitation through the solubility product data given. When `AgNO_93)` is added to the solution the minimum of the two concentrations needed to start the precipitation will be reached and thus corresponding `[Ag^(+)]` to start the precipitation increases. Its concentration eventually becomes equal to the value required for `CrO_(4)^(2-)` . At this stage practically almost the whole of `Cl^(-)` ions get precipitated . Addition of more `AgNO_(3)` causes simultaneous precipitation of both the ions together. Suppose a student of chemistry performs the following two sets of experiments in order to known the preferential precipitation of ions as well as its precipitated percentage amount. Experiment 1. He prepare a solution of anions composed of 0.1 M each of `Cl^(-)Br^(-)` and `I^(-)` . Further he adds gradually solid `AgNO_(3)` to this solution. He assumes that volume of the solution does not change after the addition of solid `AgNO_(3)(K_(sp)AgCl=1.0 xx 10^(-9)M^(2)K_(sp) AgBr=10^(-10)M^(2)` and `K_(sp) AgI=10^(-12)M^(2))` Experiment -2. He prepares a solution of cations `Cd^(2+)(0.2 M)` and `Bi^(3+) (0.3 m)` . Now he adds `S^(2-)` ions into the solution of cations in order to separate them by selective precipitation `Cd^(2+) `forms yellow precipitate of CdS and `Bi^(3+)` forms black precipitate of `Bi_(2)S_(3)` with `S^(2-)` ions respectively. `(K_(sp)Bi_(2)S_(3) = 9 xx 10^(-25) M^(5)` and `K_(sp)CdS =2 xx 10^(-20)M^(2))`.Answer the following questions ont he basis of the above write up. What `%`of the (ion) will get precipitate when the second ion start precipitating in experiment -1 ?A. `10%`B. `90%`C. `1%`D. `99%`

Accepted Answer

NA

Question 26

A monobasic acid solution has pH value of `5`. Its molarity is `0.005 M`. The degree of ionisation of the acid isA. `5 xx 10^(-3)`B. `2 xx 10^(-3)`C. `5 xx 10^(-2)`D. `2 xx 10^(-2)`

Accepted Answer

NA

Question 27

The process in which solution containing more than one typeof ions (either cation or anion) and in which one ion undergoes almst complete precipitation followed by the second ion and similarly second ion undergoes complete precipitation followed by the third ion is known as preferential precipitation. It is the solubility of the saltnot the solubility product of the salt by means of which one can predict the preferential precipitation of the salt or ion among the lot of ions in the solution. For example in a solution containing `Cl^(-) Br^(-)` and `I^(-)` ions if `Ag^(+)` ions are added then out of three the less soluble salt is precipitated first . If the addition of `Ag^(+)` ions is continued eventually a stage is reached when the next lesser soluble salt starts precipitating along with the least soluble salt and so on. If the stiochiometry of the salt is same then the salt with the minimum `K_(sp)` ( solubility product ) will have minimum solubility and will precipitate first followed by the salt of next higher solubility product and so on. If the stiochiometry of teh salt is not same then from the solubility product data alone we cannot predict which ions will precipitate first. For example a solution containing `Cl^(-)` and `CrO_(4)^(2-)`. In order to predict which ion will precipitate first we have to calculate the amount of `Ag^(+)` ions needed to start precipitation through the solubility product data given. When `AgNO_93)` is added to the solution the minimum of the two concentrations needed to start the precipitation will be reached and thus corresponding `[Ag^(+)]` to start the precipitation increases. Its concentration eventually becomes equal to the value required for `CrO_(4)^(2-)` . At this stage practically almost the whole of `Cl^(-)` ions get precipitated . Addition of more `AgNO_(3)` causes simultaneous precipitation of both the ions together. Suppose a student of chemistry performs the following two sets of experiments in order to known the preferential precipitation of ions as well as its precipitated percentage amount. Experiment 1. He prepare a solution of anions composed of 0.1 M each of `Cl^(-)Br^(-)` and `I^(-)` . Further he adds gradually solid `AgNO_(3)` to this solution. He assumes that volume of the solution does not change after the addition of solid `AgNO_(3)(K_(sp)AgCl=1.0 xx 10^(-9)M^(2)K_(sp) AgBr=10^(-10)M^(2)` and `K_(sp) AgI=10^(-12)M^(2))` Experiment -2. He prepares a solution of cations `Cd^(2+)(0.2 M)` and `Bi^(3+) (0.3 m)` . Now he adds `S^(2-)` ions into the solution of cations in order to separate them by selective precipitation `Cd^(2+) `forms yellow precipitate of CdS and `Bi^(3+)` forms black precipitate of `Bi_(2)S_(3)` with `S^(2-)` ions respectively. `(K_(sp)Bi_(2)S_(3) = 9 xx 10^(-25) M^(5)` and `K_(sp)CdS =2 xx 10^(-20)M^(2))`.Answer the following questions ont he basis of the above write up. Which of the following will precipitate first in experiment -1 ?A. `I^(-)`ionB. `Br^(-)` ionC. `Cl^(-)` ionD. cannot be predicted

Accepted Answer

NA

Question 28

How much volume of `0.1M Hac` should be added to `50mL` of `0.2M NaAc` solution to have a `pH 4.91`?

Accepted Answer

NA

Question 29

A mono basic weak acid solution has a molarity of 0.005 and pH of 5. What is its percentage ionisation in this solutino ?gtA. `2.0`B. 0.1C. 0.5D. 0.25

Accepted Answer

NA

Question 30

Some solutions are taken at specified concentration and have the following solution codes. Identify the four digit number abcd where ab = Sum of solution codes of those solutions which will have a singificant buffer action. Cd = Sum of solution codes of those solutions which for the given amount of acid or base and salt have maximum buffer capacity.

Accepted Answer

NA

Question 31

What concentartion of `Ac^(-)` ions will reduce `H_(3)O^(+)` ion to `2 xx 10^(-4) M` in `0.40 M` solution of `HAc ? [K_(a) (HAc) = 1.8 xx 10^(-5)]` Given your answer after multiplying by 1000.

Accepted Answer

NA

Question 32

Read of the following passage giving the role of `CO_(2)` buffer in controlling pH of blood. The importance of pH maintance in Blood. Maintenance of the pH in blood and intracellur fluids is absolutely crucial to the processes the occur in living ogranisms. This is primarily because the functioning of enzymes-catalysts for these processe- is sharply pH dependent. The normal pH value of blood plasma i `7.4`. Severe illness or dear can result from subtained variations fo a few tenths of pH unit. Among the factors that lead to a condition of acidosis in which there is decreas in the pH of blood are heart failure kidney failure diabetis mellitus persistent diarroheoa or a long term high protein diet temporary condition acidosis may result from proglonged intensive in ph of blood may several factors are involved in the control of the pH of blood. A particularly important one is the ratio of dissolved ` OH_(3)^(-)` to `H_(2)CO_(3).CO_(2)(g)` is moderarately soluble in water and in aqeous solution reactns only a limilated extent to product `H_(2)CrO_(3)` `CO_(2) + H_(2)O hArr H_(2)CO_(3(aq))` `H_(2)CO_(3) + H_(2)O hArr HCO_(3)^(-) + H_(3)O^(+) pK_(al) = 6.11` `HCO_(3)^(-) + H_(2)O hArr CO_(3)^(2-) + H_(3)O^(+) pK_(a2) = 10.25` In the `H_(2)CO_(3) HCO_(3)^(-)` buffer system we deal only with the first ionisation step `(K_(al)) H_(2)CO_(3)` is a wear acid and `HCO_(3)^(-)` is the conjugate base (salt). `CO_(2)` enters the blood from issues as the by product metabolic reaction. In lungs `CO_(2)(g)` is exchanged for `O_(2)(g)` which is transported throughout the body the blood. Following reaction occurs in the body If `CO_(2)` escape from the systemA. pH will decreasesB. pH will increaseC. `[H_(2)CO_(3)]` remains unchangedD. Forward reactions is promoted

Accepted Answer

NA

Question 33

How many of the following may act as buffer ? (a) `20mL 0.2 M -H_(2)CO_(3)` solution `+10 ml 0.2 M-NaOH` solution (b) `20 ml 0.2 M -H_(2)CO_(3)` solution `+20 ml 0.2 M-NaOH` solution (c) `20 mL 0.2 M-H_(2)CO_(3)` solution `+30 mL 0.2 M-NaOH` solution (d) `20 mL 0.2 M - H_(2)CO_(3)` solution `+10 mL 0.2 M-NaHCO_(3)` solution (e) `20 mL 0.2 M -H_(2)CO_(3)` solution `+10 mL 0.2 M-Na_(2)CO_(3)` solution (f) `20 mL 0.2 M -H_(2)CO_(3)` solution `+20 mL 0.2 M -Na_(2)CO_(3)` solution (g) `20 mL 0.2 M -H_(2)CO_(3)` solution `+30 ml 0.2 M-Na_(2)CO_(3)` solution

Accepted Answer

NA

Question 34

Read of the following passage giving the role of `CO_(2)` buffer in controlling pH of blood. The importance of pH maintance in Blood. Maintenance of the pH in blood and intracellur fluids is absolutely crucial to the processes the occur in living ogranisms. This is primarily because the functioning of enzymes-catalysts for these processe- is sharply pH dependent. The normal pH value of blood plasma i `7.4`. Severe illness or dear can result from subtained variations fo a few tenths of pH unit. Among the factors that lead to a condition of acidosis in which there is decreas in the pH of blood are heart failure kidney failure diabetis mellitus persistent diarroheoa or a long term high protein diet temporary condition acidosis may result from proglonged intensive in ph of blood may several factors are involved in the control of the pH of blood. A particularly important one is the ratio of dissolved ` OH_(3)^(-)` to `H_(2)CO_(3).CO_(2)(g)` is moderarately soluble in water and in aqeous solution reactns only a limilated extent to product `H_(2)CrO_(3)` `CO_(2) + H_(2)O hArr H_(2)CO_(3(aq))` `H_(2)CO_(3) + H_(2)O hArr HCO_(3)^(-) + H_(3)O^(+) pK_(al) = 6.11` `HCO_(3)^(-) + H_(2)O hArr CO_(3)^(2-) + H_(3)O^(+) pK_(a2) = 10.25` In the `H_(2)CO_(3) HCO_(3)^(-)` buffer system we deal only with the first ionisation step `(K_(al)) H_(2)CO_(3)` is a wear acid and `HCO_(3)^(-)` is the conjugate base (salt). `CO_(2)` enters the blood from issues as the by product metabolic reaction. In lungs `CO_(2)(g)` is exchanged for `O_(2)(g)` which is transported throughout the body the blood. The pH of blood system is maintained by a proper balance of `H_(2)CO_(3)` and `NaHCO_(2)` concentration. The volume of `5M NaHCO_(3)` solution should be mixed with a `10 m` sample of blood which is `2M` in `H_(2)CO_(3)` order to maitain its pHA. `40 mL`B. `38 mL`C. `50 mL`D. `78 mL`

Accepted Answer

NA

Question 35

Read of the following passage giving the role of `CO_(2)` buffer in controlling pH of blood. The importance of pH maintance in Blood. Maintenance of the pH in blood and intracellur fluids is absolutely crucial to the processes the occur in living ogranisms. This is primarily because the functioning of enzymes-catalysts for these processe- is sharply pH dependent. The normal pH value of blood plasma i `7.4`. Severe illness or dear can result from subtained variations fo a few tenths of pH unit. Among the factors that lead to a condition of acidosis in which there is decreas in the pH of blood are heart failure kidney failure diabetis mellitus persistent diarroheoa or a long term high protein diet temporary condition acidosis may result from proglonged intensive in ph of blood may several factors are involved in the control of the pH of blood. A particularly important one is the ratio of dissolved ` OH_(3)^(-)` to `H_(2)CO_(3).CO_(2)(g)` is moderarately soluble in water and in aqeous solution reactns only a limilated extent to product `H_(2)CrO_(3)` `CO_(2) + H_(2)O hArr H_(2)CO_(3(aq))` `H_(2)CO_(3) + H_(2)O hArr HCO_(3)^(-) + H_(3)O^(+) pK_(al) = 6.11` `HCO_(3)^(-) + H_(2)O hArr CO_(3)^(2-) + H_(3)O^(+) pK_(a2) = 10.25` In the `H_(2)CO_(3) HCO_(3)^(-)` buffer system we deal only with the first ionisation step `(K_(al)) H_(2)CO_(3)` is a wear acid and `HCO_(3)^(-)` is the conjugate base (salt). `CO_(2)` enters the blood from issues as the by product metabolic reaction. In lungs `CO_(2)(g)` is exchanged for `O_(2)(g)` which is transported throughout the body the blood. Important diagnostic analysis in the blood isA. B. C. D.

Accepted Answer

NA

Question 36

The pH of a solution obtaine by mixing `50 mL` of `0.4 N HCl` and `50 mL` of `0.2 N NaOH` isA. `log 2`B. `-log0.2`C. `1.0`D. `2.0`

Accepted Answer

NA

Question 37

The pH of a solution prepared by mixing 2.0 ml of HCl solution of pH 3.0 and 3.0 ml of NaOH of pH 10.0 isA. 2.5B. 3.5C. 5.5D. 6.5

Accepted Answer

NA

Question 38

The pH of a solution obtaine by mixing `50 mL` of `0.4 N HCl` and `50 mL` of `0.2 N NaOH` isA. log 2B. `1.0`C. `-log (0.05)`D. `2.0`

Accepted Answer

NA

Question 39

The pH of a 0.1 M `CH_(3)COOH` which is `2%` ionised in aqueous solution isA. `2.0`B. `1.0`C. `log((0.1)(0.001))`D. 2.7

Accepted Answer

NA

Question 40

For a hypothetical reaction `2A+B hArr C +` energy the activeation energy of forward reaction `(E_(af))` and backward reaction `(E_(ab))` are related asA. `E_(af)=E_(ab)`B. `E_(af)gtE_(ab)`C. `E_(af)+E_(ab)=0`D. `E_(af)=E_(ab)+DeltaH`

Accepted Answer

NA

Question 41

In the following reaction `HC_(2)O_(4)^(-)(aq)+PO_(4)^(3-)(aq)hArrHPO_(4)^(-2)(aq)+C_(2)O_(4)^(2-)(aq)` which are the two Bronsted bases?A. `HC_(2)O_(4)andPO_(4)^(2-)`B. `HPO_(4)^(2-)andC_(2)O_(4)^(2-)`C. `PO_(4)^(3-)andC_(2)O_(4)^(2-)`D. `HC_(2)O_(4)^(-)`and HPO_(4)^(2-)`

Accepted Answer

NA

Question 42

In the following reaction `HC_(2)O_(4)^(-)(aq)+PO_(4)^(3-)(aq)hArrHPO_(4)^(-2)(aq)+C_(2)O_(4)^(2-)(aq)` which are the two Bronsted bases?A. `HC_(2)O_(4)^(-)` and `PO_(4)^(3-)`B. `HPO_(4)^(2-)+C_(2)O_(4)^(2-)`C. `HC_(2)O_(4)^(-)` and `HPO_(4)^(2-)`D. `PO_(4)^(3-)` and `C_(2)O_(4)^(2-)`

Accepted Answer

NA

Question 43

In the following reaction `HC_2O_4^(-) (aq)+PO_4^(3-) (aq) hArr HPO_4^(2-)(aq)+C_2O_4^(2-)(aq)` which are the two Bronsted bases ?A. `HC_2O^- and PO_4^(3-)`B. `HPO_4^(2-) and C_2O_4^(2-)`C. `HC_2O_4^(-) and HPO_4^(2-)`D. `PO_4^(3-) and C_2O_4^(2-)`

Accepted Answer

NA

Question 44

`1 gH_(2)` gas `STP` is expanded so that the volume is doubled. Hence work done isA. `-22.4 L` atmB. `-11.2 `L atmC. `-144.8 ` L atmD. `-1.12 L` atm

Accepted Answer

NA

Question 45

The solubility of `Ag_(2)C_(2)O_(4)` at `25^(@)C` is `1.20 xx 10^(-11)`. A solution of `K_(2)C_(2)O_(4)` containing `0.15mol` in `500mL` water is mixed with excess of `Ag_(2)CO_(3)` till the following equilibrium is established `Ag_(2)CO_(3) + K_(2)C_(2)O_(4) hArr Ag_(2)C_(2)O_(4) + K_(2)CO_(3)` At equilibrium the solution constains `0.03 mol` of `K_(2)CO_(3)`. Assuming that the degree of dissociation of `K_(2)C_(2)O_(4)` and `K_(2)CO_(3)` to be equal calculate the solubility product of `Ag_(2)CO_(3)`. [Take `100%` ionisation of `K_(2)C_(2)O_(4)` and `K_(2)CO_(3)]`

Accepted Answer

NA

Question 46

If `alpha` is the fraction of HI dissociated at equilibrium in the reaction `2HI(g)hArrH_2(g)+I_2(g)` starting with the 2 moles of HI. Then the total number of moles of reactants and products at equilibrium areA. `1.0`B. `1+alpha`C. `2.0`D. `2+2alpha`

Accepted Answer

NA

Question 47

Which one of the following statements is not true?A. the conjugate base of `H_(2)PO_(4)^(-)` is `HPO_(4)^(2-)`B. `pH+pOH=14` for all aqueous solutions at `25^(@)C`.C. The `pH` of `1xx10^(-8)M HCl` is `8`.D. `96500` coulombs of electrcity when passed through a `CuSO_(4)` solution deposities `1` gram equivalent of copper at the cathode.

Accepted Answer

NA

Question 48

The conjugate base of `H_(2)PO_(4)^(-)` is A. `PO_(4)^(3-)`B. `P_(2)O_(5)`C. `H_(3)PO_(4)`D. `HPO_(4)^(2-)`

Accepted Answer

NA

Question 49

When rain is accompained by a thunderstorm the collected rain water will have a `pH`A. influenced by occurrence of thunderstormB. depends upon the amount of dust in waterC. slightly lower than that of rainwater without thunderstormD. slightly higher than that when thunderstorm is not there

Accepted Answer

NA

Question 50

When rain is accompained by a thunderstorm the collected rain water will have a `pH`A. Slightly lower than that of rain water without thunderstromB. uninfluenced by occurrenance of thunderstromC. which depends on the amount of dust in air .D.

Accepted Answer

NA

Explore topic-wise InterviewSolutions in .

Which of the following solutions will have pH close to 1.0 ?A. `100 mL M//10 HCl + 100 mL M//10 NaOH`B. `160 mL 0.075 M H_(2)SO_(4) + 40 mL M//10 NaOH`C. `55 mL M//10 HCl + 45 mL M//10 NaOH`D. `75 mL M//5 HCl + 25 mL M//5 NaOH`

The number of moles of `Ca(OH)_(2)` that must be dissolved to make 250 mL solution in water of pH = 10.65 isA. `5.6 xx 10^(-5)`B. `6.5 xx 10^(-5)`C. `4.5 xx 10^(-5)`D. `5.4 xx 10^(-5)`

Which of the following solutions will have pH close to 1.0A. 100 ml of (M/10) HCl + 100 ml of (M/10) NaOHB. 55 ml of (M/10) HCl + 45ml of (M/10) NaOHC. 10 ml of (M/10) HCl + 90 ml of (M/10) NaOHD. 75 ml of (M/5) HCl + 25 ml of (M/15) NaOH

The cyanide ion, `CN^(-), N_(2)` and `N_(2)` are isoelectronic. But in contrast to `CN^(-), N_(2)` is chemically inert, because ofA. Low bond energyB. Absence of bond polarityC. Unsymmetrical electron distributionD. Presence of more number of electrons in bonding orbitals

Given pH of a solution A is 3 and it is mixed with another solution B having pH 2. If both mixed then resultant pH of the solution will beA. 3.2B. 1.9C. 3.4D. 3.5

An aqueous solution of metal chloride `MCI_(2)(0.05M)` is saturated with `H_(2)S (0.1M)`. The minimum `pH` at which metal sulphide will be precipiated is `[K_(sp)MS = 5 xx 10^(-21),K_(1)(H_(2)S) = 10^(-7),K_(2)(H_(2)S) = 10^(-14)`.A. `3.25`B. `2.50`C. `1.50`D. `1.25`

On adding sold potassium cyanide to waterA. pH will increaseB. pH will decreaseC. pH will not changeD. Electrical conducting will not change

The pH of 0.0001 N solution of KOH will beA. 4B. 6C. 10D. 12

What `(H_(3)O^(+))` must be maintained in a saturated `H_(2)S` solution to precipitate `Pb^(2+)` , but not `Zn^(2+)` from a solution in which each ion is present at a concetration of `0.01 M ? (K_(SP)`for `H_(2)S= 1.1xx10^(-22), K_(SP)`for `ZnS=1.0xx10^(-21))`

An acid `HA` is `40%` dissociated in an aqueous solution. The hydronium ion concentration of its `0.2M` solution would beA. 0.08 MB. 0.4 MC. 0.2 MD. None of the above.

The hydronium ion concentration in `5xx10^(-4)` M aqueous solution of NaOH at `25^@C` , is :A. zeroB. `5xx10^(-4) M`C. `2xx10^(-11) M`D. `2xx10^(3) M`

Calculate the per cent error in hydronium ion concentration made by neglecting the ionisation of water in `1.0 xx 10^(-6) M NaOH`

A monobasic acid solution has pH value of `5`. Its molarity is `0.005 M`. The degree of ionisation of the acid isA. `5 xx 10^(-3)`B. `2 xx 10^(-3)`C. `5 xx 10^(-2)`D. `2 xx 10^(-2)`

How much volume of `0.1M Hac` should be added to `50mL` of `0.2M NaAc` solution to have a `pH 4.91`?

A mono basic weak acid solution has a molarity of 0.005 and pH of 5. What is its percentage ionisation in this solutino ?gtA. `2.0`B. 0.1C. 0.5D. 0.25

What concentartion of `Ac^(-)` ions will reduce `H_(3)O^(+)` ion to `2 xx 10^(-4) M` in `0.40 M` solution of `HAc ? [K_(a) (HAc) = 1.8 xx 10^(-5)]` Given your answer after multiplying by 1000.

The pH of a solution obtaine by mixing `50 mL` of `0.4 N HCl` and `50 mL` of `0.2 N NaOH` isA. `log 2`B. `-log0.2`C. `1.0`D. `2.0`

The pH of a solution prepared by mixing 2.0 ml of HCl solution of pH 3.0 and 3.0 ml of NaOH of pH 10.0 isA. 2.5B. 3.5C. 5.5D. 6.5

The pH of a solution obtaine by mixing `50 mL` of `0.4 N HCl` and `50 mL` of `0.2 N NaOH` isA. log 2B. `1.0`C. `-log (0.05)`D. `2.0`

The pH of a 0.1 M `CH_(3)COOH` which is `2%` ionised in aqueous solution isA. `2.0`B. `1.0`C. `log((0.1)/(0.001))`D. 2.7

For a hypothetical reaction, `2A+B hArr C +` energy , the activeation energy of forward reaction `(E_(af))` and backward reaction `(E_(ab))` are related asA. `E_(af)=E_(ab)`B. `E_(af)gtE_(ab)`C. `E_(af)+E_(ab)=0`D. `E_(af)=E_(ab)+DeltaH`

In the following reaction `HC_(2)O_(4)^(-)(aq)+PO_(4)^(3-)(aq)hArrHPO_(4)^(-2)(aq)+C_(2)O_(4)^(2-)(aq)`, which are the two Bronsted bases?A. `HC_(2)O_(4)andPO_(4)"^(2-)`B. `HPO_(4)"^(2-)andC_(2)O_(4)^(2-)`C. `PO_(4)"^(3-)andC_(2)O_(4)^(2-)`D. `HC_(2)O_(4)^(-)`and HPO_(4)^(2-)`

In the following reaction `HC_(2)O_(4)^(-)(aq)+PO_(4)^(3-)(aq)hArrHPO_(4)^(-2)(aq)+C_(2)O_(4)^(2-)(aq)`, which are the two Bronsted bases?A. `HC_(2)O_(4)^(-)` and `PO_(4)^(3-)`B. `HPO_(4)^(2-)+C_(2)O_(4)^(2-)`C. `HC_(2)O_(4)^(-)` and `HPO_(4)^(2-)`D. `PO_(4)^(3-)` and `C_(2)O_(4)^(2-)`

In the following reaction `HC_2O_4^(-) (aq)+PO_4^(3-) (aq) hArr HPO_4^(2-)(aq)+C_2O_4^(2-)(aq)`, which are the two Bronsted bases ?A. `HC_2O^- and PO_4^(3-)`B. `HPO_4^(2-) and C_2O_4^(2-)`C. `HC_2O_4^(-) and HPO_4^(2-)`D. `PO_4^(3-) and C_2O_4^(2-)`

`1 gH_(2)` gas `STP` is expanded so that the volume is doubled. Hence, work done isA. `-22.4 L` atmB. `-11.2 `L atmC. `-144.8 ` L atmD. `-1.12 L` atm

If `alpha` is the fraction of HI dissociated at equilibrium in the reaction, `2HI(g)hArrH_2(g)+I_2(g)` starting with the 2 moles of HI. Then the total number of moles of reactants and products at equilibrium areA. `1.0`B. `1+alpha`C. `2.0`D. `2+2alpha`

The conjugate base of `H_(2)PO_(4)^(-)` is :A. `PO_(4)^(3-)`B. `P_(2)O_(5)`C. `H_(3)PO_(4)`D. `HPO_(4)^(2-)`

When rain is accompained by a thunderstorm, the collected rain water will have a `pH`:A. influenced by occurrence of thunderstormB. depends upon the amount of dust in waterC. slightly lower than that of rainwater without thunderstormD. slightly higher than that when thunderstorm is not there

When rain is accompained by a thunderstorm, the collected rain water will have a `pH`:A. Slightly lower than that of rain water without thunderstromB. uninfluenced by occurrenance of thunderstromC. which depends on the amount of dust in air .D.

1501.	In the third group of qualitive analysis, the precipitating reagent is `NH_(4)CI//NH_(4)OH`. The function of `NH_(4)CI` is toA. Increase the ionization of `NH_(4)OH`B. Supress the ionization of `NH_(4)OH`C. Stabilise the hydroxides of group cationsD. Convert the ions of group third into their respective chlorides.
Answer» Correct Answer - C common ion effect

1502.	Which of the following solutions will have pH close to 1.0 ?A. `100 mL M//10 HCl + 100 mL M//10 NaOH`B. `160 mL 0.075 M H_(2)SO_(4) + 40 mL M//10 NaOH`C. `55 mL M//10 HCl + 45 mL M//10 NaOH`D. `75 mL M//5 HCl + 25 mL M//5 NaOH`
Answer» Correct Answer - B::D `pH = 1 rArr [H^(+)] = 10^(-1)` `:. [H^(+)] = 0.1 = (N_(a)V_(a) - N_(b)V_(b))/((V_(a) + V_(b))`

1503.	The number of moles of `Ca(OH)_(2)` that must be dissolved to make 250 mL solution in water of pH = 10.65 isA. `5.6 xx 10^(-5)`B. `6.5 xx 10^(-5)`C. `4.5 xx 10^(-5)`D. `5.4 xx 10^(-5)`
Answer» Correct Answer - A We know that pH + pOH = 3.35 pOH = (14-10.65) = 3.35 `[OH^(-)] = 10^(-3.35) = 4.47 xx 10^(-4) M` One molecule of `Ca(OH)_(2) = (4.47 xx 10^(-4))/(2) = 2.235 xx 10^(-4) M` No. of moles in `250 mL = (2.235 xx 10^(-4))/(4) = 5.6 xx 10^(-5)`

1504.	Which of the following solutions will have pH close to 1.0A. 100 ml of (M/10) HCl + 100 ml of (M/10) NaOHB. 55 ml of (M/10) HCl + 45ml of (M/10) NaOHC. 10 ml of (M/10) HCl + 90 ml of (M/10) NaOHD. 75 ml of (M/5) HCl + 25 ml of (M/15) NaOH
Answer» Correct Answer - D As `N_(1)V_(1) gt N_(2)V_(2)` So acid is left at the end of reaction `N_("final solution") = [H^(+)] = (N_(1)V_(1) - N_(2)V_(2))/(V_(1)+V_(2))` `= ((1)/(5)xx 75-(1)/(5)xx 25)/(75+25) ` `= (1)/(10) = 0.1 = pH = -log[H^(+)] = 1`

1505.	The cyanide ion, `CN^(-), N_(2)` and `N_(2)` are isoelectronic. But in contrast to `CN^(-), N_(2)` is chemically inert, because ofA. Low bond energyB. Absence of bond polarityC. Unsymmetrical electron distributionD. Presence of more number of electrons in bonding orbitals
Answer» Correct Answer - B Similar atoms are linked to each other and thus there is not polarity.

1506.	Given pH of a solution A is 3 and it is mixed with another solution B having pH 2. If both mixed then resultant pH of the solution will beA. 3.2B. 1.9C. 3.4D. 3.5
Answer» Correct Answer - B pH of the solution A = 3 `[H^(+)]_(B) = 10^(-3)M`. pH of the solution B = 2 `[H^(+)]_(B) = 10^(-2)M` `[H^(+)]_(B) = 10^(-3) + 10^(-2) = 10^(-3) + 10 xx 10^(-3) = 11 xx 10^(-3)`. `pH = -log(11 xx 10^(-3)) = 3 - log 11` `= 3 - 1.04 = 1.95`.

1507.	An aqueous solution of metal chloride `MCI_(2)(0.05M)` is saturated with `H_(2)S (0.1M)`. The minimum `pH` at which metal sulphide will be precipiated is `[K_(sp)MS = 5 xx 10^(-21),K_(1)(H_(2)S) = 10^(-7),K_(2)(H_(2)S) = 10^(-14)`.A. `3.25`B. `2.50`C. `1.50`D. `1.25`
Answer» Correct Answer - C `K_(sp) MS = [M^(2+)] [S^(2-)]` `5 xx 10^(-21) = (0.05) [S^(2-)]` For precipitation `Q_(sp) = K_(sp)` `:. [S^(2-)] = (5 xx 10^(-21))/(5xx10^(-2)) = 10^(-19)` For `H_(2)S:` `H_(2)S hArr H^(o+) +HS^(Theta) (K_(1))` `HS^(Theta) hArr H^(o+) +S^(2-) (K_(2))` `K_(1)xx K_(2) =([H^(o+)]^(2)[S^(2-)])/(H_(2)S)` `10^(-7) xx 10^(-14) = ([H^(o+)]^(2)[10^(-19)])/(0.1)` `[H^(o+)]^(2) = 10^(-3), rArr` [Taking negative logaritham both sides] `-2 log [H^(o+)] = 3` `pH = (3)/(2) = 1.5`

1508.	During neutralisation of an acid by a base, the end point refers for the completion of reaction. The detection of end point in acid -base neutralisation is usually made by an acid-base indicator. An acid-base indicator is itself a weak acid (Phenolphthlein) or a weak base (Mrthyl orange). At about `50%` ionisation which depends on the medium, the anion furnished by an indicator (acid) or cation furnished by indicator (basic) imparts its characteristic colour to solution at point. For example phenolphthalein, the dissociation is `underset("Colourless")(H In hArr H^(+))+underset("Pink")(In^(-)), K_(H In)= ([H^(+)][In^(-)])/([H In])` favoured in presence of alkali and pink colour of phenolphalein ion is noticed as soon as the medium changes to alkaline nature. The end point of acid-base neutralisation not necessarily coincides with equivalent point but it is closer and closer to equivalence point. Also at equivalence point of acid-base neutralisation pH is not necessarliy equal to 7. Bromophenol blue is an acid indicator having dissociation constant `5.84xx10^(-5)` . The percentage of coloured ion furnished at a pH of `4.84` is:A. `80%`B. `40%`C. `20%`D. `90%`
Answer» Correct Answer - A `HBPh hArr H^(+)+Bph^(-)` `K_(a)= ([H^(+)][BPh^(-)])/([HBPh])` `pH = 4.84` `:. [H^(+)] = 1.44xx10^(-5)` `:. 5.84xx10^(-5) = (1.44xx10^(-5).Calpha)/(C(1-alpha))` `:. alpha = 80%`

1509.	On adding sold potassium cyanide to waterA. pH will increaseB. pH will decreaseC. pH will not changeD. Electrical conducting will not change
Answer» Correct Answer - A `CN^(-) + H_(2)O rarr HCN + OH^(-)` Because `OH^(-)` concentration is increased.

1510.	The pH of 0.0001 N solution of KOH will beA. 4B. 6C. 10D. 12
Answer» Correct Answer - C `[OH^(-)] = 0.001 N, pOH = 4, pH + pOH = 14` `pH = 14 - pOH = 14 - 4 = 10`.

1528.	How much volume of `0.1M Hac` should be added to `50mL` of `0.2M NaAc` solution to have a `pH 4.91`?
Answer» It is an acidic buffer `pH = pK_(a) + "log" (["Salt"])/(["Acid"])` `4.91 = 4.76 + log [("Salt")/("Acid")]` `:. log [("salt")/("Acid")] = (4.91 - 4.76 ) = 0.15` `(["Salt"])/(["Acid"]) = "Antilog" (0.15) = 1.41` mmoles of `NaAc = 50 xx 0.2 = 10` mmoles of HAc required `= 0.1 M xxV` `[("Salt")/("Acid")] = 1.41 = (10)/(0.1xxV)` `rArr V = (10)/(0.1 xx 1.41) = 70.92mL` `:.` Volume of 0.1M HAc required `= 70.92mL`

1511.	During neutralisation of an acid by a base, the end point refers for the completion of reaction. The detection of end point in acid -base neutralisation is usually made by an acid-base indicator. An acid-base indicator is itself a weak acid (Phenolphthlein) or a weak base (Mrthyl orange). At about `50%` ionisation which depends on the medium, the anion furnished by an indicator (acid) or cation furnished by indicator (basic) imparts its characteristic colour to solution at point. For example phenolphthalein, the dissociation is `underset("Colourless")(H In hArr H^(+))+underset("Pink")(In^(-)), K_(H In)= ([H^(+)][In^(-)])/([H In])` favoured in presence of alkali and pink colour of phenolphalein ion is noticed as soon as the medium changes to alkaline nature. The end point of acid-base neutralisation not necessarily coincides with equivalent point but it is closer and closer to equivalence point. Also at equivalence point of acid-base neutralisation pH is not necessarliy equal to 7. Which of the following statements are correct ? (1) Phenolphthalein is not a good indicator for weak alkali titrations (2) Phenolphthalein does not give pink colour with weak alkalies as `NH_(4)OH` (3) Phenolphthalein is an acid indicator and imparts colour in basic medium (4) Phenolphthalein is a basic indicator and imparts colour in basic medium (5) Phenolphthalein furnishes coloured cationA. `1,2,4,5`B. `1,2,3,4`C. `1,3`D. `2,4`
Answer» Correct Answer - C `HPh hArr H^(+) + underset("Pink")(Ph^(-))`

1512.	During neutralisation of an acid by a base, the end point refers for the completion of reaction. The detection of end point in acid -base neutralisation is usually made by an acid-base indicator. An acid-base indicator is itself a weak acid (Phenolphthlein) or a weak base (Mrthyl orange). At about `50%` ionisation which depends on the medium, the anion furnished by an indicator (acid) or cation furnished by indicator (basic) imparts its characteristic colour to solution at point. For example phenolphthalein, the dissociation is `underset("Colourless")(H In hArr H^(+))+underset("Pink")(In^(-)), K_(H In)= ([H^(+)][In^(-)])/([H In])` favoured in presence of alkali and pink colour of phenolphalein ion is noticed as soon as the medium changes to alkaline nature. The end point of acid-base neutralisation not necessarily coincides with equivalent point but it is closer and closer to equivalence point. Also at equivalence point of acid-base neutralisation pH is not necessarliy equal to 7. The dissociation constant of an acid-base indicator which furnished coloured anion is `1xx10^(-5)`. The pH of solution at which indicator is `80%` in dissociated form is:A. `5.2310`B. `5.6020`C. `8.3980`D. `8.4820`
Answer» Correct Answer - B `K_(a)= ([H^(+)][In^(-)])/([H In])` `1xx10^(-5)= ([H^(+)]xx80xx100)/(100xx20)` `:. [H^(+)] = (1)/(4)xx10^(-5)` `:. pH = 5.6020`

1513.	During neutralisation of an acid by a base, the end point refers for the completion of reaction. The detection of end point in acid -base neutralisation is usually made by an acid-base indicator. An acid-base indicator is itself a weak acid (Phenolphthlein) or a weak base (Mrthyl orange). At about `50%` ionisation which depends on the medium, the anion furnished by an indicator (acid) or cation furnished by indicator (basic) imparts its characteristic colour to solution at point. For example phenolphthalein, the dissociation is `underset("Colourless")(H In hArr H^(+))+underset("Pink")(In^(-)), K_(H In)= ([H^(+)][In^(-)])/([H In])` favoured in presence of alkali and pink colour of phenolphalein ion is noticed as soon as the medium changes to alkaline nature. The end point of acid-base neutralisation not necessarily coincides with equivalent point but it is closer and closer to equivalence point. Also at equivalence point of acid-base neutralisation pH is not necessarliy equal to 7. Which among the following statements are correct ? (1) At equivalence point of NaOH and HCI, pH = 7 (2) At equivalence point to of NaOH and `CH_(3)COOH,pH gt 7` (3) At equivalence point of `NH_(4)OH` and `HCI, pH lt 7` (4) an indicator shoes best results, if equivalence point is within the pH range `pK_(a)` of In `+1` (5) At equivalence point of `NH_(4)OH` and formic acid, `pH lt 7`A. `1,2,3,4`B. `1,3,4,5`C. `1,4,5`D. `1,2,3,5`
Answer» Correct Answer - A at equivalence point, pH of (5) will be nearly 7.

1514.	During neutralisation of an acid by a base, the end point refers for the completion of reaction. The detection of end point in acid -base neutralisation is usually made by an acid-base indicator. An acid-base indicator is itself a weak acid (Phenolphthlein) or a weak base (Mrthyl orange). At about `50%` ionisation which depends on the medium, the anion furnished by an indicator (acid) or cation furnished by indicator (basic) imparts its characteristic colour to solution at point. For example phenolphthalein, the dissociation is `underset("Colourless")(H In hArr H^(+))+underset("Pink")(In^(-)), K_(H In)= ([H^(+)][In^(-)])/([H In])` favoured in presence of alkali and pink colour of phenolphalein ion is noticed as soon as the medium changes to alkaline nature. The end point of acid-base neutralisation not necessarily coincides with equivalent point but it is closer and closer to equivalence point. Also at equivalence point of acid-base neutralisation pH is not necessarliy equal to 7. The dissociation constant of an acid-base indicator which furnishes coloured cation is `1xx10^(-5)` . The pH of solution at which indicator will furnish its colour is :A. 5B. 9C. 6D. 10
Answer» Correct Answer - B `K_(b) = ([OH^(-)][In^(+)])/([H In])`, The colour is provide when indicator `505` dissoczted. Alsoi indicator giving coloured cation is base. In OH, `:. -OH^(-)]=K_(b)=1xx10^(-5)` `:. pOH = 5` , `:. pH = 9`

1515.	What `(H_(3)O^(+))` must be maintained in a saturated `H_(2)S` solution to precipitate `Pb^(2+)` , but not `Zn^(2+)` from a solution in which each ion is present at a concetration of `0.01 M ? (K_(SP)`for `H_(2)S= 1.1xx10^(-22), K_(SP)`for `ZnS=1.0xx10^(-21))`
Answer» For ZnS not to be precipitated from a solution of `Zn^(2+)` and `Pb^(2+)`. `[Zn^(2+)][S^(2-)]lt K_(SP)of ZnS` `[10^(-2)][S^(2-)]lt 1.0xx10^(-21)` or themaximum `[S^(2-)]=10^(-19)` at which ZnS will begin to precipitate or upto this concentration, to precipitation will occur. `H_(2)ShArr2H^(+)+S^(2+)` `:. [H^(+)]^(2)[S^(2-)]=1.1xx10^(-22)` `:. [H^(+)]^(2) [10^(-19)]=1.1xx10^(-22)` `:. [H^(+)]^(2)=11xx10^(-4)` `:. [H^(+)]=3.3xx10^(-2)M` Thus, if `[H^(+)]=3.3xx10^(-2)` or slightly higher, the precipitation of ZnS will not take place and only PbS will precipitate.

1516.	During neutralisation of an acid by a base, the end point refers for the completion of reaction. The detection of end point in acid -base neutralisation is usually made by an acid-base indicator. An acid-base indicator is itself a weak acid (Phenolphthlein) or a weak base (Mrthyl orange). At about `50%` ionisation which depends on the medium, the anion furnished by an indicator (acid) or cation furnished by indicator (basic) imparts its characteristic colour to solution at point. For example phenolphthalein, the dissociation is `underset("Colourless")(H In hArr H^(+))+underset("Pink")(In^(-)), K_(H In)= ([H^(+)][In^(-)])/([H In])` favoured in presence of alkali and pink colour of phenolphalein ion is noticed as soon as the medium changes to alkaline nature. The end point of acid-base neutralisation not necessarily coincides with equivalent point but it is closer and closer to equivalence point. Also at equivalence point of acid-base neutralisation pH is not necessarliy equal to 7. The indicator phenolphalein is a tautomeric mixture of two forms as given below: Which of the following statements are correct ? (1) The form I is referred as quinonoid form and is deeper in colour (2) The form I is referred as quinonoid form and is lighter in colour (3) The form II is more stable in alkaline solution (4) The change in pH form acidic to alkaline solution brings in the more and more conversion of I form to II form (5) The form I is more stable in acidic mediumA. `1,2,3,4`B. `1,3,4,5`C. `3,4,5`D. `2,3,4,5`
Answer» Correct Answer - B All are facts.

1517.	A sample of AgCI was treated with `5.00mL` of `1.5M` `Na_(2)CO_(3)` solubility to give `Ag_(2)CO_(3)` . The remaining solution contained `0.0026g of CI^(-)` per litre. Calculate the solubility product of AgCI. `(K_(SP)for Ag_(2)CO_(3)=8.2xx10^(-12))`A. `1.1 xx 10^(-2)`B. `1.71 xx 10^(-11)`C. `1.71 xx 10^(-10)`D. `1.32 xx 10^(-9)`
Answer» Correct Answer - C `[Ag^(+)] = sqrt((K_(sp))/([CO_(3)^(2-)])) = 2.34 xx 10^(-6)m` `K_(spAgCl) = [Ag^(+)][Al^(-)] = 1.71 xx 10^(-10)`

1518.	An acid `HA` is `40%` dissociated in an aqueous solution. The hydronium ion concentration of its `0.2M` solution would beA. 0.08 MB. 0.4 MC. 0.2 MD. None of the above.
Answer» Correct Answer - A `[H_(3)O^(+)]=C alpha =0.2 xx 0.4 =0.08 M`

1519.	The hydronium ion concentration in `5xx10^(-4)` M aqueous solution of NaOH at `25^@C` , is :A. zeroB. `5xx10^(-4) M`C. `2xx10^(-11) M`D. `2xx10^(3) M`
Answer» Correct Answer - c

1520.	Calculate the per cent error in hydronium ion concentration made by neglecting the ionisation of water in `1.0 xx 10^(-6) M NaOH`
Answer» Correct Answer - 1 Because `[H^(+)] = 1.0 xx 10^(-8) =10 xx 10^(-9)` Which is very small and so negligible if ionisation of `H_(2)O` is not neglected, then `{:(H_(2)O,hArr,H^(+)+,H^(-)),(,,a,(10^(-6)+a)):}` `:. a xx (10^(-6) + a) = 10^(-14) " " a = 9.9 xx 10^(-9)` `:.` % error `= (10 xx 10^(-9) - 9.9 xx 10^(-9))/(9.9 xx 10^(-9)) xx 100 = 1%`

1521.	The process in which solution containing more than one typeof ions (either cation or anion) and in which one ion undergoes almst complete precipitation followed by the second ion and similarly second ion undergoes complete precipitation followed by the third ion is known as preferential precipitation. It is the solubility of the salt,not the solubility product of the salt, by means of which one can predict the preferential precipitation of the salt or ion among the lot of ions in the solution. For example , in a solution containing `Cl^(-), Br^(-)` and `I^(-)` ions, if `Ag^(+)` ions are added, then out of three , the less soluble salt is precipitated first . If the addition of `Ag^(+)` ions is continued, eventually , a stage is reached when the next lesser soluble salt starts precipitating along with the least soluble salt and so on. If the stiochiometry of the salt is same, then the salt with the minimum `K_(sp)` ( solubility product ) will have minimum solubility and will precipitate first, followed by the salt of next higher solubility product and so on. If the stiochiometry of teh salt is not same, then from the solubility product data alone, we cannot predict which ions will precipitate first. For example, a solution containing `Cl^(-)` and `CrO_(4)^(2-)`. In order to predict which ion will precipitate first, we have to calculate the amount of `Ag^(+)` ions needed to start precipitation through the solubility product data given. When `AgNO_93)` is added to the solution, the minimum of the two concentrations needed to start the precipitation will be reached and thus corresponding `[Ag^(+)]` to start the precipitation increases. Its concentration eventually becomes equal to the value required for `CrO_(4)^(2-)` . At this stage, practically almost the whole of `Cl^(-)` ions get precipitated . Addition of more `AgNO_(3)` causes simultaneous precipitation of both the ions together. Suppose a student of chemistry performs the following two sets of experiments in order to known the preferential precipitation of ions as well as its precipitated percentage amount. Experiment 1. He prepare a solution of anions composed of 0.1 M each of `Cl^(-),Br^(-)` and `I^(-)` . Further he adds gradually solid `AgNO_(3)` to this solution. He assumes that volume of the solution does not change after the addition of solid `AgNO_(3)(K_(sp)AgCl=1.0 xx 10^(-9)M^(2),K_(sp) AgBr=10^(-10)M^(2)` and `K_(sp) AgI=10^(-12)M^(2))` Experiment -2. He prepares a solution of cations `Cd^(2+)(0.2 M)` and `Bi^(3+) (0.3 m)` . Now he adds `S^(2-)` ions into the solution of cations in order to separate them by selective precipitation `Cd^(2+) `forms yellow precipitate of CdS and `Bi^(3+)` forms black precipitate of `Bi_(2)S_(3)` with `S^(2-)` ions respectively. `(K_(sp)Bi_(2)S_(3) = 9 xx 10^(-25) M^(5)` and `K_(sp)CdS =2 xx 10^(-20)M^(2))`.Answer the following questions ont he basis of the above write up. In experiment -2 what `%` of the metal ions get precipitated at which `S^(2-)` get saturated with another ion ?A. `10%`B. `80%`C. `20%`D. `90%`
Answer» Correct Answer - A

1522.	Which of the following statement(s) is (are) correct?A. The `p^(H)` of `10^(-8)` M solution of `HCl` is `8`B. The conjugate base of `H_(2)PO_(4)^(-)` is `HPO_(4)^(2-)`C. Auto protolysis constant of water increases with temperatureD. When a solution of weak monoprotic acid is titrated against a strong base, at half of neutraliastion point `p^(H) = 1//2p^(Ka)`
Answer» Correct Answer - B::C

1523.	The process in which solution containing more than one typeof ions (either cation or anion) and in which one ion undergoes almst complete precipitation followed by the second ion and similarly second ion undergoes complete precipitation followed by the third ion is known as preferential precipitation. It is the solubility of the salt,not the solubility product of the salt, by means of which one can predict the preferential precipitation of the salt or ion among the lot of ions in the solution. For example , in a solution containing `Cl^(-), Br^(-)` and `I^(-)` ions, if `Ag^(+)` ions are added, then out of three , the less soluble salt is precipitated first . If the addition of `Ag^(+)` ions is continued, eventually , a stage is reached when the next lesser soluble salt starts precipitating along with the least soluble salt and so on. If the stiochiometry of the salt is same, then the salt with the minimum `K_(sp)` ( solubility product ) will have minimum solubility and will precipitate first, followed by the salt of next higher solubility product and so on. If the stiochiometry of teh salt is not same, then from the solubility product data alone, we cannot predict which ions will precipitate first. For example, a solution containing `Cl^(-)` and `CrO_(4)^(2-)`. In order to predict which ion will precipitate first, we have to calculate the amount of `Ag^(+)` ions needed to start precipitation through the solubility product data given. When `AgNO_93)` is added to the solution, the minimum of the two concentrations needed to start the precipitation will be reached and thus corresponding `[Ag^(+)]` to start the precipitation increases. Its concentration eventually becomes equal to the value required for `CrO_(4)^(2-)` . At this stage, practically almost the whole of `Cl^(-)` ions get precipitated . Addition of more `AgNO_(3)` causes simultaneous precipitation of both the ions together. Suppose a student of chemistry performs the following two sets of experiments in order to known the preferential precipitation of ions as well as its precipitated percentage amount. Experiment 1. He prepare a solution of anions composed of 0.1 M each of `Cl^(-),Br^(-)` and `I^(-)` . Further he adds gradually solid `AgNO_(3)` to this solution. He assumes that volume of the solution does not change after the addition of solid `AgNO_(3)(K_(sp)AgCl=1.0 xx 10^(-9)M^(2),K_(sp) AgBr=10^(-10)M^(2)` and `K_(sp) AgI=10^(-12)M^(2))` Experiment -2. He prepares a solution of cations `Cd^(2+)(0.2 M)` and `Bi^(3+) (0.3 m)` . Now he adds `S^(2-)` ions into the solution of cations in order to separate them by selective precipitation `Cd^(2+) `forms yellow precipitate of CdS and `Bi^(3+)` forms black precipitate of `Bi_(2)S_(3)` with `S^(2-)` ions respectively. `(K_(sp)Bi_(2)S_(3) = 9 xx 10^(-25) M^(5)` and `K_(sp)CdS =2 xx 10^(-20)M^(2))`.Answer the following questions ont he basis of the above write up. In experiment 02, what is the maximum concentration of `S^(2-)` ion at which one of the two metal ions gets maximum precipitation ?A. `10^(-8)` MB. `2 xx 10^(-7)M`C. `2 xx 10^(-9)M`D. None
Answer» Correct Answer - A

1524.	The process in which solution containing more than one typeof ions (either cation or anion) and in which one ion undergoes almst complete precipitation followed by the second ion and similarly second ion undergoes complete precipitation followed by the third ion is known as preferential precipitation. It is the solubility of the salt,not the solubility product of the salt, by means of which one can predict the preferential precipitation of the salt or ion among the lot of ions in the solution. For example , in a solution containing `Cl^(-), Br^(-)` and `I^(-)` ions, if `Ag^(+)` ions are added, then out of three , the less soluble salt is precipitated first . If the addition of `Ag^(+)` ions is continued, eventually , a stage is reached when the next lesser soluble salt starts precipitating along with the least soluble salt and so on. If the stiochiometry of the salt is same, then the salt with the minimum `K_(sp)` ( solubility product ) will have minimum solubility and will precipitate first, followed by the salt of next higher solubility product and so on. If the stiochiometry of teh salt is not same, then from the solubility product data alone, we cannot predict which ions will precipitate first. For example, a solution containing `Cl^(-)` and `CrO_(4)^(2-)`. In order to predict which ion will precipitate first, we have to calculate the amount of `Ag^(+)` ions needed to start precipitation through the solubility product data given. When `AgNO_93)` is added to the solution, the minimum of the two concentrations needed to start the precipitation will be reached and thus corresponding `[Ag^(+)]` to start the precipitation increases. Its concentration eventually becomes equal to the value required for `CrO_(4)^(2-)` . At this stage, practically almost the whole of `Cl^(-)` ions get precipitated . Addition of more `AgNO_(3)` causes simultaneous precipitation of both the ions together. Suppose a student of chemistry performs the following two sets of experiments in order to known the preferential precipitation of ions as well as its precipitated percentage amount. Experiment 1. He prepare a solution of anions composed of 0.1 M each of `Cl^(-),Br^(-)` and `I^(-)` . Further he adds gradually solid `AgNO_(3)` to this solution. He assumes that volume of the solution does not change after the addition of solid `AgNO_(3)(K_(sp)AgCl=1.0 xx 10^(-9)M^(2),K_(sp) AgBr=10^(-10)M^(2)` and `K_(sp) AgI=10^(-12)M^(2))` Experiment -2. He prepares a solution of cations `Cd^(2+)(0.2 M)` and `Bi^(3+) (0.3 m)` . Now he adds `S^(2-)` ions into the solution of cations in order to separate them by selective precipitation `Cd^(2+) `forms yellow precipitate of CdS and `Bi^(3+)` forms black precipitate of `Bi_(2)S_(3)` with `S^(2-)` ions respectively. `(K_(sp)Bi_(2)S_(3) = 9 xx 10^(-25) M^(5)` and `K_(sp)CdS =2 xx 10^(-20)M^(2))`.Answer the following questions ont he basis of the above write up. What percent of the anions `Br^(-)` and `I^(-)` get precipitated respectively when the third ion starts precipitating in experiment -1 ?A. `90%,99.9%`B. `99.9%,90%`C. `80%,90%`D. `90%,80%`
Answer» Correct Answer - A

1525.	The process in which solution containing more than one typeof ions (either cation or anion) and in which one ion undergoes almst complete precipitation followed by the second ion and similarly second ion undergoes complete precipitation followed by the third ion is known as preferential precipitation. It is the solubility of the salt,not the solubility product of the salt, by means of which one can predict the preferential precipitation of the salt or ion among the lot of ions in the solution. For example , in a solution containing `Cl^(-), Br^(-)` and `I^(-)` ions, if `Ag^(+)` ions are added, then out of three , the less soluble salt is precipitated first . If the addition of `Ag^(+)` ions is continued, eventually , a stage is reached when the next lesser soluble salt starts precipitating along with the least soluble salt and so on. If the stiochiometry of the salt is same, then the salt with the minimum `K_(sp)` ( solubility product ) will have minimum solubility and will precipitate first, followed by the salt of next higher solubility product and so on. If the stiochiometry of teh salt is not same, then from the solubility product data alone, we cannot predict which ions will precipitate first. For example, a solution containing `Cl^(-)` and `CrO_(4)^(2-)`. In order to predict which ion will precipitate first, we have to calculate the amount of `Ag^(+)` ions needed to start precipitation through the solubility product data given. When `AgNO_93)` is added to the solution, the minimum of the two concentrations needed to start the precipitation will be reached and thus corresponding `[Ag^(+)]` to start the precipitation increases. Its concentration eventually becomes equal to the value required for `CrO_(4)^(2-)` . At this stage, practically almost the whole of `Cl^(-)` ions get precipitated . Addition of more `AgNO_(3)` causes simultaneous precipitation of both the ions together. Suppose a student of chemistry performs the following two sets of experiments in order to known the preferential precipitation of ions as well as its precipitated percentage amount. Experiment 1. He prepare a solution of anions composed of 0.1 M each of `Cl^(-),Br^(-)` and `I^(-)` . Further he adds gradually solid `AgNO_(3)` to this solution. He assumes that volume of the solution does not change after the addition of solid `AgNO_(3)(K_(sp)AgCl=1.0 xx 10^(-9)M^(2),K_(sp) AgBr=10^(-10)M^(2)` and `K_(sp) AgI=10^(-12)M^(2))` Experiment -2. He prepares a solution of cations `Cd^(2+)(0.2 M)` and `Bi^(3+) (0.3 m)` . Now he adds `S^(2-)` ions into the solution of cations in order to separate them by selective precipitation `Cd^(2+) `forms yellow precipitate of CdS and `Bi^(3+)` forms black precipitate of `Bi_(2)S_(3)` with `S^(2-)` ions respectively. `(K_(sp)Bi_(2)S_(3) = 9 xx 10^(-25) M^(5)` and `K_(sp)CdS =2 xx 10^(-20)M^(2))`.Answer the following questions ont he basis of the above write up. What `%`of the (ion) will get precipitate when the second ion start precipitating in experiment -1 ?A. `10%`B. `90%`C. `1%`D. `99%`
Answer» Correct Answer - D

1526.	A monobasic acid solution has pH value of `5`. Its molarity is `0.005 M`. The degree of ionisation of the acid isA. `5 xx 10^(-3)`B. `2 xx 10^(-3)`C. `5 xx 10^(-2)`D. `2 xx 10^(-2)`
Answer» Correct Answer - B `alpha = ([H^(+)])/(C) = (10^(-5))/(5 xx 10^(-3))` `= 2 xx 10^(-3)`

1527.	The process in which solution containing more than one typeof ions (either cation or anion) and in which one ion undergoes almst complete precipitation followed by the second ion and similarly second ion undergoes complete precipitation followed by the third ion is known as preferential precipitation. It is the solubility of the salt,not the solubility product of the salt, by means of which one can predict the preferential precipitation of the salt or ion among the lot of ions in the solution. For example , in a solution containing `Cl^(-), Br^(-)` and `I^(-)` ions, if `Ag^(+)` ions are added, then out of three , the less soluble salt is precipitated first . If the addition of `Ag^(+)` ions is continued, eventually , a stage is reached when the next lesser soluble salt starts precipitating along with the least soluble salt and so on. If the stiochiometry of the salt is same, then the salt with the minimum `K_(sp)` ( solubility product ) will have minimum solubility and will precipitate first, followed by the salt of next higher solubility product and so on. If the stiochiometry of teh salt is not same, then from the solubility product data alone, we cannot predict which ions will precipitate first. For example, a solution containing `Cl^(-)` and `CrO_(4)^(2-)`. In order to predict which ion will precipitate first, we have to calculate the amount of `Ag^(+)` ions needed to start precipitation through the solubility product data given. When `AgNO_93)` is added to the solution, the minimum of the two concentrations needed to start the precipitation will be reached and thus corresponding `[Ag^(+)]` to start the precipitation increases. Its concentration eventually becomes equal to the value required for `CrO_(4)^(2-)` . At this stage, practically almost the whole of `Cl^(-)` ions get precipitated . Addition of more `AgNO_(3)` causes simultaneous precipitation of both the ions together. Suppose a student of chemistry performs the following two sets of experiments in order to known the preferential precipitation of ions as well as its precipitated percentage amount. Experiment 1. He prepare a solution of anions composed of 0.1 M each of `Cl^(-),Br^(-)` and `I^(-)` . Further he adds gradually solid `AgNO_(3)` to this solution. He assumes that volume of the solution does not change after the addition of solid `AgNO_(3)(K_(sp)AgCl=1.0 xx 10^(-9)M^(2),K_(sp) AgBr=10^(-10)M^(2)` and `K_(sp) AgI=10^(-12)M^(2))` Experiment -2. He prepares a solution of cations `Cd^(2+)(0.2 M)` and `Bi^(3+) (0.3 m)` . Now he adds `S^(2-)` ions into the solution of cations in order to separate them by selective precipitation `Cd^(2+) `forms yellow precipitate of CdS and `Bi^(3+)` forms black precipitate of `Bi_(2)S_(3)` with `S^(2-)` ions respectively. `(K_(sp)Bi_(2)S_(3) = 9 xx 10^(-25) M^(5)` and `K_(sp)CdS =2 xx 10^(-20)M^(2))`.Answer the following questions ont he basis of the above write up. Which of the following will precipitate first in experiment -1 ?A. `I^(-)`ionB. `Br^(-)` ionC. `Cl^(-)` ionD. cannot be predicted
Answer» Correct Answer - A

1529.	A mono basic weak acid solution has a molarity of 0.005 and pH of 5. What is its percentage ionisation in this solutino ?gtA. `2.0`B. 0.1C. 0.5D. 0.25
Answer» Correct Answer - B

1530.	Some solutions are taken at specified concentration and have the following solution codes. Identify the four digit number abcd where ab = Sum of solution codes of those solutions, which will have a singificant buffer action. Cd = Sum of solution codes of those solutions, which for the given amount of acid or base and salt, have maximum buffer capacity.
Answer» Correct Answer - 8877

1531.	What concentartion of `Ac^(-)` ions will reduce `H_(3)O^(+)` ion to `2 xx 10^(-4) M` in `0.40 M` solution of `HAc ? [K_(a) (HAc) = 1.8 xx 10^(-5)]` Given your answer after multiplying by 1000.
Answer» Correct Answer - 36