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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
At `30^(@) C` the solubility of `Ag_(2)CO_(3) (K_(SP)=8xx10^(-12))` would be gretest in one litre of:A. `0.05M Na_(2)CO_(3)`B. `0.05M AgNO_(3)`C. pure waterD. `0.05M K_(2)CO_(3)` |
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Answer» Correct Answer - C Presence of common ion decreases the solubility of salt. |
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| 52. |
At `30^(@) C` the solubility of `Ag_(2)CO_(3) (K_(SP)=8xx10^(-12))` would be gretest in one litre of:A. 0.05M `Na_(2)CO_(3)`B. 0.05M `AgNO_(3)`C. pure waterD. 0.05M `K_(2)CO_(3)` |
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Answer» Correct Answer - c |
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| 53. |
The concentration of `Ag^(+)` ions in a given saturated solution of AgCl at `25^(@)C` is `1.06 xx 10^(-5) g ` ion per litre. The solubility product of AgCl isA. `0.353 xx 10^(-10)`B. `0.530 xx 10^(-10)`C. `1.12 xx 10^(-10)`D. `2.12 xx 10^(-10)` |
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Answer» Correct Answer - C Solubility of AgCl` =(1.06 xx 10^(-5))^(2)` `=1.12 xx 10^(-10)` |
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| 54. |
Assertion : Solubility of `AgCl` in `NH_(3)(aq)` is greater than in pure water. Reason : When `AgCl` dissolve in `NH_(3)(aq)`, complex ion [`Ag(NH_(3))_(2)^(+)`] formation takes place and solubility equilibrium of `AgCl_(3)` shifted in forward direction.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-11B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is FalseD. Statement-1 and Statement-2 is True |
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Answer» Correct Answer - a |
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| 55. |
Number of `H^(+)` ions present in 250 ml of lemon juice of pH = 3 isA. `1.506 xx 10^(22)`B. `1.506 xx 10^(23)`C. `1.506 xx 10^(20)`D. `3.012 xx 10^(21)` |
| Answer» Correct Answer - C | |
| 56. |
When `2.5 mL` of `2//5M` weak monoacidic base `(K_(b) = 1 xx 10^(-12) at 25^(@)C)` is titrated with `2//15 M HCI` in water at `25^(@)C` the concentration of `H^(o+)` at equivalence point is `(K_(w) = 1 xx 10^(-14) at 25^(@)C)`A. `3.7 xx 10^(-13)M`B. `3.2 xx 10^(-7)M`C. `3.2 xx 10^(-2)M`D. `2.7 xx 10^(-2)M` |
| Answer» Correct Answer - D | |
| 57. |
Acid base indicators are either weak organic acids or weak organic bases. Indicator change colour in dilute solution when the hydronium ion concentration reaches a particular value. For example, phenopthalein is a colour less substance in any aqueous solution with a pH less than `8.3` in between the pH range `8.3` to `10`, transaction of colour (colourless to pink) takes place and if pH of solution is greater than `10`, then solution colour is dark pin. Considering an acid indicator HIn, base `In^(-)` can be represented as : `HIn hArr H^(+) + In^(-)` acidic form basic form pH of solution can be computed as : `pH = pK_(In) + "log" ([In^(-)])/([HIn])` In general, transition of colour takes place in between the pH range `pK_(In) +-1`Which of the following indicator is most suitable for titration of HBr with strong base:A. Phenolphthalein `(8.3 - 10)`B. Bromothymol blue `(6-7.6)`C. Methyl red `(4.2-6.3)`D. Malachite green `(11.4-13)` |
| Answer» Correct Answer - A | |
| 58. |
Acid base indicators are either weak organic acids or weak organic bases. Indicator change colour in dilute solution when the hydronium ion concentration reaches a particular value. For example, phenopthalein is a colour less substance in any aqueous solution with a pH less than `8.3` in between the pH range `8.3` to `10`, transaction of colour (colourless to pink) takes place and if pH of solution is greater than `10`, then solution colour is dark pin. Considering an acid indicator HIn, base `In^(-)` can be represented as : `HIn hArr H^(+) + In^(-)` acidic form basic form pH of solution can be computed as : `pH = pK_(In) + "log" ([In^(-)])/([HIn])` In general, transition of colour takes place in between the pH range `pK_(In) +-1` An indicator is a weak acid and pH range is `4.0` to `6.0`. If indicator is `50%` ionized in a given solution then what is ionization consatant of the acid?A. `10^(-4)`B. `10^(-5)`C. `10^(-6)`D. `10^(-7)` |
| Answer» Correct Answer - B | |
| 59. |
Acid base indicators are either weak organic acids or weak organic bases. Indicator change colour in dilute solution when the hydronium ion concentration reaches a particular value. For example, phenopthalein is a colour less substance in any aqueous solution with a pH less than `8.3` in between the pH range `8.3` to `10`, transaction of colour (colourless to pink) takes place and if pH of solution is greater than `10`, then solution colour is dark pin. Considering an acid indicator HIn, base `In^(-)` can be represented as : `HIn hArr H^(+) + In^(-)` acidic form basic form pH of solution can be computed as : `pH = pK_(In) + "log" ([In^(-)])/([HIn])` In general, transition of colour takes place in between the pH range `pK_(In) +-1` Select the correct statement among the followingA. In the complete ionisation of indicator its `pH = pK_(In)`B. Methyl orange (working range : `3.1` to `4.4`) is a suitable indicator for weak acid and strong baseC. Bromothymol blue (working range of `6.0` to `7.6`) is a good indicator for titration of `HCl` and `NaOH`D. Thymal blue (working range `1.2` to `2.8` ) is good indicator for titration of `100 ml` of `0.1 M NH_(4)OH(pK_(b) = 4.74)` and `0.1 M HCl`. |
| Answer» Correct Answer - C | |
| 60. |
The solubility product of a soluble salt `A_(x)B_(y)` is given by : `K_(sp)=[A^(y+)]^(z)[B^(x-)]^(y)`. As soon as prodcut of concentration of `A^(y+)` and `B_(x-)`. Becomes more than its `K_(sp)`, the salt starts precipitation. It may practically be noticed that AgCI is more soluble in water and its solubility decreases dramatically in 0.1m NaCI or 0.1 m `AgNO_(3)` solution. It can be concluded that in presence of a common ion the solubility of salt decreases. `K_(sp)` of `SrF_(2)` in water is `8xx10^(-10)`. The solubility of `SrF_(2)` in 0.1MNaF aqueous solution is :A. `8xx10^(-10)`B. `2xx10^(-3)`C. `2.71xx10^(-10)`D. `8xx10^(-8)` |
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Answer» Correct Answer - d |
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| 61. |
The solubility product of a soluble salt `A_(x)B_(y)` is given by : `K_(sp)=[A^(y+)]^(z)[B^(x-)]^(y)`. As soon as prodcut of concentration of `A^(y+)` and `B_(x-)`. Becomes more than its `K_(sp)`, the salt starts precipitation. It may practically be noticed that AgCI is more soluble in water and its solubility decreases dramatically in 0.1m NaCI or 0.1 m `AgNO_(3)` solution. It can be concluded that in presence of a common ion the solubility of salt decreases. The salting out action of RCOONa (soap) in presence of NaCI is based on :A. Buffering actionB. Hydrolysis of saltC. Solubility productD. Complex formation |
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Answer» Correct Answer - c |
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| 62. |
Calculate `[OH^(-)]` in `0.20M` solution of `NH_(3)`, if `K_(b)` for `NH_(3)` is `1.8xx10^(-5)`. |
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Answer» `underset(1-x)(NH_(3))H_(2)OhArr underset(x)(NH_(4)^(+))+underset(x)(OH^(-))` `K_(b)=([NH_(4)^(+)][OH^(-)])/([NH_(3)])= (X^(2))/((1-X))=X^(2)` , `(.: 1-X ~~1)` `:. X= [OH^(-)]= sqrt(K_(b))=sqrt(1.8xx10^(-5))` `= 4.24xx10^(-3)M` |
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| 63. |
The solubility product of a soluble salt `A_(x)B_(y)` is given by: `K_(SP)= [A^(y+)]^(x) [B^(x-)]^(y)`. As soon as the product of concentration of `A^(y+)` and `B^(x-)` increases than its `K_(SP)`, the salt start precipitation. It may practically be noticed that AgCI is more soluble in water and its solublity decreases dramatically in `0.1M NaCI` or `0.1M AgNO_(3)` solution. It may therefore be conncluded that in presence of a common ion, the solubiolity of salt decreases. The volume of water neede to dissolve 1g `BaSO_(4)(K_(SP)= 1xx10^(-10))` is:A. 230 litreB. 429 litreC. 500 litreD. 320 litre |
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Answer» Correct Answer - B `S= sqrt(K_(SP))=sqrt(1xx10^(-10))=10^(-5)` `:. 10^(-5)"mole" BaSO_(4)` present in 1 litre `:. (1)/(233) "mole" BaSO_(4)` present in `(1)/(233)xx(1)/(10^(-5))=429` litre |
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| 64. |
A solubility contains one mole each of HA and HB (both are weak acids) in one litre solution. Now 1 mole of NaOH is added to this solution so that both the acids are partly neutralised. Heat of neutralisation for HA and HB are- `11.8` and- `12.4` kcal respectively and the heat produced during partial neutralisation of HA and HB is- `12.25` kcal. Mole ratio of neutralisation of HA and HB is:A. `1:4`B. `1:2`C. `1:3`D. `1:5` |
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Answer» Correct Answer - C Let a mole of HA and b mole of HB are neutralised. Thus, since both are monobasic acid, so `axx11.8+bxx12.4=12.25` Also, 1 mole of NaOH used only 1 mole of of both acids, thus `a+b=1` `axx11.8+(1-a)xx12.4=12.25` `:. a=0.25` and `b= 0.75` Thus mole ratio of neutralisation of HA and HB `= (0.25)/(0.75)=(1)/(3)` |
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| 65. |
150 mL of 0.0008 M ammonium sulphate is mixed with 50 mL of 0.04 M calcium nitrate. The ionic product of `CaSO_(4)` will be : `(K_(sp)=2.4xx10^(-5) for CaSO_(4))`A. `ltK_(sp)`B. `gtK_(sp)`C. `~~K_(sp)`D. None of these |
| Answer» Correct Answer - 1 | |
| 66. |
Calculate K for the reaction, `A^(-)+H_(3)^(+)O hArr HA+H_(2)O` if `K_(a)` value for the acid HA is `1.0xx10^(-6)`.A. `1xx10^(-6)`B. `1xx10^(8)`C. `1xx10^(-8)`D. `1xx10^(5)` |
| Answer» `K_(a)=10^(-6)` for `Ha+H_(2)OhArrH_(3)O^(+)+A^(-)` Thus `K` for reversing reaction is `(1)/(10^(-6))=10^(6)` | |
| 67. |
Calculate the dissociation constant of `NH_(4)OH` at `25^(@)C`. If `Delta H^(@)` and `Delta S^(@)` for the given changes are as follows: `NH_(3)+H^(+)hArrNH_(4)^(+),` , `DeltaH^(@)= -52.2kJ mol^(-1),` `DeltaS^(@)= +1.67JK^(-1)mol^(-1)` `H_(2)OhArrH^(+)+OH^(-), DeltaH^(@)=56.6kJ mol^(-1),` `DeltaS^(@)= -78.2JK^(-1)mol^(-1)` |
| Answer» Correct Answer - `1.7xx10^(-5);` | |
| 68. |
The self ionisation constant for pure formic acid `K=[HCOOH_(2)^(+)][HCOO^(-)]` has been estimated as `10^(-6)` at room temperature .The density of formic acid is `1.15g//cm^(3)` .What percentage of formic acid molecules in pure fomic acid are converted to formation ion?A. `0.002%`B. `0.004%`C. `0.006%`D. `0.008%` |
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Answer» Given density of formic acid =`1.15g//cm^(3)` :.Weight of formic acid in `1` litre solution =`1.15xx10^(3)g` thus `[HCOOH]=(1.15xx10^(3))/(46)=25M` since in case of auto ionisation `[HCOOH_(2)^(+)]=[HCOO^(-)]` and `[HCOO^(-)][HCOOH_(2)^(+)]=10^(-6)rArr [HCOO^(-)]=10^(-3)` Now `%` dissociation of `HCOOH=([HCOO^(-)]xx100)/([HCOOH])=(10^(-3))/(25)xx100=0.004%` |
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| 69. |
The self ionisation constant for pure `HCOOH, K = [HCOO overset(o+)H_(2)] [HCOO^(Θ)]` is `10^(-6)` at room temperature. What percentage of `HCOOH` molecules are converted to `HCOO^(Θ)` ions. The density of `HCOOH` iws `1.22 g cm^(-3)`. |
| Answer» Correct Answer - `0.004%;` | |
| 70. |
For the self-ionisation (protonation) process liquid formic acid at `20^(@)`C. `HCOOH+HCOOHto HCOOH_(2)^(+)+HCOO^(-)` the equilibrium constant, `K=([HCOOH_(2)^(+)][HCOO^(-)])/([HCOOH]^(2))=10^(-10)` At `20^(@)`C, the density of liquid formic acid is 0.92 gm/mL. How many formate ions `(HCOO^(-))` are present in each 5ml of liq.formic acid at `20^(@)` C? `(N_(A)=6xx10^(23))`A. `2xx10^(-4)`B. `1.2xx10^(20)`C. `6xx10^(17)`D. `3xx10^(21)` |
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Answer» Correct Answer - c |
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| 71. |
For the self-ionisation (protonation) process liquid formic acid at `20^(@)`C. `HCOOH+HCOOHto HCOOH_(2)^(+)+HCOO^(-)` the equilibrium constant, `K=([HCOOH_(2)^(+)][HCOO^(-)])/([HCOOH]^(2))=10^(-10)` At `20^(@)`C, the density of liquid formic acid is 0.92 gm/mL. What percentage of formic acid molecules are self ionised?A. `0.02%`B. `0.001%`C. `0.005%`D. `0.002%` |
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Answer» Correct Answer - d |
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| 72. |
The `K_(sp)` of `Mg(OH)_(2)` is `1xx10^(-12). 0.01M Mg^(2+)` will precipitate at the limiting pH ofA. `8`B. `9`C. `10`D. `12` |
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Answer» Correct Answer - B `{:(Mg(OH)_(2)rarr,Mg^(2+)+,2overset(Theta)OH,),(,S,2S,):}` Let `S_(1)` is the solubility in `0.1M MgCI_(2)` (common ion `Mg^(2+))` `:. [Mg^(2+)] = (S_(1) + 0.001) ~~ 0.01` `[overset(Theta)OH] = 2S_(1)` `K_(sp) = (0.01) (2S_(1))^(2)` `S_(1) = ((K_(sp))/(4xx0.01))^(1//2) = ((10^(-12))/(4xx10^(-2)))^(1//2) = 0.5xx10^(-5)M` `:. [overset(Theta)OH] = 2S_(1) = 2xx 0.5 xx 10^(-5) = 10^(-5)M` `pOH = 5, pH = 14 - 5 = 9` |
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| 73. |
The concentration of hydrogen ion `[H^(+)]` in 0.01 M HCl isA. `10^(12)`B. `10^(-2)`C. `10^(-1)`D. `10^(-12)` |
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Answer» Correct Answer - B `[H^(+)]` Concentration in 0.01 M HCl is `10^(-2)` M because 0.01 M HCl have only `H^(+)`. `Hcl H^(+) + Cl^(-)`. |
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| 74. |
At `25^(@)C`, the `[H^(+)]` of a solution is `2 xx 10^(-9) M`, the nature of the solution isA. NeutralB. AcidicC. BasicD. Can not be predicted |
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Answer» Correct Answer - C `[H^(+)] lt 10^(-7)` therefore basic |
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| 75. |
The `K_(sp)` of `Mg(OH)_(2)` is `1 xx 10^(-12), 0.01 M Mg(OH)_(2)` will precipitate at the limiting pHA. 3B. 9C. 5D. 8 |
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Answer» Correct Answer - B `Mg(OH)_(2) Mg^(2+) + 2OH^(-)` `K_(sp) = [Mg^(2+)] [OH^(-)]^(2)` `1 xx 10^(-12) = 0.01 [OH^(-)]^(2)` `[OH^(-)]^(2) = 1 xx 10^(-10) rArr [OH^(-)] = 10^(-5)` `[H^(+) = 10^(-14)//10^(-5) = 10^(-9)` `pH = -log[H^(+)] = -log[10^(-9)] = 9`. |
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| 76. |
an acid with molecular formula `C_(7)H_(6)O_(3)` forms three types of sodium salt i.e., `C_(7)H_(5)O_(3)Na, C_(7)H_(4)O_(3)Na_(2)` and `C_(7)H_(3)O_(3)Na_(3)`. The basicity of the acid:A. oneB. twoC. threeD. four |
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Answer» Correct Answer - C The molecule shows three H atoms are replaceable, i.e., basicity of acid. |
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| 77. |
Amonst the following, the total number of compounds whose aqueous solution turns red litmus paper blue is: `{:(KCN,K_(2)SO_(4),(NH_(4))_(2)C_(2)O_(4),NaCI),(Zn(NO_(3))_(2),FeCI_(3),K_(2)CO_(3),NH_(4)NO_(3)),(LiCN, , ,):}` |
| Answer» Correct Answer - 3 | |
| 78. |
Which one is more acidic in aqueous solution ?A. `NiCI_(2)`B. `FeCI_(3)`C. `AICI_(3)`D. `BeCI_(2)` |
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Answer» Correct Answer - C `AICI_(3)` on hydrolysis gives weak base and strong acid among all. |
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| 79. |
`pOH` of `H_(2)O` is `7.0` at `298K`. If water is heated at `350 K`, which of the following statement should be true?A. pOH will decreaseB. pOH will increaseC. pOH will remain sevenD. Concentration of `H^(+)` ions will increase but that of `OH^(-)` will decrease. |
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Answer» Correct Answer - A At 298 K , for pure water `Ph=Poh =7.` On increasing the temperature, the self ionisation of water increases thereby increasing the concentrations of `H_93)O^(+)` as well as `OH^(-)` ions. Consequenctly, at high temperature, the pH as well as pOH will be `lt7`. |
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| 80. |
The `pH` value of `0.001M` aqueous solution of `NaCI` isA. `7`B. `4`C. `11`D. Unpredictable |
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Answer» Correct Answer - A `NaCI` is salt of `S_(A)//S_(B)`. So it is neutral and `pH = 7`. |
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| 81. |
Which pair will show common ion effect?A. `AgNO_(3) + KNO_(3)`B. `NaCl + HCl`C. `NH_(4)OH + NH_(4)Cl`D. `BaCl_(2) + Ba(NO_(3))_(2)` |
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Answer» Correct Answer - C Weak base its salt of strong acid |
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| 82. |
a. Will a precipitate of`Mg(OH)_(2)` be formed in a `0.001M` solution of `Mg(NO_(3))_(2)` if the `pH` of solution is adjusted to `9.K_(sp)` of `Mg(OH)_(2) = 8.9 xx 10^(-12)`. b. Calculate `pH` at which `Mg(OH)_(2)` begin to precipitae form a solution containing `0.1M Mg^(2+)` ions. `K_(sp) of Mg(OH)_(2) =1 xx 10^(-11)`. c. Calculate `[overset(Theta)OH]` of a solution after `100mL` of `0.1M MgC1_(2)` is added to `100mL` of `0.2M NaOH. K_(sp) Mg(OH)_(2) = 1.2 xx 10^(-11)`. |
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Answer» a. `pH = 9 :. [H^(o+)] = 10^(-9)M or [OH^(Theta)] = 10^(-5)m` Now if `Mg(NO_(3))_(2)` is present in a solution of `[OH^(Theta)] = 10^(-5)M`, then Product of ionic concentration `= [Mg^(2+)] [OH^(Theta)]^(2)` `= [0.001][10^(-5)]^(2)` `10^(-13)` lesser than `K_(sp) of Mg (OH)_(2)` (i.e., `8.9 xx 10^(-12))` `Mg(OH)_(2)` will not precipitate. b. When `Mg(OH)_(2)` starts precipitation, then, `[Mg^(2+)] [OH^(Theta)]^(2) = K_(sp) of Mg (OH)_(2)` `[0.1] [OH^(Theta)]^(2) = 1 xx 10^(-11)` `:. [OH^(Theta)] = 10^(-5)M :. pOH =5` `:. pOH = 14 - pOH` `pH = 14 - 5 = 9` c. `{:(,MgCI_(2)+,2NaOhrarr,Mg(OH)_(2)+,2NaCI),("mmoles before reaction",rArr10,20,0,0),("mmoles after reaction",rArr0,0,10,20):}` Thus, `10mmol` of `Mg(OH)_(2)` are formed. The product of `[Mg^(2)] [OH^(Theta)]^(2)` is, therefore, `[(10)/(200)]xx[(20)/(200)]^(2) = 5 xx 10^(-4)` which is more than `K_(sp)` of `Mg(OH)_(2)`. Now solubility `(S)` of `Mg(OH)_(2)` can be derived by `K_(sp) = 4S^(3)` `:. S = 3sqrt((K_(sp))/(4)) = 3sqrt((1.2 xx 10^(-11))/(4)) = 1.4 xx 10^(-4)` `:. [OH^(Theta)] = 2S =2.8 xx 10^(-4)`. |
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| 83. |
Which pair will show common ion effect ?A. `BaCl_(2) +Ba(NO_(3))_(2)`B. `NaCl+ HCl`C. `NH_(4)OH +NH_(4)Cl`D. `AgCN +KCN` |
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Answer» Correct Answer - C Since `NH_(4)OH` is a weak electrolyte whereas `NH_(4)Cl` is a strong electrolyte. |
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| 84. |
The `pH` value of `0.001M` aqueous solution of `NaCI` isA. 7B. 4C. 11D. unpredictable |
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Answer» Correct Answer - A NaCl does not hydrolyse, hence it produces neutral solution |
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| 85. |
A solution of an acid has `pH= 4.70`. Find out the concentration of `OH^(-)` ions `(pK_(w)=14)`.A. `5 xx 10^(-10)`B. `4 xx 10^(-10)`C. `2 xx 10^(-5)`D. `9 xx 10^(-4)` |
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Answer» Correct Answer - A `pH=4.70 implies -log [H^(+)]=4.70` `:. log[H^(+)]=-4.7 =bar(5).30` or `[H^(+)]=1.999xx10^(-5)~~2 xx 10^(-5)` `pK_(w)=14 implies [H^(+)]xx[OH^(-)]=10^(-14)` `:. [OH^(-)]=(10^(-14))/(2 xx 10^(-5))=0.5 xx 10^(-9) = 5 xx 10^(-10)` |
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| 86. |
The units of ionic product of water `(K_(w))` areA. `mol^(-1) L^(-1)`B. `mol ^(-2) L^(-2)`C. `mol^(-2) L^(-1)`D. `mol^(2)L^(-2)` |
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Answer» Correct Answer - D `K_(w)=[H^(+)][OH^(-)]` Units of `K_(w)=(mol L^(-1))(mol L^(-1))` `=mol^(2)L^(-2)` |
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| 87. |
The value of `pK_(w)` of waterA. increases with increase in temperatureB. decreases with rise in temperatureC. does not change with variation in temperatureD. increases till `50^(@)C` and there after decreases. |
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Answer» Correct Answer - B With increase of temperature `K_(w)` of water increases while `pK_(w)` of water decreases. |
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| 88. |
Hydrolysis of the salt of a strong acid and weak base willA. Increase with increase in temperatureB. Decrease with increase in temperatureC. Remains unaffected with change in temperatureD. Remains uneffected with change in concentration of the salt |
| Answer» Correct Answer - A | |
| 89. |
Which of the following salt undergoes hydrolysis?A. `CH_(3)COOK`B. `NaNO_(3)`C. `KCI`D. `K_(2)SO_(4)` |
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Answer» Correct Answer - A Similarly like in problem `102`(above), `NaNO_(3), KCI` and `K_(2)SO_(4)` are salts of `S_(A)//S_(B)` and do not hydrolyse. But `CH_(3)COOk` is a salt of `W_(A)//S_(B)` and hydrolyses. |
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| 90. |
`10 mL` of `10^(-6) M HCl` solution is mixed with `90mL H_(2)O`. `pH` will change approximately:A. by one unitB. by 0.3 unitC. by 0.7 unitD. by 0.1 unit |
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Answer» Correct Answer - c |
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| 91. |
`K_(sp)` of `Mg(OH)_(2)` is `1.8 xx 10^(-11)` at `30^(@)C` . Its molar solubility is `…..........` at pH `=5`A. `1.8 xx 10^(-11)M`B. `1.8 xx 10^(-9) M`C. `1.34 xx 10^(-54)M`D. `1.8 xx 10^(-7)M` |
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Answer» Correct Answer - D `Mg(OH)_(2) hArr Mg^(2+)+2OH^(-)` `pH =12 implies pOH =2` `:. [OH^(-)]=10^(-2)M` `K_(sp)=[Mg^(2+)][OH^(-)]^(2)` `1.8 xx 10^(-11) =s xx (10^(-2))^(2)` `:. s=1.8 xx 10^(-7) M` |
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| 92. |
The maximum pH of a solution which is `0.10 M` is `Mg^(2+)` from which `Mg(OH)_(2)` is not precipitated is [`K_(sp)` of `Mg(OH)_(2) = 1.2 xx 10^(-11) M^(3)`]A. `4.96`B. `6.96`C. `7.04`D. `9.04` |
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Answer» Correct Answer - D For `Mg(OH)_(2)` not to be precipitated `[OH^(-)] lt [(K_(sp)(Mg(OH)_(2)))/([Mg^(2+)])]^(1//2)` `[OH^(-)] lt [(1.2x10^(-11)M^(3))/(0.10M)]^(1//2) lt 1.035x10^(-5)M` `pOH lt 4.36,pH gt 14-4.36 = 9.04` |
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| 93. |
Fear or exitement, generally cause one to breathe rapidaly and it results in the decrease of concentration of `CO_(2)` in blood. In what way it will change pH of blood ?A. pH will increseB. pH will decreaseC. No changeD. pH will be 7 |
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Answer» Correct Answer - C Blood buffer do not allow a change in pH. It contains serum protein which act as buffer. |
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| 94. |
A monoprotic acid `(HA)` is `1%` ionised in its aqueous solution of `0.1M` strength. Its `pOH` will beA. `11`B. `3`C. `10`D. `2` |
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Answer» Correct Answer - A `{:(,HA+,H_(2)OhArr,H_(3)O^(o+)+,A^(Theta)),("Initial",1,0,0,0),("Final",C(1-alpha),,Calpha,Calpha):}` `[H_(3)O^(o+)] = Calpha = 0.1 xx 0.01 = 10^(-3)`. `pH = 3, pOH = 11`. |
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| 95. |
A monoprotic acid `(HA)` is `1%` ionised in its aqueous solution of `0.1M` strength. Its `pOH` will beA. 11B. 3C. 10D. 2 |
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Answer» Correct Answer - A `[H_(2)O^(+)]=c alpha =0.1 xx (0.01)=1xx10^(-3)` `:. pH=3` . Hence, `pOH =14-3=11`. |
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| 96. |
Hydrolysis constant of `NH_(4)^(+)` is `5.55 xx 10^(-10)`. The ionisation constant of `NH_(4)^(+)` isA. `1.8xx 10^(9)`B. `5.55xx10^(-10)`C. `5.55xx10^(4)`D. `1.8xx10^(-5)` |
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Answer» Correct Answer - B `NH_(4)^(+)+H_(2)O hArr NH_(4)OH +H^(+)` `K_(h)=([NH_(4)OH][H^(+)])/([NH_(4)^(+)])` Since `[NH_(4)OH]=[H^(+)]` `:. K_(h)=([H^(+)]^(2))/ ([NH_(4)^(+)])` `NH_(4)^(+)` ionises as follows , `NH_(4)^(+) hArr NH_(3) +H^(+)` `K_(a)=([NH_(3)][H^(+)])/([NH_(4)^(+)])` Since `[NH_(3)]=[H^(+)]` `:. K_(a)=([H^(+)]^(2))/([NH_(4)^(+)])` `:. K_(a)=K_(h)=5.55 xx 10^(-10)` |
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| 97. |
The `pH` of blood is 7.4 . What is the ratio of `[(HPO_(4)^(2-))/(H_(2)PO_(4)^(-))]` in the blood. `pK_(a)(H_(2)PO_(4)^(-))=7.1`A. `2:1`B. `1:2`C. `3:1`D. `1:3` |
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Answer» Correct Answer - 1 `{:H_(2)PO_(4)^(-),hArr,HPO_(4)^(2-),+,H^(+):}` `8xx10^(-8)=((HPO_(4)^(2-))/(H_(2)PO_(4)^(2-)))xx4xx10^(-8)" "rArr" "((HPO_(4)^(2-))/(H_(2)PO_(4)^(2-)))=(2)/(1)` |
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| 98. |
The pH of `1 M PO_(4)^(3-) (aq)` solution is, [ Given `pK_(b)=(PO_(4)^(3-))=1.62]`A. 13.19B. 1.62C. 8.1D. 4.86 |
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Answer» Correct Answer - A `PO_(4)^(3-)+H_(2)O hArr HPO_(4)^(2-)+OH^(-)` `[OH^(-)]=sqrt(K_(b)C)` `pOH =(1)/(2)[pK_(b)-log C]` `=(1)/(2)[1.62-log 1 ] =0.81` `pH =14-pOH =14-0.81=13.19` |
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| 99. |
pH of `10^(-9)` M HCl isA. 8B. 7C. `lt7`D. `gt7` |
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Answer» Correct Answer - C `lt7` |
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| 100. |
Calculate the amount of `NH_(3)` and `NH_(4)CI` required to prepare a buffer solution of pH `9.0` when total concentration of buffering reagents is `0.6 mol L^(-1)`. `(pK_(b)for NH_(3)=4.7,log 2=0.30)` |
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Answer» `pOH=-logK_(b)=log""(["Salt"])/(["Base"])` `5=4.7+log""(a)/(b)` `(a)/(b)=2 :.a=2b` given`a+b=0.6` `2b+b=0.6` `3b=0.6` ,brgt `b=0.2mol e` and `a=0.4mol e` thus `["Salt"]=0.4M`and `[Base]=0.2M` |
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