InterviewSolution
Saved Bookmarks
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.
| 701. |
Calculate the `pH` of solution obtained by mixing `10 ml` of `0.1 M HCl` and `40 ml` of `0.2 M H_(2)SO_(4)`A. 0.74B. 7.4C. 4.68D. 0.486 |
| Answer» Correct Answer - 4 | |
| 702. |
Calculate the `pH` of solution obtained by mixing `10 ml` of `0.1 M HCl` and `40 ml` of `0.2 M H_(2)SO_(4)` |
|
Answer» Correct Answer - D Milli equivalent of `H^(o+)` from `HC1 = 10 xx 0.1 = 1` Milli equivalent of `H^(o+)` from `H_(2)SO_(4) = 40 xx 0.2 xx2 = 16` `:.` Total `mEq` of `H^(o+)` in solution `= 1 + 16 = 17` `:. [H^(o+)] = (17)/(50) = 3.4 xx 10^(-1) (because [H^(o+)] = (mEq)/(V_("in mL")))` `:. pH =- log [H^(o+)] =- log (0.34)` `pH = 0.4685` |
|
| 703. |
The `pH` of a solution increased from `3` to `6`. Its `[H^(o+)]` will beA. Reduced by `1000` timesB. Increased to `1000` timesC. DoubledD. Reduced to half |
|
Answer» Correct Answer - A `pH = 3, [H^(o+)] = 10^(-3)` `pH = 6, [H^(o+)] = 10^(-6)` `[H^(o+)]` decreases, which is only possible if the solution is diluted to `1000` times, i.e., |
|
| 704. |
For pure water,A. Both `pH` and `pOH` decrease with increase in temperature.B. Both `pH` and `pOH` increase with increase in temperature.C. `pH` decreases and `pOH` increases with increase in temperature.D. `pH` increase and `pOH`decreases with increase in temperature. |
|
Answer» Correct Answer - A With increase in temperature the ionisation of `H_(2)O` increases and hence the `[H^(o+)]` and `[overset(Theta)OH]` increases. Therefore, `pH` and `pOH` decreases with increase of temperature. |
|
| 705. |
Which of the following is the strongest acid?A. `CIO_(3)(OH)`B. `CIO_(2)(OH)`C. `SO(OH)_(2)`D. `SO_(2)(OH)_(2)` |
|
Answer» Correct Answer - A `CIO_(3)(OH)` or `HCIO_(4)` is the strongest acid. In oxy acids of halogens, the higher the oxidation state of the hsalogen, the stronger the acid. |
|
| 706. |
If `Ag^(+)+NH_(3)hArr[Ag(NH_(3))]^(+)`, `K_(1)=3.5xx10^(-3)` and `[Ag(NH_(3))]^(+)+NH_(3)hArr[Ag(NH_(3))_(2)]^(+)`, `K_(2)=1.74xx10^(-3)`. The formation constant of `[Ag(NH_(3))_(2)]^(+)` is :A. `6.08 xx 10^(-6)`B. `6.08 xx 10^(6)`C. `6.08 xx 10^(-9)`D. None of these |
|
Answer» Correct Answer - A If a reaction occurs in two steps, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants for the individual steps. `Ag^(o+) +NH_(3) hArr [Ag(NH_(3))]^(o+), K_(1) = 3.5 xx 10^(-3)` `[Ag(NH)_(3)]^(o+) +NH_(3) hArr [Ag(NH_(3))_(2)]^(o+), K_(2) = 1.7 xx 10^(-3)` `Ag^(o+) +2NH_(3) hArr [Ag(NH_(3))_(2)]^(o+)`, Over all `K = K_(1) xx K_(2)` `= 3.5 xx 10^(-3) xx 1.7 xx 10^(-3)` `= 5.95 xx 10^(-6)` |
|
| 707. |
The temperature of `1` mole helium gas is increased by `1^@C`. Find the increase in internal energy.A. 7 calB. 5 calC. 3.5 calD. 3 cal |
|
Answer» Correct Answer - D `((DeltaE)/(DeltaT))_(V)=C_(v)` `:. Delta E =C_(v) DeltaT` For a monoatomic gas, Here, `Delta T =1` `:. Delta E =3//2 R xx 1` `=3//2 xx2 cal =3 cal` |
|
| 708. |
`100 mL` of `0.2 N NaOH` is mixed with `100` mL `0.1 NHCl` and the solution is made 1L. The pH of the solution is :A. 4B. 8C. 10D. 12 |
|
Answer» Correct Answer - D `[OH^(-)] = (100 xx 0.2-100 xx 0.01)/(1000)` `= 10^(-2) rArr pOH = 2 rArr pH = 12` |
|
| 709. |
Three reactions involving `H_(2)PO_(4)^(-)` are given below `I. H_(3)PO_(4)+H_(2)OrarrH_(3)O^(+)+H_(2)PO_(4)^(-)` `II. H_(2)PO_(4)^(-)+H_(2)OrarrHPO_(4)^(2-)+H_(3)O^(+)` `III. H_(2)PO_(4)^(-)+OH^(-)rarrH_(3)PO_(4)+O^(2+)` In which of the above does `H_(2)PO_(4)^(-)` act as an acid?A. (ii) onlyB. (i) and (ii)C. (iii) onlyD. (i) only |
|
Answer» Correct Answer - A (i) `{:(H_(3)PO_(4)+H_(2)OhArrH_(3)O^(+)+,H_(2)PO_(4)^(-)),("acid","conjugate base"):}` In reverse reaction `H_(2)PO_(4)^(-)` is accepting a proton, so it base. (ii) `{:(H_(2)PO_(4)^(-)+H_(2)OhArr,HPO_(4)^(2-),+H_(3)O^(+)),("acid","conjugate base",):}` `H_(2)PO_(4)^(-)` is donating a proton, which is accepting by `H_(2)O`, so it is an acid. (iii) `{:(H_(2)PO_(4)^(-)+,OH^(-)hArr,H_(3)PO_(4),+O^(2-)),("acid","acid",conjugate base",):}` `H_(2)PO_(4)^(-)` is a accepting a proton, so it is base. Only in reaction (ii) `H_(2)PO_(4)^(-)` acids as an acid. |
|
| 710. |
pH of 0.01 M `(NH_(4))_(2) SO_(4)` and 0.02 M `NH_(4)OH` buffer `(pK_(a)` of `NH_(4)^(+)=9.26)` isA. `4.74+log 2 `B. `4.74 -log2`C. `9.26+log 2`D. `9.26+log 1` |
|
Answer» Correct Answer - D `pK_(a)(NH_(4)^(+))=9.26` `pK_(b)(NH_(3))=14-9.26=4.74` `[NH_(4)]=2xx[(NH_(4))_(2)SO_(4)]` `=2xx0.01=0.02 M` `pOH =pK_(b)+log . ([NH_(4)^(+)])/([NH_(4)OH])` `pOH =4.74 +log . (0.02)/(0.02)=4.74-log 1` `pH =14-pOH =1404.74+log 1` `=9.26+log 1` |
|
| 711. |
What is the pH at the equivalence point in a titration of `0.2 M NH_(3) (aq)` with `0.02 M HBr` (aq). For the ammonia `K_(b) = 1.8 xx 10^(-5)`A. `5.48`B. `5.6`C. `7.0`D. `8.5` |
|
Answer» Correct Answer - B pH `= 7 - (1)/(2)pKb - 1/2 "log"C` |
|
| 712. |
A sample of `100 mL` of `0.10 M` weak acid, HA `(K_(a) = 1.0 xx 10^(-5))` is tirated with standard `0.10 M KOH`. What volume of `KOH` must have been addeed when the pH in the titration flask is 5A. `10`B. `25`C. `50`D. `100` |
|
Answer» Correct Answer - C In the titration of weak acid and strong solution acts as acidic buffer, before the equivalence point. `:. pH - pK_(a) + "log" (["salt"])/(["base"])` Given pH `= pK_(a) = 5` `:. ["salt"] = ["acid"]` `:. 0.1 xx V_(a) = 10 - (0.1 xx V_(B))` `:. V_(B) = 50 mL` |
|
| 713. |
Three reactions involving `H_(2)PO_(4)^(-)` are given below (i) `H_(3)PO_(4) + H_(2)O rarr H_(3)O^(+) + H_(2)PO_(4)^(-)` (ii) `H_(2)PO_(4)^(-) + H_(2)O rarr HPO_(4)^(2-) + H_(3)O^(+)` (iii) `H_(2)PO_(4)^(-) + OH^(-) rarr H_(3)PO_(4) + O^(2-)` In which of the above does `H_(2)PO_(4)^(-)` act as an acidA. (i) onlyB. (ii) onlyC. (i) and (ii)D. (iii) only |
| Answer» Correct Answer - B | |
| 714. |
What is the pH of a 0.15 M solution of formic acid, HCOOH ? `{:("Formic Acid",K_a),(HCOOH,1.9xx10^(-4)):}`A. 1.49B. 2.27C. 3.72D. 4.55 |
|
Answer» Correct Answer - d |
|
| 715. |
When one drop of a concentrated `HC1` is added to `1L` of pure water at `25^(@)C`, the `pH` drops suddenly from `7` to `4`. When the second drop of the same acid is added, the `pH` of the solution further drops to aboutA. `0`B. `1.0`C. `2.0`D. `3.7` |
|
Answer» Correct Answer - D When `pH = 7, [H^(o+)]` initial `=10^(-7)M`. When `pH = 4, [H^(o+)]` Total `=10^(-4)M`. `:. [H^(o+)]` added from `HC1 = (10^(-4) - 10^(-7))` `=10^(-4) (1-0.001)~~ 10^(-4)M`. When another drop of `HC1` is added: When `pH = 4, [H^(o+)] "initial" = 10^(-4)M` `:. [H^(o+)] added ~~ 10^(-4)` `[H^(o+)] Total = (10^(-4)+10^(-4)) = 2 xx 10^(-4) M` `:.pH =- log (2 xx 10^(-4)) =- 0.3 + 4 = 3.7` Note: Change in volume on adding one drop of `HC1` is assumed to be negligible. |
|
| 716. |
Calculate the amount of acetic acid presnt in `1L` of solution having `alpha = 1%` and `K_(a) = 1.8 xx 10^(-5)`. |
|
Answer» Correct Answer - A `{:(CH_(3)COOH hArr,CH_(3)COO^(Theta)+,H^(o+),,),(1,0,0,,),(1-alpha,alpha,alpha,,):}` where `alpha` is degree of dissociation of acid, if `C mol L^(-1)` is concentration of acid, then `[H^(o+)] = C alpha, [CH_(3)COO^(Theta)] = C alpha, [CH_(3)COOH] = C(1-alpha)` Also `K_(a) = ([H^(o+)][CH_(3)COO^(Theta)])/([CH_(3)COOH]) = (C alpha.C alpha)/(C(1-alpha)) = C alpha^(2)` [`:. K_(a)` is small `:. alpha` will also be small and thus `1-alpha =1]` or `1.8 xx 10^(-5) = X xx ((1)/(100))^(2) :. C = 0.18` `:. 1L` solution contains = 0.18 mole of `CH_(3)COOH` `:. 1L` solution contains `= 0.18 xx 60 = 10.8 gCH_(3) COOH` |
|
| 717. |
The dissociation constant of acetic acid at a given temperature is `1.69xx10^(-5)`.The degree of dissociation of 0.01 M acetic acid in presence of 0.01 M HCl is equal to :A. 0.41B. 0.13C. `1.69xx10^(-3)`D. `0.013` |
|
Answer» Correct Answer - c |
|
| 718. |
The mass of acetic acid present in `500 ml` of solution in which it is `1%` ionides (Ka of ` CH_(3)COOH = 1.8 xx 10^(-4)`)A. `5.4`B. `12.6`C. `6.4`D. `10.8` |
|
Answer» Correct Answer - A Normality `= (Ka)/(alpha^(2)) = (1.8 xx 10^(-5))/((10^(-2))^(2)) = 1.8 xx 10^(-1) N` `W = (1.8 xx 10^(-1) xx 60 xx 500)/(1000) = 5.4 g` |
|
| 719. |
Identify Bronsted -Lowry acids in the reactions given . `[Al(H_(2)O)_(6)]^(+3) + HCO_(3)^(-) hArr` `underset(C)([Al(H_(2)O)_(5)(OH^(-))]^(2+)) + underset(D)(H_(2)CO_(3))` The correct AnswerA. `A,C`B. `B,D`C. `A,D`D. `B,C` |
|
Answer» Correct Answer - C Proton donor is bronsted-lowry acid |
|
| 720. |
A solution consist of `0.2 M NH_(4)OH` and `0.2 M NH_(4)Cl`. If `K_(b)` of `NH_(4)OH` is `1.8 xx 10^(-5)`, the `[OH^(-)]` of the resulting isA. `0.9 xx 10^(-5) M`B. `1.8 xx 10^(-5) M`C. `3.2 xx 10^(-5) M`D. `3.6 xx 10^(-5) M` |
|
Answer» Correct Answer - B `[OH^(-)] = (K_(b)["Base"])/(["Salt"])` `= 1.8 xx 10^(-5) xx (0.2)/(0.2), = 1.8 xx 10^(-5)` |
|
| 721. |
The strongest base of the following species isA. `NH_(2^(-))`B. `OH^(-)`C. `O^(-2)`D. `S^(2-)` |
|
Answer» Correct Answer - A Weak acid conjugate base is the strongest base |
|
| 722. |
Which anions is strongest base?A. `C_(2)H_(5)O^(-)`B. `NO_(2^(-))`C. `Cl^(-)`D. `CH_(3)COOH^(-)` |
|
Answer» Correct Answer - A Weak acid conjugate base is the strongest base |
|
| 723. |
When `Na_(2)CO_(3)` solution is titrated against HCl solution, the indicator used isA. PhenolphthaleinB. Methyl OrangeC. Methyl redD. Starch |
|
Answer» Correct Answer - B Methyl orange indicator is used. |
|
| 724. |
`50 ml` of `0.05 M Na_(2)CO_(3)` is titrated against `0.1 M HCl`. On adding `40 ml` of `HCl, pH ` of the resulting solution will be `[{:(H_(2)CO_(3):pK_(a_(1))=6.35,pK_(a_(2))=10.33),(log3=0.477,log2=0.30):}]`A. `6.35`B. `6.526`C. `8.34`D. `6.173` |
|
Answer» Correct Answer - D `CO_(3)^(2-) + H^(-) rArr HCO_(3)^(-)` Initial millies moles `50 xx 0.05 40 xx 0.1` Final milli moles `1.5 , 2.5` `{:(,HCO_(3)^(-)+,H^(+),rarr,H_(2)CO_(3)),("Initial milli mles",2.5,1.5,,"__"),("Final milli moles",1,"___",,1.5):}` `pH = pK_(a_(1)) + "log"([HCO_(3)^(-)])/([H_(2)CO_(3)]) = 6.173` |
|
| 725. |
Concentration of `NH_(4)Cl` and `NH_(4)OH` in a buffer solution are in the ratio of `1 : 10`, Kb for `NH_(4)OH` is `10^(-10)`. The pH of the buffer isA. `4`B. `5`C. `9`D. `11` |
|
Answer» Correct Answer - B `P^(OH) = 10 + "log" (1)/(10)` `P^(OH) = 9 , P_(H) = 5` |
|
| 726. |
Assertion: In the titration of `Na_(2)CO_(3)` with `HCl` using methyl orange indicator, the volume of acid required is twice that of the acid required using phenolphthalein as indicaton. Reason: Two moles of `HCl` are required for the complete neutralisation of one mole of `Na_(2)CO_(3)`.A. If both (A) and (R) are correc, and (R) is the correct explanation of (A).B. If both (A) and (R) are correct but (R) is not the correct explanation of (A).C. If (A) is correct, but (R) is incorrect.D. If (A) is incorrect, but (R) is correct. |
|
Answer» Correct Answer - B Both `(A)` and `(R)` are correct but `(R)` is not the correct explanation of `(A)`. Correct `R: Na_(2)CO_(3)` reacts with `HCI` upto the stage of `NaHCO_(3)` in the presence of phenolphthalein. `Na_(2)CO_(3) + HCI rarr NaHCO_(3) + NaCI` |
|
| 727. |
When `Na_(2)CO_(3)` solution is titrated against HCl solution, the indicator used isA. PhenophthaleinB. Dilute `H_(2)SO_(4)`C. Methyl orangeD. None |
|
Answer» Correct Answer - C Factual question. |
|
| 728. |
The pH of a solution of `H_(2)O_(2)` is 6.0. Some chlorine gas is bubbled into this solution. Which of the following is correct ?A. The pH of the resulting solution becomes 8B. Hydrogen gas is liberated from the resultant solution.C. The pH of the resultant solution becomes less than 6.0 and oxygent gas is liberatedD. `Cl_(2)O` is formed in the resultant solution |
|
Answer» Correct Answer - C `{:(,H_(2)O_(2),rarr,H_(2)O,+,O),(Cl_(2)+,H_(2)O+O,rarr,2HCl,+,O_(2)):}` `ul (bar(Cl_(2)+H_(2)O_(2)rarr 2HCl +O_(2)))` Due to the formation of stronga cid HCl, the pH of the solution will decrease along with the formation of `O_(2)` gas |
|
| 729. |
The dissoctiation of water of `25^(@)C` is `1.9 xx 10^(-7)%` and the density of water is `1g//cm^(3)` the ionisation constant of water isA. `3.42 xx 10^(-6)`B. `3.42 xx 10^(-8)`C. `1.00 xx 10^(-14)`D. `2.00 xx 10^(-16)` |
|
Answer» Correct Answer - D `alpha = 1.9 xx 10^(-7) % = (1.9 xx 10^(-7))/(100) , C= (1000)/(8)` ltbr gt `K_(a) = ([H^(-)][OH^(-)])/([H_(2)O])= Calpha^(2)` `= 1.9 xx 10^(-9) xx 1.9 xx 10^(-9) xx (1000)/(18) = 2.0 x 10^(-16)` |
|
| 730. |
Higher the amount of acid or base used to product a deinite change of pH in a buffer solution, higher will be its buffe capacity. Buffer capcity of solution is maximum under the following conditions [Salt] = [Acid](in acid buffer) [Salt] = [Base] (in base capacity] pH of buffer solution lies inthe range given below `pH = pH_(a) +-1` In other words,any buffer solution can be used as buffer up to two pH units only, depending upon the value of `pK_(a)` or `pK_(b)`. A buffer is said to be efficient when `pH = pK_(a)` or `pOH = pK_(b)` Acifity of `BF_(3)` can be explained on the basis of which of the following conceptsA. Arrehenius conceptB. Bronsted Lowry conceptC. Lewis conceptD. Bronsted Lowry as well as Lewis concept |
| Answer» Correct Answer - C | |
| 731. |
Higher the amount of acid or base used to product a deinite change of pH in a buffer solution, higher will be its buffe capacity. Buffer capcity of solution is maximum under the following conditions [Salt] = [Acid](in acid buffer) [Salt] = [Base] (in base capacity] pH of buffer solution lies inthe range given below `pH = pH_(a) +-1` In other words,any buffer solution can be used as buffer up to two pH units only, depending upon the value of `pK_(a)` or `pK_(b)`. A buffer is said to be efficient when `pH = pK_(a)` or `pOH = pK_(b)`Buffer capacity is maximum whenA. One mole of `NH_(4)Cl` is added to two moles of `NH_(4)OH`B. One mole of `NH_(4)Cl` is added to one moles of `NH_(4)OH`C. One mole of `NH_(4)Cl` is added to one moles of `NaOH`D. One mole of `NHCl` is added to one moles of `NaOH` |
| Answer» Correct Answer - B | |
| 732. |
The correct statementsA. All are correctB. A & C correctC. B,C & D correctD. D is correct |
|
Answer» Correct Answer - B A) `P^(H)` of water decreases with increase in temperature due to decrease concentration of `H^(+)` and `OH^(-)` ions C) `P^(H)` of water decreases with addition of acid |
|
| 733. |
Which of the following statement (s) is/are correctA. pOH of an acidic buffer decreases if less salt is addedB. pOH of a basic buffer decreases if less salt is addedC. At the saturation point the ionic product is equal to solubility productD. The term solubility product is only for sparingly soluble salt |
|
Answer» Correct Answer - A::B::C `pH = pK_(a) + log.(["Salt"])/(["Acid"]), pOH = pK_(b) + log.(["Salt"])/(["Base"])` Thus as salt concentration decreases for acidic buffers, so pH decreases. Statement (a) is correct, Thus as salt concentration decreases for basic buffers, pOH decreases, Choice (b) is correct. For any salt when `Q = K_(sp)` Hence the solutions is saturated. Hence choice (c) is correct. The solubility product is for all salts, choice (d) is incorrect. |
|
| 734. |
Choose the correct statement/s:A. pH of acidic buffer solution decreases if more salt is addedB. pH of acidic buffer solution increase if more salt is addedC. pH of basic buffer decrases if more salt is addedD. pH of basic buffer incrases if more salt is added |
| Answer» Correct Answer - B::C | |
| 735. |
At any temperature for a neutral solutionA. `pH gt P^(OH)`B. `pH = P^(OH) = 7`C. `pH = P^(OH)`D. `pH lt P^(OH)` |
|
Answer» Correct Answer - C At any temperature `P^(H) = P^(OH)` |
|
| 736. |
If you have a saturated solution of `CaF_(2)`, thenA. `[Ca^(2+)] = (k_(sp)//4)^(1//3)`B. `2xx[Ca^(2+)]=[F^(-)]`C. `[Ca^(2+)] = 2[F^(-)]`D. `[Ca^(2+)] = sqrt(K_(sp))` |
| Answer» Correct Answer - A::B | |
| 737. |
At `50^(0)C, pH + pOH` isA. Less than `14`B. More than `14`C. Equal to `14`D. Equal to `4` |
|
Answer» Correct Answer - A At `50^(0)c P^(H) + P^(OH) lt 14` |
|
| 738. |
pH of `HCl(10^(-12)M)` isA. 12B. `-12`C. `~~7`D. 14 |
|
Answer» Correct Answer - C AS the solution is acidic `pH lt 7`. This is because `[H^(+)]` from `H_(2)O[10^(-7)]` cannot be neglected in comparison to `10^(-12) M`. |
|
| 739. |
Which one is buffer solutionA. `[PO_(4)^(-)] [HPO_(4)^(-)]`B. `[PO_(3)^(3-)] [[H_(2)PO_(4)^(-)]`C. `[HPO_(4)^(-)] [H_(2)PO_(4)^(-)]`D. All of these |
|
Answer» Correct Answer - B [Normal salt + acidic salt] is a buffer solution. |
|
| 740. |
The solubility of `Ag_(2)CO_(3)` in water at `25^(@)` is `1xx10^(-4)` mole /litre. What is its solubility in 0.01 M `Na_(2)CO_(3)` solution? Assume no hydrolysis of `CO_(3)^(2-)` ion.A. `6xx10^(-6)` mole/litreB. `4xx10^(-5)` mole/litreC. `10^(-5)` mole/litreD. `2xx10^(-5)` mole/litre |
|
Answer» Correct Answer - c |
|
| 741. |
In which of the following acid-base titration, the pH is greater than 8 at the equivalence point ?A. Aceitic acid vs ammoniaB. Acetic acid vs sodium hydroxideC. Hydrochloric acid vs ammoniaD. Hydrochloric acid vs sodium hydroxide |
|
Answer» Correct Answer - B Titration of `W_(A)//S_(B).pH` at equivalence point is `gt8`. |
|
| 742. |
The solubility of` PbCl_(2)` at `25^(@)C` is `6.3 xx 10^(-3)` mole/litre. Its solubility product at that temperature isA. `(6.3 xx 10^(-3)) xx (6.3 xx 10^(-3))`B. `(6.3 xx 10^(-3)) xx (12.6 xx 10^(-3))`C. `(6.3 xx 10^(-3)) xx (12.6 xx 10^(-3))^(2)`D. `(12.6 xx 10^(-3)) xx (12.6 xx 10^(-3))` |
|
Answer» Correct Answer - C `PbCl_(2) rarr Punderset(S)(b^(++)) + underset(2S)(2Cl^(-))` `K_(sp) = S xx (2S)^(2) = [6.3 xx 10^(-3)] xx [12.6 xx 10^(-3)]^(2)`. |
|
| 743. |
select the correcty statement (s)A. If pOH of aq solution at `70^(@)` C is 7, then solution will be acidic.B. If`[H^(+)]` concentration is decreased 100 times, pH of solution increases by 2 unitsC. If solution of `CH_(3)COON(aq)` is diluted then pH increasesD. Buffer capacity remains constant with dilution |
|
Answer» Correct Answer - a,b |
|
| 744. |
For pure water:A. pH increases with increase in themperatureB. pH decreases with increase in temperatureC. pH=7 at temperature of `25^(@)C`D. pH increases as temperature decreases but decreases as stemperature increases |
|
Answer» Correct Answer - b,c,d |
|
| 745. |
If the equilibrium for reaction of `HCN` with `NaOH` is `10^(10)` then calculate `pH` of `10^(-3)MNaCN` solution of `25^(@)C` |
|
Answer» `{:(,HCN+NaOH,hArr,H_(2)O,+,NaCN,K=10^(10)),(rArr,CN^(-)+ H_(2)O,hArr,HCN,+,OH^(-),K=10^(10)),(t=0,10^(-3)M,,0,,0,),(at eq.,10^(-3)(1-h),,10^(-3)h,,10^(-3)h,):}` `K_(h)=10^(-10)=(10^(-3)hxx10^(-3)h)/(10^(-3)(1-h))rArrsqrt(K_(h)/(c ))=sqrt(10^(-7))( lt0.1)` `pH=7-(1)/(2)log10^(-10)+(1)/(2)log10^(-10)=7+5-(3)/(2)=10.5` |
|
| 746. |
In the following reaction : `[AI](H_(2)O)_(6)]^(3)+H_(2)O hArr H_(3)O+[AI(H_(2)O)_(5)OH]^(+2)`A. A is an acid, B is a baseB. A is a base, B is an acidC. C is conjugate acid of B and D is conjugate base of AD. C is conjugate base of B and D conjugate acid of A |
|
Answer» Correct Answer - a,c |
|
| 747. |
Calculate degree of hydrolysis(h) and pH of solution obtained by dissolving `0.1` moles of `CH_(3)COOHNa` in water to get `100L` of solution .Take `K_(a)` of acetic acid `=2xx10^(-5)` at `25^(@)C`. |
|
Answer» `c=(0.1)/(100)=1xx10^(-3)M` `K_(h)=(K_W)/(K_(a))=(10^(-14))/(2xx10^(-5))=5xx10^(-10)rArr h=sqrt((K_(h))/(c))=sqrt((5xx10^(-10))/(2xx10^(-5)))=5xx10^(-3)=0.5%` `,.pH=(1)/(2)[pK_(w)+pK_(a)+logc]=(1)/(2)[14+5-log2+log10^(-3)]=(1)/(2)[15.7]=7.85` |
|
| 748. |
In the reaction `[A (H_(2)O)_(6)]^(3+)+H_(2)O hArr [Al(H_(2)O)_(5)OH]^(2+)+H_(3)O^(+)`A. `[Al(H_(2)O)^(6)]^(3+)` is an acidB. `[Al(H_(2)O)^(6)]^(3+)` is a baseC. Both (A) and (B)D. None |
|
Answer» Correct Answer - A As `[Al(H_(2)O)_(6)]^(3+)` gives `H^(+)` to `H_(2)O` , it acts as an acid. |
|
| 749. |
What is true about zwitter ion. `H_(3)N^(+)CH_(2)COO?`A. `NH_(2)CH_(2)COO^(-)` is its conjugate baseB. `H_(3)N ^(+)CH_(2)COOH` is its conjugate acidC. Both (A) and (B)D. None of these |
|
Answer» Correct Answer - C `underset("acid")(H_(3)overset(+)(N)H_(2)COO^(-))hArr underset(" Conjugate base")(H_(3)NCH_(2)COO^(-))+H^(+)` `underset("Base")(H_(3)overset(+)(N)CH_(2)COO^(-))+H^(+) hArr underset(" Conjugate acid")(H_(3)overset(+)(N)CHOOH)` |
|
| 750. |
The correct relation between hydrolysis constant `(K_(b))` and degree of hydrolysis `(alpha)` for the following equilibrium isA. `alpha = sqrt((K_(w).C)/(K_(a)))`B. `alpha = sqrt((K_(w))/(K_(a).C))`C. `alpha = sqrt((K_(a).C)/(K_(w))`D. `alpha = sqrt((K_(a))/(K_(w).C))` |
|
Answer» Correct Answer - B Salt of weak acid and strong base. `K_(h) = (K_(w))/(K_(a))` `h = sqrt((K_(h))/(c))` `h = sqrt((K_(w))/(K_(a)c))` |
|