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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.
| 351. |
The aqueous solution of aluminium chloride is acidic due toA. Cation hydrolysisB. Anion hydrolysisC. Hydrolysis of both anion and cationD. Dissociation |
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Answer» Correct Answer - A `Al^(+++)` of `AlCl_(3)` undergoes hydrolysis. |
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| 352. |
Which buffer solution out of the following will have `pH gt 7`A. `CH_(3)COOH + CH_(3)COONa`B. `HCOOH + HCOOK`C. `CH_(3)COONH_(4)`D. `NH_(4)OH + NH_(4)Cl` |
| Answer» Correct Answer - D | |
| 353. |
Conjugate base of `NH_(3)` isA. `NH_(4)^(o+)`B. `NH_(2)^(o+)`C. `NH_(2)^(o+)`D. `N_(2)` |
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Answer» Correct Answer - C `NH_(3) hArr NH_(2)^(-) + H^(+)` |
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| 354. |
The conjugate base of `NH_(2)^(-)` isA. `NH_(3)`B. `NH^(2-)`C. `NH_(4)^(+)`D. `N_(3)^(-)` |
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Answer» Correct Answer - B `NH_(2)^(-) hArr NH^(-2) + H^(+)` Conjugate acid, base pair. |
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| 355. |
Glycine exists as the zwittr ion, `overset(+)NH_(3)CH_(2)COO^(-)`. Its conjugate base isA. `NH_(2)CH_(2)COOH^(2^(+))`B. `NH_(2)CH_(2)COO^(-)`C. `overset(+)NH_(3)CH_(2)COOH`D. `NH_(3)CH_(2)COO overset(+)H` |
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Answer» Correct Answer - B `NH_(3)^(+)CH_(2)COO^(-) overset(-H^(+))rarr NH_(2)CH_(2)COO^(-)` |
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| 356. |
Which of the following is the conjugate base of `[C_(2)H_(5)NH_(3)]^(+)`?A. `[C_(2)H_(5)NH]^(-)`B. `[C_(6)H_(5)NH_(3)]OH`C. `C_(2)H_(5)NH_(2)`D. `C_(2)H_(5)NH_(2)` |
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Answer» Correct Answer - C If removal proton from species, the resulting species is called conjugate base |
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| 357. |
The conjugate base of `NH_(2)^(-)` isA. `NH_(3)`B. `NH^(-2)`C. `NH_(4)^(+)`D. `N_(3)^(-1)` |
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Answer» Correct Answer - B `NH_(2)^(-)rarr underset("Conjugate base ")(NH^(2-))+H^(+)` |
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| 358. |
What is the maximum molarity of `Co^(+2)` ions in `0.1M HC1` saturated with `0.1M H_(2)S. (K_(a) = 4 xx 10^(-21))`. Given: `K_(sp)` of `CoS = 2xx10^(-21)`.A. `0.10M`B. `1.00M`C. `4.48 xx 10^(-11)M`D. `0.50M` |
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Answer» Correct Answer - D `H_(2)S hArr 2H^(o+) +S^(2-) (K_(a) = ([H^(o+)]^(2)[S^(2-)])/([H_(2)S]))` `[S^(2-)] = (K_(a)[H_(2)S])/([H^(o+)]^(2)) = (4 xx 10^(-21)xx0.1)/((0.1)^(2)) = 4 xx 10^(-20)M` For precipitation, `Q_(sp) = K_(sp)` `[Co^(2+)] = (K_(sp)of CoS)/([S^(2-)]) = (2xx 10^(-21))/(4xx10^(-20)) = 0.05M` |
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| 359. |
An aqueous solution contains `10%` amonia by mass and has a density of `0.99 g cm^(-3)`. Calculate hydroxy`1` and hydrogen ion concentration in this solution `K_(a)` for `NH_(4)^(o+) = 5.0 xx 10^(-10)M`. |
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Answer» Given, `("Weight of" NH_(3))/("Weight of solution") = 10//100` `100g` solution contains `10gm NH_(3)` `M_(NH_(3)) = (10 xx 1000) (17 xx (100//0.99)]` `= 5.82(V = mass//density)` Now `{:(NH_(3)+H_(2)Orarr,NH_(4)OHhArr,NH_(4)^(o+)+,overset(Θ)OH,),("Before dissociation",1,0,0,),("After dessociation",(1-alpha),alpha,alpha,):}` `:. [overset(Θ)OH] = Calpha = C sqrt((K_(b)//C)) = sqrt((K_(b)C))` `[C = 5.82M` and `K_(b) = K_(w)//K_(a) = 10^(-14)//(5xx10^(-10)) = 2 xx 10^(-5))` `[overset(Θ)OH] = sqrt(2xx10^(-5) xx 5.82) = 1.07 xx 10^(-2)M` `:. [H^(o+)] = 10^(-14)//1.07 xx 10^(-2) = 0.9268 xx 10^(-12)M` `:. pH =- log [H^(o+)] =- log 0.9268 xx 10^(-12)M = 12.033` |
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| 360. |
`k_(b_(1))` of `N_(2)H_(4)` is `4xx10^(-6)` .Then what is the acid dissociation constant of `N_(2)H_(5)^(+)` and `N_(2)H_(6)^(+)` respectively?A. data insufficient `4xx10^(-6)`B. data insufficient `2.5xx10^(-8)`C. `2.5xx10^(-9)` Data insufficientD. `2.5xx10^(-9),4xx10^(-6)` |
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Answer» `K_(a_(1))=(K_(w))/(K_(b_(1)))=(10^(-14))/(4xx10^(-6))=2.5xx10^(-9)` `K_(b_(2))` data not given so `K_(b_(2))` cannot be determined. |
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| 361. |
Which equilibrium constant(s) or ratio of equilibrium contants should be used to calculate the `pH` of `1.00L` of each of the following solutions? a. `KOH` b. `NH_(3)` c. `HC_(2)H_(3)O_(2)` d. `HC_(2)H_(3)O_(2) +NaC_(2)H_(3)O_(2)` e. `KC_(2)H_(3)O_(2)` f. `0.01 mol HC_(2)H_(3)O_(2) + 0.050 mol NaOh` g. `H_(2)S` h. `0.01NH_(4)CI + 0.50mol NaOH` i. `0.010 mol HC_(2)H_(3)O_(2) + 0.10mol NaOH` |
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Answer» a. `K_(w)(S_(B))` b. `K_(b)` and `K_(W)(W_(B))` c. `K_(a) (W_(A))` d. `K_(A)` (Acidic buffer) e. `K_(h) = K_(W)//K_(a)` (Salt of `W_(A)//S_(B))` f. `K_(a)` (The solution is the same as `0.05mol HC_(2)H_(3)O_(2) + 0.05 mol NaC_(2)H_(3)O_(2))` (It forms acidic buffer) g. `K_(1)` h. `K_(b)` and `K_(w)` (The reaction `NH_(4)^(o+) + overset(Θ)OH rarr H_(2)O + NH_(3)` produces a basic buffer solution of `NH_(3)` with the excess of `NH_(4)^(o+))` i. `K_(h) = K_(w)//K_(a)` (The neutralised acid results in the formation of salt of `W_(A)//S_(B)`. |
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| 362. |
An aqueous solution contains `10%` ammonia by mass and has a density of `0.09g cm^(-3)`. Calculate hydroxyl and hydrogen ion concentration in this solution. `(K_(a)for NH_(4)^(+)=5.0xx10^(-10)M)` |
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Answer» Given, `(wt.of NH_(3))/("wt.of solution")=(10)/(100)` `.: 100g "solution contains "10 g NH_(3)` `:. M_(NH_(3))=(10xx1000)//[17xx100//0.99)]=5.82` `(.: V= "Mass"//"Density")` Now `{:( NH_(3)+H_(2)Orarr, NH_(4)OH hArr NH_(4)^(+),+OH^(-)),("Before dissociation",1,0,0),("After dissociation",(1-alpha),alpha,alpha):}` `:. [OH^(-)]=C.alpha=Csqrt((K_(b)//C))=sqrt((K_(b).C))` `[.: C = 5.82 M and K_(b)=K_(w)//K_(a)=10^(-14)//(5xx10^(-10))=2xx10^(-5)]` `:. [OH^(-)]=sqrt([2xx10^(-5)xx5.82])=1.07xx10^(-2)M` `:. [H^(+)]=10^(-14)//1.07xx10^(-2)` `=.09268xx10^(-12)M` `:. pH= -log [H^(+)]= - log 0.9268xx10^(-12)` `=12.0330` |
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| 363. |
The dissociation constant of a weak acid `HA si 4.9 xx 10^(-8)`. After making the necessary approximations, calculate i. Percentage ionisation ii. `pH` iii. `overset(Theta)OH` concentration in a decimolar solution of the acid. Water has a `pH of 7`. |
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Answer» Correct Answer - A::C::D The dissociation constant of a weak acid `HA` is `4.9 xx 10^(-8)`. i. For the weak acid, `alpha = sqrt(((K_(a))/(C))) =sqrt(((4.9xx10^(-8))/(1//10)))` `= 7 xx 10^(-4) = 0.07 %` ii. `pH = - log [H^(o+)]` `=- log [C alpha] = [(1)/(10) xx7 xx 10^(-4)]` `=- log [7 xx 10^(-5)] = 4.15` iii. `[overset(Theta)OH] [H^(o+)] = 10^(-14)` or `[overset(Theta)OH] = (10^(-14))/([H^(o+)])` `= (10^(-14))/([Calpha]) = (10^(-14))/(7xx10^(-5)) =1.43 xx 10^(-10)` |
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| 364. |
Nicotinic acid `(K_(a)=1.4xx10^(-5))` is repersented by the formula HNiC. Calculate its per cent dissociation in a solution, which contains `0.10` mole of nicotinic acid per `2.0` litre of solution. |
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Answer» Given, `{:(HNiC hArr, H^(+)+,NiC^(-)),(1,0,0),((1-alpha),alpha,alpha):}` Also `C=(0.1)/(2)=5xx10^(-2)"mol litre"^(-1)`, `K_(a)=1.4xx10^(-5)` `:. K_(a)=(Calpha^(2))/((1-alpha))=Calpha^(2) , (.: 1-alpha=~~1)` `:. alpha=sqrt(((K_(a))/(C )))=sqrt(((1.4xx10^(-5))/(5xx10^(-2))))` `=1.67xx10^(-2)` or `1.67%` |
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| 365. |
Nicotinic acid `(K_(a) = 1.4 xx 10^(-5))` si represented by the formula `HNiC`. Calculate its percent dissociation in a solution which contains `0.10` moles of nictinic acid per `2.0L` of solution. |
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Answer» Given, `{:(HNiC,hArr,H^(o+),+,NiC^(Θ)),(1,,0,,0),(1-alpha,,alpha,,alpha):}` Also `C = 0.1//2 = 5 xx 10^(-2) mol L^(-1), K_(a) = 1.4 xx 10^(-5)` `:. K_(a) = (C alpha^(2))/((1-alpha)) = Calpha^(2)` `alpha = sqrt(K_(a)//C) = sqrt(1.4 xx 10^(-5))/(5xx10^(-2)) = 1.67 xx10^(-2) or 1.67%` |
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| 366. |
Statement-1: Aqueous solution of ammonium carbonate is basic. Statement-2: Acidic/basic nature of a salt solution of a salt of weak acid and weak base depends on `K_(a)` and `K_(b)` value of the acid and the base forming it.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-21B. 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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| 367. |
Assertion : Salting out action of sodium soap in presence of `NaCl` is based on common ion effect. Reason : Salting out action of soap is based on the fact that as the concentration of `Na^(+)` increases, the `RCOONa` shows precipitation because `[RCOO^(-)][Na^(+)]gtK_(sp)`.A. S is correct but E is wrong.B. S is wrong but E is correc.C. Both S and E are correct and E is correct explanation of S.D. Both S and E are correct but E is correct explanation of S. |
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Answer» Correct Answer - B `RCOONararrRCOO^(-)+Na^(+)`, In presence of `NaCI, [Na^(+)]` increases and `[RCOO^(-)][Na^(+)]` exceeds than `K_(SP)` of RCOONa. |
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| 368. |
The enthalpy of neutralisation of HCl by NaOH IS `-55.9 kJ ` and that of HCN by NaOH is `-12.1 kJ mol^(-1)`. The enthalpy of ionisation of HCN isA. `-68.0kJ`B. `-43.8 kJ`C. `68.0kJ`D. `43.8kJ`. |
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Answer» Correct Answer - D `Delta H_("ionisation")= - 12.1(-55.9)=43.8kJ` |
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| 369. |
100ml of 0.01 M solution of NaOH is diluted to `1dm^(3)`. What is the pH of the diluted solutionA. 12B. 11C. 2D. 3 |
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Answer» Correct Answer - B 100 ml 0.01 M NaOH solution is diluted to `1 dm^(3)` (i.e., 10 times dluted) hence, the resultant solution will be 0.001 M. `:. [OH^(-)] = 0.001 = 10^(-3)` `[H^(+)] = (10^(-14))/([OH^(-)]) = (10^(-14))/(10^(-3)) = 10^(-11)` `because pH = - log[H^(+)] = -log[10^(-11)]` pH = 11. |
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| 370. |
`10^(-6) M NaOH` is diluted by `100` times. The `pH` of diluted base isA. Between 6 and 7B. Between 10 and 11C. Between 7 and 8D. Between 5 and 6 |
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Answer» Correct Answer - C On dilution of 100 times concentration of `[OH^(-)]` becomes `10^(-8)M` . Thus pH is between 7 and 8. |
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| 371. |
Give the decreasing order of the acidic properties of oxides. a. `ZnO` , b. `KO_(2)` , c. `P_(2)O_(5)` , d. `MgO` |
| Answer» `Na_(2)O_(2) lt MgO lt ZnO lt P_(2)O_(5)` | |
| 372. |
The `K_(sp)` of `Ca(OH)_(2)` is `4.42 xx 10^(-5)` at `25^(@)C`. A 500 ml of saturated solution of `Ca(OH)_(2)` is mixed with an equal volume of `0.4M NaOH`. How much `Ca(OH)_(2)` in mg is precipitated ? |
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Answer» Correct Answer - `0743` |
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| 373. |
A 500 mL saturated solution of `MgCO_(3)` (M=84) is reduced to 120mL by evaporation. What mass of solid `MgCO_(3)` is formed? [`K_(sp)=4.0xx10^(-5)`}A. 0.013 gB. 0.064 gC. 0.20 gD. 0.27g |
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Answer» Correct Answer - c |
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| 374. |
At `pH lt3.1`, the indicator methyl red is coloured red ,at `pH gt 6.3` , it is yellow and at the intermediate values of the `pH` ,it is orange .What will the colour of indicator be in a `0.1 M` solution of `NH_(4)Br`? Take `pK_(b)(NH_(4)OH)=4.74`.A. RedB. YellowC. OrangeD. `pH=3.1` so colour cannot be predicted |
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Answer» `NH_(4)Br rarr` Salt of `SAWB` `pH=7-(1)/(2)pK_(b)-(1)/(2)logC=7-(4.74)/(2)log10^(-1)=7-2.37+(1)/(2)=5.13` |
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| 375. |
A certain indicator (an organic dye) has `pK_(a)=5`. For which of the following titrations may it be suitableA. Acetic acid against `NaOH`B. Aniline hydrochloride against `NaOH`C. Sodium carbonate against `HCl`D. Barium hydroxide against oxalice acid |
| Answer» Correct Answer - C | |
| 376. |
Equal volumes of 0.25 M `HNO_(2)` and 0.25 M `HNO_(3)` are titrated separtely with 0.25 M KOH. Which would be the same for both titrations ?A. Initial pHB. pH halfway to the equivalence pointC. pH at the equivalence point.D. pH when 5 mL excess KOH has been added |
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Answer» Correct Answer - d |
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| 377. |
A certain indicator (an organic dye) has `pK_(a)=5`. For which of the following titrations may it be suitableA. Acetic acid against NaOHB. Aniline hydrochloride against NaOHC. Sodium carbonate against HCID. Barium hydroxide against oxalic acid |
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Answer» Correct Answer - c |
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| 378. |
What indicators will be suitable for the following acid-base titrations: a. `HCOOH` against `NaOH` b. `HBr` against `KOH` c. `NH_(4)OH` with `HNO_(3)` |
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Answer» a. Phenolphthalein and thymolphthalein are suitable for the titration of `W_(A)` and `S_(B)`. b. Bromothymol blue, phenolphthalein, methy`1` orange, and thymolphthalein are suitable for the titration of `S_(A)` and `S_(B)`. c. Methylorange, bromocresol green, and methy`1` red are suitable for the titration of `W_(B)` and S_(A)`. |
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| 379. |
Statement: Hydrolysis of salt is an exothermic phenomenon. Explanation: It involves breaking up of water molecule to produce acids and base respectively.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-15B. 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 - d |
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| 380. |
In a saturated solution of the spatingly soluble strong electrolyte `AgIO_(3)` (molecular mass `=283`) the equilibrium which sets in is `AgIO_(3) (s)hArrAg^(+)(aq)+IO_(3)^(-)(aq)` If the solubility product constant `K_(SP)` of `AgIO_(3)` at a given temperature is `1.0xx10^(-8)`, what is the mass of `AgIO_(3)` cotained in `100 mL` of its saturated solution?A. `1.0xx10^(-7)g`B. `1.0xx10^(-4) g`C. `28.3xx10^(-2) g`D. `2.83xx10^(-3) g` |
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Answer» Correct Answer - d |
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| 381. |
In a saturated solution of the spatingly soluble strong electrolyte `AgIO_(3)` (molecular mass `=283`) the equilibrium which sets in is `AgIO_(3) (s)hArrAg^(+)(aq)+IO_(3)^(-)(aq)` If the solubility product constant `K_(SP)` of `AgIO_(3)` at a given temperature is `1.0xx10^(-8)`, what is the mass of `AgIO_(3)` cotained in `100 mL` of its saturated solution?A. `1.0 xx 10^(-4)g`B. `28.3 xx 10^(-2)g`C. `2.83 xx 10^(-3)g`D. `1.0 xx 10^(-7)g` |
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Answer» Correct Answer - C `AgIO_(3)hArr underset(s)(Ag^(+))+underset(s)(IO_(3)^(-))` `K_(sp)=s^(2)` or `s=sqrt(K_(sp))=sqrt(10^(-8))-10^(-4)mol L^(-1)` `=10^(-4) xx 283 g L^(-1) =2.83 xx 10^(-2) g L^(-1)` Thus 1000 mL of the saturated solution will contain `AgIO_(3)=2.83 xx 10^(-2)g` `:. `100 mL of the saturated solution will contain `AgIO_(3)=2.83 xx10^(-3)g` |
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| 382. |
In a saturated solution of the spatingly soluble strong electrolyte `AgIO_(3)` (molecular mass `=283`) the equilibrium which sets in is `AgIO_(3) (s)hArrAg^(+)(aq)+IO_(3)^(-)(aq)` If the solubility product constant `K_(SP)` of `AgIO_(3)` at a given temperature is `1.0xx10^(-8)`, what is the mass of `AgIO_(3)` cotained in `100 mL` of its saturated solution?A. `28.3 xx 10^(-2) g`B. `2.83 xx 10^(-3) g`C. `1.0 xx 10^(-7) g`D. `1.0 xx 10^(-4) g` |
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Answer» Correct Answer - B `AgIO_(3)(s) hArr Ag^(+)(aq) + IO_(3)^(-)(aq)` Let the solubility of `AgIO_(3)` be s `K_(sp) = [Ag^(+)][IO_(3)^(-)]` `1.0 xx 10^(-8) = S^(2)` `S = 10^(-4) "mol"//"lit"` `= (10^(-4) xx283)/(1000) xx 100 = 283 xx 10^(-5)` `= 2.83 xx 10^(-3) g//100 ml` |
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| 383. |
According to the reaction `PbCl_(2) = Pb^(2) + 2Cl^(-)`, the solubility coefficient of `PbCl_(2)` isA. `[Pb^(2+)] [Cl^(-)]^(2)`B. `[Pb^(2+)] [Cl^(-)]`C. `[Pb^(2+)]^(2) [Cl^(-)]`D. None of these |
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Answer» Correct Answer - A Solubility coefficient `= [Pb^(2+)] [Cl^(-)]^(2)`. |
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| 384. |
Aqueous solution of `CuSO_(4).5H_(2)O` changes blue litmus paper to red due toA. Presence of `Cu^(++)` ionsB. Presence of `SO_(4)^(--)` ionsC. Hydrolysis taking placeD. Reduction taking place |
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Answer» Correct Answer - C Hydrolysis of `Cu^(2+)` produces `H^(+)` ions in solution. |
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| 385. |
Which solution will not show change in pH of dilution ?A. `0.1M NaH_(2)PO_(4)`B. `0.1M CH_(3)COONH_(4)`C. `0.1M CH_(3)COONH_(4)`D. `0.1M NH_(4)OH + 0.01M NH_(4)Cl` |
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Answer» Correct Answer - A::B:C:D pH of salts (other than weak acid + weak base, strong acid + strong base) depends upon the salt concentration. |
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| 386. |
Calculate the `pH` at equilibrium point when a solution of `10^(-6) M CH_(3)COOH` is titrated with a solution of `10^(-6)M NaOH. K_(a)` for acid `2 xx 10^(-5) (pK_(a) = 4.7)` (Answer given in whole number). |
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Answer» Salt of `W_(A)//S_(B) (CH_(3)COONa)` is formed. If we calculate `pH` by using hydrolysis formula for salt of `W_(A)//S_(B)` `pH = (1)/(2) (pK_(w) + pK_(a) + logC)]` Salt of `S_(A)` `[["Salt"]=(10^(-6))/(2)(underset("doubled")("Volume is"))]` `= (1)/(2) (14 +4.7 + "log" (10^(-6))/(2))` `= (1)/(2) (18.7 - 6- 0.3) = 6.2` But `pH` of salt of `W_(A)//S_(B)` is always `gt 7`. So for such dilute solutions of `CH_(3)COOH` and `NaOH`. Contribution of `overset(Theta)OH` ions from `H_(2)O` must be considered. `{:(CH_(3)COOH +, NaOH rarr, CH_(3)COONa +H_(2)O),(10^(-6)M xxV, 10^(-6)M xxV, 10^(-6)M xxV):}` `[CH_(3)COONa] = (10^(-6)M xx V)/(2V) = 0.5 xx 10^(-6)` `= 5 xx 10^(-7) lt 10^(-6)M` `[overset(Theta)OH] = ((K_(w)xxC)/(K_(a)))^(1//2) = ((10^(-14)xx5xx10^(-7))/(2xx10^(-5)))^((1)/(2))` `=(2.5)^((1)/(2)) xx 10^(-8)M = 1.58 xx 10^(-8)M lt 10^(-6)M` Total `[overset(Theta)OH] = [1.58 xx 10^(-8) +10^(-7)` (from `H_(2)O)]` `= 10^(-7) (1.58 xx 10^(-1) +1) = 1.158 xx 10^(-7)`. `pOH =- log (1.158 xx 10^(-7)) = 6.93` `pH = 14 - 6.93 = 7.07 ~~7`. |
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| 387. |
`K_(a)` for `HCN` is `5 xx 10^^(-10)` at `25^(@)C`. For maintaining a constant `pH` of `9.0`, the volume of `5M KCN` solution required to be added to `10mL` of `2M HCN` solution isA. `5 ml`B. `2 ml`C. `6.95 ml`D. `10.2 ml` |
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Answer» Correct Answer - B `P^(H) = P^(ka) + "log"(N_(S)V_(S))/(N_(a)V_(a))` `9 = 9.3 + "log" (5 xx V_(S))/(2 xx 10)` `:. V_(S) = 2 ml` |
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| 388. |
`K_(a)` for HCN is `5.0xx10^(-10)` at `25^(@)C`. For maintaining a constant pH of 9. Calculate the volume of `5.0M KCN` solution required to be added to 10 mL of `2.0M HCN` solution.A. 2 mLB. 5 mLC. 3mLD. 4mL |
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Answer» Correct Answer - A `pH=pK_(a)+log.(["Salt"])/(["Acid"])` `9=-log (5xx10^(-10))+log.(["Salt"])/(["Acid"])` or `log. (["Salt"])/(["Acid"])=9+log (5xx10^(-10))=1.6990` or `(["Salt"])/(["Acid"])=`antilog`(1.6990)=0.5` Ph `=-log K_(a)+log.(["Salt"])/(["Acid"])` |
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| 389. |
`K_(a)` for `HCN` is `5 xx 10^^(-10)` at `25^(@)C`. For maintaining a constant `pH` of `9.0`, the volume of `5M KCN` solution required to be added to `10mL` of `2M HCN` solution isA. `9.3 mL`B. `7.95 mL`C. `4mL`D. `2mL` |
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Answer» Correct Answer - D `pK_(a) =- log(5 xx 10^(-10))` `=- 0.7 + 10 =9.3` Acidic buffer will be formed. `pH = pK_(a) + "log" ([KCN])/([HCN])` Let `V mL` of `KCN` is added Total volume `= (V +10)mL` mmol of `KCN = 5 xx V` `:. [KCN] = (5 xx V "mmol")/((V + 10)mL)` mmol of `HCN = 2 xx 10 = 20` `:. [HCN] = (20 "mmol")/((V +10)mL)` Thus, from equation (i), `9 = 9.3 + log[(5V//(V+10))/(20//(V+10))]` `- 0.3 = log ((5V)(20))` `log (V)/(4) = - 0.3` `(V)/(4) = "Antilog" (-0.3 + 1-1)` `= "Antilog" (bar(1).7) = 5 xx 10^(-1)` `V = 20 xx 10^(-1) = 2 mL` |
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| 390. |
A certain mixture of `HCl` and `CH_(3)-COOH` is `0.1 M` in each of the acids. `20 ml` of this solution is titrated against `0.1 M NaOH`. By how many units does the `pH` change from the start to the stage when the `HCl` is almost completely neutralized and acidic acid remains unreacted? `K_(a)` for acetic acid `=2xx10^(-5)`.A. `1.5`B. 3C. 2D. `3.25` |
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Answer» Correct Answer - c |
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| 391. |
Buffer solutions can be prepared form mixtures ofA. `HCI` and `NaCI`B. `NaH_(2)PO_(4)` and `Na_(2)HPO_(4)`C. `CH_(3)COOH +NaCI`D. `NH_(4)OH+NH_(3)` |
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Answer» Correct Answer - B Buffer solution is a mixture of `W_(A) + salt of W_(A)//S_(B)` or a mixture of `W_(B) + "Salt of" W_(B)//S_(A)`. So `(NaH_(2)PO_(4)+Na_(2)HPO_(4))` is a buffer. `{:(a.[HCI+NaCI,],,S_(A)+"Salt of"S_(A)//S_(B)("not buffer"),,),(b.[NaH_(2)PO_(4)+Na_(2)HPO_(4)],,W_(A)+ "Salt of" W_(A)//S_(B).("It is a buffer"),,),(c.[CH_(3)COOH+NaCI],,W_(A)+"Salt of"S_(A)//S_(B)("not buffer"),,),(d.[NH_(4)OH+NH_(3)],,W_(B)+W_(B)("not buffer"),,):}` |
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| 392. |
`1 M `benzoic acid `(pK_(a)=4.2)` and `1 M C_(6)H_(5)COONa` solutions are given separately. What is the volume of benzoic acid required to prepare a `93mL` buffer solution of `pH=4.5`? |
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Answer» `pH = pK_(a) + (log)([HCO_(3)^(-)])/([C_(6)H_(5)CO_(3)]) rArr 4.5 = 4.2 + log ((V_(2))/(V_(1))) rArr (V_(2))/(V_(1)) = 2` `:.` volume of `C_(6)H_(5)COONa` required `= V_(@) = 62 mL` volume of `C_(6)H_(5)COOH` required `= V_(1) = 31 mL`. |
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| 393. |
The `pK_a` of HCN is 9.30. The pH of a solution prepared by mixing 2.5 moles of KCN and 2.5 moles of HCN in water and making up the total volume of 500 mL is :A. `9.30`B. `7.30`C. `10.30`D. `8.30` |
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Answer» Correct Answer - a |
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| 394. |
What will happen to the pH of a buffer solution when a small amount of a strong base is added ? The pH will :A. increase slightlyB. decrease slightlyC. remain exactly the sameD. become 7.0 |
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Answer» Correct Answer - a |
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| 395. |
The pH of a solution obtained by mixing 100 mL of 0.2 M `CH_3COOH` with 100 mL of 0.2 M NaOH would be :(`pK_a` for `CH_3COOH`=4.74)A. 4.74B. 8.87C. `9.10`D. `8.57` |
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Answer» Correct Answer - b |
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| 396. |
Which of these mixtures constitute buffer solutions ? Mixture 1 : 25 mL of 0.10 M `HNO_3` and 25 mL of 0.010 M `NaNO_3` Mixture 2: 25 mL of 0.10 M `HC_2H_3O_2` and 25 mL of 0.10 M NaOHA. 1 onlyB. 2 onlyC. Both 1 and 2D. Neither 1 nor 2 |
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Answer» Correct Answer - d |
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| 397. |
For neutralisation of `HF+NaOH rarr NaF+H_(2)O`, heat released during neutralisation is:A. `gt13.7 kcal`B. `lt13.7 kcal`C. `=13.7 kcal`D. none of these |
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Answer» Correct Answer - A The `F^(-)` being smallest in size and thus releases more heat of hydration during neutralisation which gives higher value. |
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| 398. |
Which mixtures form buffer solution ? (P)100 mL of 0.200 M HF+200 mL of 0.200 M NaF (Q) 200 mL of 0.200 M HCl+200 mL of 0.400 M `CH_3CO_2Na` (R) 300 mL of 0.100 M `CH_3CO_2H`+100 mL of 0.300 M `CH_3CO_2Na`A. P onlyB. R onlyC. Q and R onlyD. P,Q and R |
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Answer» Correct Answer - d |
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| 399. |
1 M benzoic acid `(pK_a=4.20)` and 1 M `C_6H_5COONa` solutions are given separately What is the volume of benzoic acid required to prepare a 300 ml buffer solution of pH =4.5 ? [log 2 =0.3]A. 200 mLB. 150 mLC. 100 mLD. 50 mL |
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Answer» Correct Answer - c |
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| 400. |
What is the pH of a 0.010 M solution of a weak acid HA that is 4.0 % ionized ?A. `0.60`B. `0.80`C. `2.80`D. `3.40` |
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Answer» Correct Answer - d |
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