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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.
| 151. |
At `25^(@)C`, the solubility product of `Hg_(2)Cl_(2)` in water is `3.2 xx 10^(-17) mol^(3) dm^(-9)`. What is the solubility of `Hg_(2)Cl_(2)` in water `25^(@)C`A. `1.2 xx 10^(-12)M`B. `3.0 xx 10^(-6) M`C. `2 xx 10^(-6) M`D. `1.2 xx 10^(-16) M` |
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Answer» Correct Answer - C Here `Hg_(2)Cl_(2) rarr Hg_(2)^(2+) + 2Cl^(-)` Let the concentration of `Hg_(2)^(2+)` be = x Now, for each `Hg_(2)^(2+)` ion, two `Cl^(-)` ions are produced `:.` Concentration of `Cl^(-)` ions = 2x `K_(sp) = [Hg_(2)^(2+)][Cl^(-)]^(2)` `x(2x)^(2) = 3.2 xx 10^(-17) rArr 4x^(3) = 32 xx 10^(-18)` `rArr x^(3) = (32)/(4) xx 10^(-18) :. x = 2 xx 10^(-6) M` |
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| 152. |
At ` 25^(@)` C, the solubility product of `Hg_(2)CI_(2)` in water is `3.2xx10^(-17)mol^(3)dm^(-9)` what is the solubility of `Hg_(2)CI_(2)` in water at `25^(@)`C ?A. `1.2 xx 10^(-12)M`B. `3.0 xx 10^(-6) M`C. `2 xx 10^(-6) M`D. ` 1.2 xx 10^(-16)M` |
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Answer» Correct Answer - C `underset(x)(Hg_(2)Cl_(2)) hArr underset(s)(Hg_(2)^(2+))+underset(2s)(2Cl^(-))` `K_(sp)=s(2s)^(2)=4s^(3)` `:. s=(K_(sp)//4)^(1//3)=(3.2 xx 10^(-17))^(1//3)` `=(8 xx 10^(-18))^(1//3)=2 xx 10^(-6) M`. |
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| 153. |
Consider the reaction `:` `SO_(2)(g)+(1)/(2)O_(2)(g) rarr SO_(3)(g),DeltaH=-98.3kJ` If the enthalpy of formationof `SO_(3)(g)` is `-395.4kJ` , then the enthalpy of formation of `SO_92)(g)` isA. `297.1kJ`B. `493.7kJ`C. `-493.7kJ`D. `-297.1kJ` |
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Answer» Correct Answer - D For formation of `SO_(3)(g)`. `S+3//2O_(2) rarr SO_(3), Delta H =395.4 kJ` Subtract the given equation from the above and hence `Delta H =395.4+98.3= - 297.1 kJ` |
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| 154. |
Calculate the concentration of `overset(Θ)OH` in the solution of base with `pH` a. `10.4771` b. `12.301` c. `11.8451` |
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Answer» Since the solution are basic, so first calculate `pOH` of the solution and then calculate `[overset(Θ)OH]` accordingly. a. `pH = 10.4771` `pOH = 14 - 10.4771 3.5229` `pOH =- log [overset(Θ)OH] = 3.5229` `:. log [overset(Θ)OH] =- 3.5229 =- 3 - 0.5229 +1 - 1` `= bar(4).4771` `:. [overset(Θ)OH] = "Antilog" (bar(4).4771)` `= 3 xx 10^(-4) = 0.3 xx 10^(-3) N` c. `pH = 12.301` `pOH = 14 - 12.301 = 1.699` `pOH =- log [overset(Θ)OH] = 1.699` `:. log [overset(Θ)OH] =- 1.699 =- 1 - 0.699 +1 - 1` `= bar(2).301` `:. [overset(Θ)OH] = "Antilog" (bar(2).301)` `= 2 xx 10^(-2)N = 0.2 xx 10^(-1)N` d. `pH = 11.8451` `pOH = 14 - 11.8461 = 2.1549` `pOH =- log [overset(Θ)OH] =- 2.1549 =- 2 - 0.1549 +1 - 1` `= bar(3).8451` `:. [overset(Θ)OH] = "Antilog" (bar(3).8451)` `= 7 xx 10^(-3) N = 0.7 xx 10^(-2) N` |
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| 155. |
The `pH` of blood isA. `gt 10`B. Between `8` and `0`C. Between `7` and `8`D. `lt6` |
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Answer» Correct Answer - C Blood is basic with `pH = 7.4`. |
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| 156. |
The species `H_(2)O, HCO_(3)^(Θ), HSO_(4)^(Θ), NH_(3)` can act both as Brddotosted acis and bases. For each case give the corresponding conjugate acid and conjugate base. |
| Answer» `{:(,"Species",C_(A),C_(B)),("i".,H_(2)O,H_(3)O^(oplus),overset(Theta)(O)H),("ii".,HCO_(3)^(Theta),H_(2)CO_(3),CO_(3)^(2-)),("iii".,HSO_(4)^(Theta),H_(2)SO_(4),SO_(4)^(2-)),("iv".,NH_(3),overset(oplus)(N)H_(4),overset(Theta)(N)H_(2)):}` | |
| 157. |
A buffer solution is prepared in which the concentration of `NH_(3)` is 0.30 M and the concentration of `NH_(4)^(+)` is 0.20 M. If the equilibrium constant, `K_(b)` for `NH_(3)` equals `1.8 xx 10^(-5)`, what is the pH of this solution (log 2.7 = 043)A. 8.73B. 9.08C. 9.43D. 11.72 |
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Answer» Correct Answer - C `pOH = pK_(b) log.(["Salt"])/(["Base"])` `= 4.74 + log.(0.20)/(0.30) = 4.74 + (0.301 - 0.477)` `= 4.74 - 0.176 = 4.56 :. pH = 14- 4.56 = 9.44`. |
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| 158. |
Find the concentration of `H^(o+), HCO_(3)^(Θ)`, and `CO_(3)^(-2)` in a `0.01M` solution of carbonic acid if the `pH` of solution is `4.18`. `K_(1) = 4.45 xx 10^(-7),K_(2) = 4.69 xx 10^(-11)` |
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Answer» Given, `pH = 4.18 =- log [H^(o+)]` `:. [H^(o+)] = 6.61 xx 10^(-5) molL^(-1)` `H_(2)CO_(3) hArr H^(oplus)+HCO_(3)^(Theta)` `K_(1) = ([H^(o+)][HCO_(3)^( Θ)])/([H_(2)CO_(3)])` or `4.45 xx 10^(-7) = ([6.61xx10^(-5)][HCO_(3)^(Θ)])/([0.01])` or `[HCO_(3)^(Θ)] = 6.73 xx 10^(-5) molL^(-1)` Again for dissociation of `HCO_(3)^(Θ)`, we have `[HCO_(3)^(Theta)]hArr H^(oplus)+CO_(3)^(2-)` `K_(2)=([H^(o+)][CO_(3)^(2-)])/([HCO_(3)^(Θ)])` or `4.69 xx 10^(-11) = ([6.61 xx 10^(-5)][CO_(3)^(2)])/([6.73xx10^(-5)])` `[CO_(3)^(2-)] = 4.78 xx 10^(-11) molL^(-1)` |
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| 159. |
The `pH` of blood is `7,4`. If the buffer in blood constitute `CO_(2)` and `HCO_(3)^(Theta)` ions, calculate the ratio of conjugate base of acid `(H_(2)CO_(3))` to maintain the `pH` of blood. Given `K_(1)` of `H_(2)CO_(3) = 4.5 xx 10^(-7)`.A. `11.25`B. `10.0`C. `8.5`D. None |
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Answer» Correct Answer - A `CO_(2) + H_(2)O hArr H_(2)CO_(3) hArr H^(o+) + HCO_(3)^(Theta)` `K_(1) = ([H^(o+)][HCO_(3)^(Theta)])/([H_(2)CO_(3)]) = ([H^(o+)][HCO_(3)^(Theta)])/([CO_(2)])` `(K_(1))/([H^(o+)]) = ([HCO_(3)^(Theta)])/([CO_(2)]) = (["Conjugate base"])/(["Acid"])` `pH = 7.4, log [H^(o+)] =- 7 - 0.4 + 1- 1 = bar(8).6` `:. [H^(o+)] = "Antilog" (bar(8).6) = 4 xx 10^(-8)` `:. ([HCO_(3)^(Theta)])/([CO_(2)]) = (K_(1))/([H^(o+)]) = (4.5 xx 10^(-7))/(4xx10^(-8)) = 11.25` |
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| 160. |
The conjugate acid of `HPO_(3)^(2-)` isA. `H_(3)PO_(4)`B. `H_(3)PO_(3)`C. `H_(2)PO_(3)^(-)`D. `PO_(4)^(3-)` |
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Answer» Correct Answer - C `underset("conjugate acid")(H_(2)PO_(3)^(-) hArr H^(+) + HPO_(3)^(2-)` action. |
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| 161. |
A buffer solution is prepared in which the concentration of `NH_(3)` is `0.30 M` and the concentration of `NH_(4)^(+)` is `0.20 M`. If the equilibrium constant, `K_(b)` for `NH_(3)` equals `1.8xx10^(-5)`, what is the `pH` of this solution? (`log 2.7=0.43`)A. 9.08B. 9.43C. 11.72D. 8.73 |
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Answer» Correct Answer - B `pOH=pK_(b)+log.(|NH_(3)^(+)|)/(|NH_(3)|)` `=-log (1.8 xx 10^(-5))+log. (0.2)/(0.3)` `=4.74 +log .(0.2)/(0.2)=4.56` `pH ==14-4.56=9.44` |
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| 162. |
The pH of the a solution containing `0.4 M HCO_(3)^(-)` is : `[K_(a_(1)) (H_(2)CO_(3)) = 4 xx 10^(-7), K_(a_(2)) (HCO_(3)^(-)) = 4 xx 10^(-11)]`A. `10.4`B. `10.1`C. `6.1`D. `8.4` |
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Answer» Correct Answer - D `P^(H) = (P^(ka_(1)) + P^(ka_(2)))/(2)` |
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| 163. |
In the reaction `SnCl_(2)+2Cl^(-) rarr SnCl_(4)`, Lewis acid isA. `SnCl_(2)`B. `Cl^(-)`C. `SnCl_(4)`D. None of these |
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Answer» Correct Answer - A Because `SnCl_(2)` is a electron acceptor according to Lewis concept. |
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| 164. |
In the reaction `SnCl_(2) + 2Cl^(-) rarr SnCl_(4) + 2e^(-)` the Lewis acid isA. `SnCl`B. `SnCl_(3)`C. `SnCl_(2)`D. `SnCl_(4)` |
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Answer» Correct Answer - C Lone pair aceptor is lewis acid |
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| 165. |
A weak acid of dissociation constant `10^(-5)` is being titrated with aqueous NaOH solution. The pH at the point of one-third neutralization of the acid will beA. `5 + log 2 - log 3`B. 5 - log 2C. 5 - log 3D. 5 - log 6 |
| Answer» Correct Answer - B | |
| 166. |
A weak acid of dissociation constant `10^(-5)` is being titrated with aqueous NaOH solution . The pH at the point of one third of neutralization of the acid will beA. `5 + log - log 3`B. `5- log 2`C. `5- log 3`D. `5- log 6` |
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Answer» Correct Answer - B On partial neutralization of weak acid, salt is formed. Hence it becomes buffer `:. pH =pK_(a)+log. (|"Salt"|)/(|"Acid"|).` `1//3` rd neutralization of the acid means out of 1 mole of the acid, salt formed is `= 1//3 ` mole of acid left `=2//3` mole `:. pH =-log (10^(-5))+log.(1//3)/(2//3)` `=5+log.(1)/(2)=5-log 2` |
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| 167. |
0.1 M formic acid solution is titrated against 0.1 M NaOH solution. What would be the difference in pH between 1/5 and 4/5 stages of neutralization of acid?A. `2 "log" 3//4`B. `2 "log" 1//5`C. `"log" 1//3`D. `2 "log" 4` |
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Answer» Correct Answer - D Solution is acidic buffer at both stages. Difference in `P^(H)` (at` 4//5` and `1//5`) `= log 4 - "log" (1)/(4)` `= 2 log 4` |
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| 168. |
When `15mL` of `0.05M AgNO_(3)` is mixed with `45.0mL` of `0.03M K_(2)CrO_(4)`, predict whether precipitation of `Ag_(2)CrO_(4)` occurs or not? `K_(sp)` of `Ag_(2)CrO_(4) = 1.9 xx 10^(-12)` |
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Answer» First find the concentrations of `Ag^(o+)` and `CrO_(4)6(2-)` ions in the resulting mixture. `[Ag^(o+)] = (15 xx 0.05)/(15+45) = 1.25 xx 10^(-2)M` `[CrO_(4)^(2-)] = (45 xx 0.03)/(15+45) = 2.25 xx 10^(-2)M` The ionic product for `Ag_(2)CrO_(4)` is given as follows: `Ag_(2)CrO_(4) hArr 2Ag^(oplus)+CrO_(4)^(2-)` Ionic product `= [Ag^(o+)]^(2) [CrO_(4)^(2-)]` `= (1.25 xx 10^(-2))^(2) (2.15 xx 10^(-2))` `= 3.51 xx 10^(-6) gt K_(sp)` Hence, precipiation occurs. |
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| 169. |
`[Cl^(-)]` in a mixture of 200mL of 0.01 M HCl and 100 mL of 0.01 M `BaCl_(2)` isA. 0.01M `NaHCO_(3)^(-)`B. 0.0122MC. 0.03MD. 0.02M |
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Answer» Correct Answer - B `BaCl_(2) rarr Ba^(2+)+2Cl^(-)` ltbr. `[Cl^(-)]` in 0.01 M `BaCl_(2)=0.02 M` `[Cl^(-)]` in the mixture `=(200xx0.01 +100xx0.02)/(300)` ,brgt `=0.0133M` |
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| 170. |
The pH of a solution obtained by mixing 50 ml of 0.4 N HCl and 50 mL of 0.2 N NaOH is :A. 13B. 12C. `1.0`D. `2.0` |
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Answer» Correct Answer - c |
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| 171. |
Which of the following will decreases the pH of a 50 ml solution of 0.01 M HClA. Addition of 5 ml of 1 M HCLB. Addition of 50 ml of 0.01 M HClC. Addition of 50 mL of 0.002 M HClD. Addition of Mg |
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Answer» Correct Answer - A Addition of 5ml of 1M HCl will decrease the pH, because only this will increase the `H^(+)` ion concentration in the solution. |
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| 172. |
The pH of a solution obtained by mixing 50 ml of 1 N HCl and 30 ml of 1 N NaOH is [log 2.5 = 0.3979]A. 3.979B. 0.6021C. 12.042D. 1.2042 |
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Answer» Correct Answer - B No. of milli equivalent of `NaOH = 30 xx 1 = 30` No. of milli equivalent of `HCl = 50 xx 1 = 50` `:.` No. of milli equivalent of HCl left after titration = 50 - 30 = 20. Total volume of the mixture = 50 + 30 = 80 ml i.e. 20 milli equivalent or 0.02 equivalent of HCl are present in 80 ml. `:.` 250 milli equivalent or 0.25 equivalent of HCl are present in 1000 ml or 1 litre. i.e., `0.25 N HCl ~~ 0.25 N NaOH` (Monobasic) So, `[H^(+)] = -log_(10)[2.5 xx 10^(-1)] rArr pH = 1 - log_(10)[0.25]` `pH = -log_(10)[2.5 xx 10^(-2) rArr pH = 1 - log_(10)2.5` `pH = 1 - 0.3979 rArr pH = 0.6021`. |
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| 173. |
The `pK_(a)` of a weak acid, `HA`, is `4.80`. The `pK_(b)` of a weak base, `BOH`, is `4.78`. The `pH` of an aqueous solution of the corresponding salt, `BA`, will be:A. `8.58`B. `4.79`C. `7.01`D. `9.22` |
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Answer» Correct Answer - C It is a salt of weak acid and weak base. `pH= 7+(1)/(2)(pK_(a)-pK_(b))` `= 7+(1)/(2)(4.8-4.78)` `= 7.01` |
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| 174. |
The `pK_(a)` of a weak acid, `HA`, is `4.80`. The `pK_(b)` of a weak base, `BOH`, is `4.78`. The `pH` of an aqueous solution of the corresponding salt, `BA`, will be:A. `4.79`B. `7.01`C. `9.22`D. `9.58` |
| Answer» `pH=7+(1)/(2)pK_(a)-(1)/(2)pK_(b)=7+(4.8)/(2)-(4.78)/(2)=7.01` | |
| 175. |
The `pK_(a)` of a weak acid, `HA`, is `4.80`. The `pK_(b)` of a weak base, `BOH`, is `4.78`. The `pH` of an aqueous solution of the corresponding salt, `BA`, will be:A. 4.79B. 7.01C. 9.22D. 9.58 |
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Answer» Correct Answer - b |
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| 176. |
Equal volumes of `0.02M AgNO_(3)` and `0.01M HCN` are mixed. Calculate `[Ag^(o+)]` in solution after attaining equilibrium. `K_(a) HCN = 6.2 xx 10^(-10)` and `K_(sp)` of `AgCN = 2.2 xx 10^(-16)`. |
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Answer» Solution: `{:(,Ag^(o+)+,HCNrarr,AgCN+,H^(o+)),("Initial",(0.02xxV)/(2V),(0.02xxV)/(2V),0,0),("Final",0,0,0.01,0.01):}` `AgCN hArr Ag^(o+) +CN^(Theta)` `:. K_(sp) = 2.2 xx 10^(-16) = [Ag^(o+)] [CN^(Theta)]` `HCN hArr H^(o+) + CN^(Theta) K_(a) = 6.2 xx 10^(-10) = ([H^(o+)][CN^(Theta)])/([HCN])` Now, soluble `CN^(Theta)` formed will hydrolyses as `CN^(Theta) + H_(2)O hArr HCN + overset(Theta)OH` `:. [Ag^(o+)]` of soluble `AgCN = CN^(Theta)` left after hydrolysis `+HCN` formed after hydrolysis `:. [(2.2 xx 10^(-10))/([CN^(Theta)])] = [CN^(Theta)] + ([H^(o+)][CN^(Theta)])/(6.2 xx 10^(-10))` `= [CN^(Theta)] + (10^(-2)[CN^(Theta)])/(6.2 xx 10^(-10)) [because CN^(Theta)lt lt lt (10^(-2)[CN^(Theta)])/(6.2xx10^(-10))]` or `[CN^(Theta)]^(2) = (2.2xx10^(-16)xx6.2xx10^(-10))/(10^(-2))` `:. [CN^(Theta)] = 3.7 xx 10^(-12)` `:. [Ag^(o+)] = (K_(sp))/([CN^(Theta)]) = (2.2 xx 10^(-16))/(3.7xx10^(-12)) = 5.96 xx 10^(-5)M` |
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| 177. |
The `pk_(a)` of a weak acid, HA is 4.8. The `pK_(b)` of a weak base, BOH is 4.78. The pH of an aqueous solution of the corresponding salt, BA will beA. 4.79B. 7.01C. 9.22D. 9.58 |
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Answer» Correct Answer - B `pK_(a) = 4.8 , pK_(b) = 4.78` `pH = 7+(1)/(2)(pK_(a) - pK_(b)) = 7 + (1)/(2)(4.8 - 4.78) = 7.01`. |
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| 178. |
Calculate `[H^(+)], [CH_(3)COO^(-)]` and `[C_(7)H_(8)O_(2)^(-)]` in a solution that is `0.02M` in acetic acid and `0.01M` in benzoic acid. `(K_(a_(A A))= 1.8xx10^(-5), K_(a_(BA)) = 6.4xx10^(-5))` |
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Answer» Correct Answer - `[H^(+)]=10^(-3)M, [CH_(3)COO^(-)]=3.6xx10^(-4)M`, `[C_(7)H_(5)O_(2)^(-)]=6.4xx10^(-4)M`; |
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| 179. |
A solution containing `0.2` mole of dicholoracetice acid `(K_(a)=5xx10^(-2))` and0.1 mole sodium dicholoroacetate in one litre solution has `[H^(+)]` : |
| Answer» Correct Answer - `0.05`; | |
| 180. |
Calcualate `pH,[H^(+)],[OH^(-)],[CH_(3)COOH],[SH^(-)],[H_(2)S],[S^(2-)]` in a solution obtained by mixing equal volume of `0.2M H_(2)S &0.02M` acetic acid .Given that `K_(a)(CH_(3)COOH)=2xx10^(-5),K_(a_(1))(H_(2)S)=10^(-7),K_(a_(2))(H_(2)S)=10^(-14)` Take `log21=-1.32,(1)/(sqrt(21))=0.218` |
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Answer» Now `[H_(2)S]=0.1M,[CH_(3)COOH]=0.01M` after mixing For `pH` calculation ,considering only first `[H^(+)]` of `H_(2)S` the system becomes similar to a mixture of two weak monoprotic acids. `2^(nd)H^(+)` coming form `H_(2)S` would be negligible because of very low value of `K_(a_(2))` & also because of common ion effect exerted by `H^(+)` from `CH_(3)COOH` `[H^(+)]=sqrt((10^(-1)xx10^(-7))+(10^(-2)xx2xx10^(-5)))=sqrt((0.1+2)xx10^(-7))=sqrt(21)xx10^(-4)M` `pH=4-(1)/(2)log21=3.34` `[OH^(-)]=(K_(W))/([H^(+)])=2.18xx10^(-11)M` For acetic acid `K_(a_(1))=([H^(+)][HS^(-)])/([H_(2)S])=(sqrt(21)xx10^(-4)xx[CH_(3)COO^(-)])/(0.01)` `rArr [CH_(3)COOH^(-)]=4.36xx10^(-4)M` `rArr [CH_(3)COOH]=0.1M` For `H_(2)S,K_(a_(1))=([H^(+)][HS^(-)])/([H_(2)S])=(sqrt(21)xx10^(-4)xx[s^(2-)])/(0.1)rArr [HS^(-)]=2.18xx10^(-5)M` For `HS^(-)K_(a_(2))=([H^(+)][S^(2-)])/([HS^(-)])=(sqrt(21)xx10^(-4)xx[S^(2-)])/(2.18xx10^(-5))rArr [S^(2-)]=4.76xx10^(-16)M` |
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| 181. |
Calculate the `pH` of the following mixtures given `(pK_(a) = pK_(b) = 4.7447)`: a. `50mL 0.1 M NaOH +50mL 0.1MCH_(3)COOH` b. `50mL 0.1 m NaOH +50mL 0.05 M CH_(3)COOH` c. `50mL 0.05M NaOH +50mL 0.1 M CH_(3)COOH` d. `50mL 0.1M NH_(4)OH +50mL 0.05 MHCI` e. `50mL 0.05M NH_(4)OH +50mL 0.1 MHCI` f. `50mL 0.05 M NH_(4)OH + 50mL 0.05 M CH_(3)COOH` |
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Answer» Correct Answer - (a) `12.3979`, (b) `4.7447`, (c ) `8.7218`, (d)`9.2553`, (e ) `1.6021`, (f) `5.2782`, (g) 7 ; |
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| 182. |
Calculate the `pH` of a solution obtained by mixing equal volume of `0.02MHOCl &0.2MCH_(3)COOH` solutions Given that `K_(a)(HOCl)=2xx10^(-4),K_(a)(CH_(3)COOH)=2xx10^(-5)`Also calculate `[OH^(-)],[OCl^(-)],[CH_(3)COOH]` at equilibrium .Take `log2=0.3` |
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Answer» Volume of final solutions becomes double . So , concentrated becomes half so after mixing: `C_(1)=0.01M,C_(2)=0.1M` `[H^(+)]=sqrt(C_(1)K_(a_(1))+C_(2)K_(a_(2)))=sqrt(2xx10^(-4)xx0.01+2xx10^(-5)xx0.1)=sqrt(2xx10^(-6)+2xx10^(-6))=2xx10^(-3)M` `:.pH=3-log2=2.7` `[OCl^(-)]=(0.01xx2xx10^(-4))/(2xx10^(-3))=1xx10^(-3)M,[CH_(3)COO^(-)]=(0.1xx2xx10^(-4))/(2xx10^(-3))=1xx10^(-3)M`, `[OH^(-)]=(K_(w))/([H^(+)]=(10^(-14))/(2xx10^(-3))=5xx10^(-12)M` |
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| 183. |
`P^(H)` of `CH_(3)COOH` and `CH_(3)COONa` buffer is `4.8` in Which of the following conc. Conditions, the buffer capacity will be maximum `[CH_(3)COOH]` & `[CH_(3)COONa]`A. `0.1 M, 02 M`B. `0.2M, 0.1 M`C. `0.34 M, 0.34 M`D. `0.34M , 03 M` |
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Answer» Correct Answer - C `pH = pka + "log" ([s])/([A])` Buffer capacity is maximum when `P^(H) = P^(ka), :. [s] = [A]` |
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| 184. |
`pK_(a)` of `CH_(3)COOH` is 4.74 . The pH of 0.01 M `CH_(3)COONa` ISA. 8.37B. 4.37C. 4.74D. 0.474 |
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Answer» Correct Answer - A For a salt of a weak acid and strong base `pH =7+(1)/(2)[pK_(a)+log C]` `=7+(1)/(2)[4.74 +log 10^(-2)]` `=7 +(1)/(2)[4.74 -2]` `=7+(2.74)/(2)=71.37 =8.37` |
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| 185. |
Calculate the buffer capacity of `1L` solution of : `(i)0.1MCH_(3)COOH` and `0.1MCH_(3)COONa` `(ii) 0.2MCH_(3)COOH` and `0.2MCH_(3)COONa` Given:`pK_(a)(CH_(3)COOH)=4.74` Which will be a better buffer? |
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Answer» Buffer capcaity =`(2.303(a+x)(b-x))/(a+b)~~(2.303ab)/(a+b)xltlta,b` Buffer capacity =`(0.1xx0.1xx2.303)/(0.1+0.1)=0.11515` Buffer capacity `=(0.2xx0.2xx2.303)/(0.2+0.2)=0.2303` Second buffer solution (having greater buffer capcity )can be called butter buffer. |
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| 186. |
6 gm of `CH_(3)COOH, 6 gm` of `NaOH` and `6.3 gm` of `HNO_(3)` are dissolved in 1L of water. If `K_(a)` of `CH_(3)COOH` is `10^(-5)` then calculate pH of resulting solution. |
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Answer» Correct Answer - 5 |
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| 187. |
The `pH` of a solution containing `0.1mol` of `CH_(3)COOH, 0.2 mol` of `CH_(3)COONa`,and `0.05 mol` of `NaOH` in `1L. (pK_(a) of CH_(3)COOH = 4.74)` is:A. `5.44`B. `5.20`C. `5.04`D. `4.74` |
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Answer» Correct Answer - A `{:(,CH_(3)COOH+,NaOHrarr,CH_(3)COONa+,H_(2)O),("Initial mol",0.1,0.05,0,0),("Final mol",(0.1-0.05),(0.05-0.05),0.05,-),(,=0.05,=0,,):}` Total moles of `Ch_(3)COONa = 0.2 + 0.05 = 0.25`. `:. [CH_(3)COONa] = (0.25mol)/(1L) = 0.25M` Moles of `CH_(3)COOH` left `= (0.1 -0.05) = 0.05` `:. [CH_(3)COOH] = (0.05mol)/(1L) = 0.05M` Thus, acidic buffer is formed `pH = pK_(a) + "log"(["Salt"])/(["Acid"])` `= 4.74 + log ((0.25)/(0.05))` `= 4.74 + log 5 = 4.74 + 0.7 = 5.44`. |
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| 188. |
In above question concentration of Triethyl ammonium ion `[C_(6)NH_(16)]^(+)` in resulting solution will be :A. `100K_(b)`B. `200K_(b)`C. `10K_(b)`D. `K_(b)` |
| Answer» `K_(b)=([OH^(-)][C_(6)NH_(16)^(+)])/([C_(6)NH_(15)])rArr ((10^(-2))/(2)[C_(6)NH_(16)^(+)])/((10^(-2))/(2))rArr[C_(6)NH_(16)^(+)]=K_(b)` | |
| 189. |
The pH of a buffer solution containing equimolar concentrations of solution acetate and acetric acid is equal to:A. `K_(a)of CH_(3)COOH`B. `pK_(a)of CH_(3)COOH`C. 14D. `log (1/(K_(a)))of CH_(3)COOH` |
| Answer» Correct Answer - B::D | |
| 190. |
The `pH` of NaOH solution is `13`. What is the amount in grams of `NaOH` present in one litre of a solution?A. `40`B. `4`C. `0.4`D. `20` |
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Answer» Correct Answer - B `P^(OH) = 14 - P^(H) = 14 - 12 = 2` `[OH^(-)] = N = 10^(-POH) = 10^(-2)N` `N = (W)/(EW) xx (1)/(V("lit"))` `10^(-2) = (w)/(eq.w) xx (1)/(v)` `w = 10^(-2) xx 40 xx 1 = 0.4` |
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| 191. |
How many grams of `NaOH` must be dissolved in `1L^(-1)` of the solution to given it a `pH` value of `12`?A. `0.20 gL^(-1)`B. `0.40 gL^(-1)`C. `0.10 gL^(-1)`D. `1.2 gL^(-1)` |
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Answer» Correct Answer - B `pH = 12, pOH = 2, [overset(Theta)OH] = 10^(-2)M`. `[overset(Theta)OH] = 10^(-2)M = 10^(-2) xx 40 g L^(-1) = 0.4 gL^(-1)` |
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| 192. |
Calculate the amount of `(NH_(4))SO_(4)` in grams which must be added to 500 ml of 0.200 M `NH_(3)` to yield a solution with pH = 9.35 (`K_(b)` for `NH_(3) = 1.78 xx 10^(-5)`)A. 10.56 gmB. 15 gmC. 12.74 gmD. 16.25 gm |
| Answer» Correct Answer - A | |
| 193. |
The concentration of which ion is to be decreased, when `NH_(3)` solution is addedA. `OH^(-)`B. `NH_(4)^(+)`C. `H_(3)O^(+)`D. `O_(2)^(-)` |
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Answer» Correct Answer - C Because `NH_(3)` acts as Lewis acid any they give electron pair to `H_(3)O^(+)` ion. `H_(3)O^(+)` is a Lewis base. Which accept the electron pair from `NH_(3)`. |
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| 194. |
The `P^(H)` of a solution is `6`. Its `H^(+)` concentrations is decreased by `1000` times. Its `P^(H)` will beA. 9B. 6.96C. 7.04D. 8 |
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Answer» Correct Answer - A If the `H^(+)` ion concentration decreases by `1000` times than `P^(H)` increases by three units |
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| 195. |
What is minimum concentration of `SO_(4)^(2-)` required to precipitate `BaSO_(4)` in solution containing `1 xx 10^(-4)` mole of `Ba^(2+)` ? (`K_(sp)` of `BaSO_(4) = 4 xx 10^(-10)`)A. `4xx10^(-10) M`B. `2xx10^(-10) M`C. `4xx10^(-6) M`D. `2xx10^(-3) M` |
| Answer» Correct Answer - 3 | |
| 196. |
What is the minimum concentration of `SO_(4)^(2-)` required to precipitate `BaSO_(4)` in a solution containing `1.0 xx 10^(-4)` mol `Ba^(2+)` (`K_(sp)` for `BaSO_(4)` is `4 xx 10^(-10)`)A. `4 xx 10^(-10)M`B. `2 xx 10^(-7)M`C. `4 xx 10^(-6)M`D. `2 xx 10^(-3) M` |
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Answer» Correct Answer - C `BaSO_(4) hArr Ba^(2+) + SO_(4)^(2-)` `K_(sp) = S^(2) rArr S = sqrt(K_(sp)) , K_(sp) = [Ba^(2+)] xx [SO_(4)^(2-)]` `4 xx 10^(-10) = [1 xx 10^(-4)] xx [SO_(4)^(2-)]` `[SO_(4)^(2-)] = (4 xx 10^(-10))/(1 xx 10^(4)) = 4 xx 10^(-6)`. |
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| 197. |
What is minimum concentration of `SO_(4)^(2-)` required to precipitate `BaSO_(4)` in solution containing `1 xx 10^(-4)` mole of `Ba^(2+)` ? (`K_(sp)` of `BaSO_(4) = 4 xx 10^(-10)`)A. `2 xx 10^(-3) M`B. `2 xx 10^(-5) M`C. `4 xx 10^(-10) M`D. `4 xx 10^(-6)` |
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Answer» Correct Answer - D to get precpitate Ionic product is `gt K_(sp)` The minimum Conc. Of. `SO_(4^(-2))` to get Precipitate `= (K_(sp))/([Ba^(+2)])` `= (4 xx 10^(-10))/(1 xx 10^(-4)) = 4 xx 10^(-6)` |
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| 198. |
The `pH` of `NaOH` solution is `12`. What is the amount in grams of `NaOH` present in one litre of a solution?A. `40`B. `4`C. `0.4`D. `20` |
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Answer» Correct Answer - C `P^(H) = 12, P^(OH) = 2` `[OH^(-)] = 10^(-2)` `N = (w)/(Gmw) xx (1000)/(v)` `10^(-2) = (w)/(40) xx (1000)/(1000)` `w = 40 xx 10^(-2) = 0.4` |
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| 199. |
The pH of a solution is 5. Its `H^(+)` ion concentration is decreased by `100` times, then the nature of the solution formed isA. AcidicB. BasicC. NeutralD. Amphoteric |
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Answer» Correct Answer - C What ever the volume of water added, the `P^(H)` of acid is always `lt 7` at `25^(@)C` |
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| 200. |
The solubility of `PbSO_(4)` in `0.1M Na_(2)SO_(4)` solution is (`K_(sp)` of `PbSO_(4)` is `1.25 xx 10^(-9)`) |
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Answer» `PbSO_(4) hArr Pb^(+2) + SO_(4)^(-2)` `S , S + 1` `K_(sp) = S(S + 0.1) = 0.1 xx S (S lt lt 0.1)` `:. S = (K_(SP))/(0.1) = 1.25 xx 10^(-8) M` |
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