MCQ Questions for Class 11 Medical Chemistry Equilibrium Quiz 11

Assertion:
Solubility of

$AgCl$ in water decreases if

$NaCl$ is added to it.
Reason:

$NaCl$ is soluble freely in water but

$AgCl$ is sparingly soluble.

0%
Both Assertion and Reason are correct and Reason is the correct explanation of Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion
0%
Assertion is correct but Reason is not correct
0%
Assertion is not correct but Reason is correct
0%
Both Assertion and Reason are not correct

Explanation

$NaCl$ is a strong electrolyte. It completely dissociates to form ions. Hence, it is highly soluble in water.

$AgCl$ dissociates to little extent. Hence, it is sparingly soluble in water.
When

$NaCl$ is added to a solution of

$AgCl$ , due to common ion effect (chloride ion is the common ion), the dissociation of

$AgCl$ is suppressed. Also as the concentration of chloride ion (from

$NaCl$ ) increases, the ionic product of

$AgCl$ exceeds its solubility product and precipitation occurs.

By adding which of the following in 1 L 0.1 M solution of HA,

$(Ka=10^{-5})$ , the degree of dissociation of HA decreases appreciably?

0%
${10^{-3} M \: HCl, 1\; L}$
0%
${0.5 M \: HX (Ka=2\times 10^{-6}), 1 \: L}$
0%
${0.1 M \: HNO_{3}, 1\: L}$
0%
All of these

Explanation

$HA, \alpha =\sqrt{\displaystyle\frac{10^{-5}}{0.1}}=0.01$
(A) On adding

$HCl, \left [ HA \right ]=\frac{0.1}{2},\left [ HCl \right ]=\displaystyle\frac{10^{-3}}{2}$

$HCl \rightarrow H^{+}+Cl^{-}$

$HA \rightleftharpoons H^{+} +A^{-}$

$\displaystyle\frac{0.1}{2}\left ( 1- \alpha \right )\left (\displaystyle\frac{10^{-3}}{2}+\displaystyle\frac{0.1 \alpha }{2} \right ) \displaystyle\frac{0.1 \alpha }{2}$

$\Rightarrow 10^{-5}=\displaystyle\frac{\displaystyle\frac{0.1\alpha }{2} \times \left (\displaystyle\frac{10^{-3}}{2}+\displaystyle\frac{0.1 \alpha }{2} \right ) }{\displaystyle\frac{0.1}{2}\left ( 1-\alpha \right )}$
Neglecting

$\alpha$

$10^{-5}=\alpha \times \left ( \displaystyle\frac{10^{-3}}{2}+\displaystyle\frac{0.1\alpha }{2} \right )$

$\Rightarrow 0.1 \alpha ^{2}+10^{-3} \alpha -2\times 10^{-5}=0$

$\Rightarrow \alpha ^{2}+10^{-2}\alpha -2\times 10^{-4}=0$

$\Rightarrow \alpha =\displaystyle\frac{-10^{-2}+\sqrt{10^{-4}+8\times 10^{-4}}}{2}=\displaystyle\frac{2\times 10^{-2}} {2}=2$
Which is similiar to initial.
(B) In case of two weak acids

$\left [\mathrm H^{+} \right ]=\sqrt{10^{-5}\times \frac{0.1}{2}+2\times 10^{-6}\times \displaystyle\frac{0.5}{2}}$

$=\sqrt{\displaystyle\frac{10^{-6}}{2}+\displaystyle\frac{10^{-6}}{2}}=10^{-3}$

$\alpha =\displaystyle\frac{10^{-5}}{10^{-3}}=10^{-2}$
which is same as earlier
(C)

$HNO_{3}\rightarrow H^{+}+NO_{3}^{-}$

$\displaystyle\frac{0.1}{2} \displaystyle\frac{0.1M}{2}$

$HA \rightleftharpoons H^{+} + A^{-}$

$\displaystyle\frac{0.1M}{2}\left ( 1-\alpha \right ) \displaystyle\frac{0.1}{2}+\displaystyle\frac{0.1}{2}\alpha \displaystyle\frac{0.1}{2}\alpha$

$\Rightarrow 10^{-5}=\displaystyle\frac{\frac{0.1\alpha }{2}\times \displaystyle\frac{0.1 }{2} \left (1+\alpha \right ) }{\displaystyle\frac{0.1}{2}\left (1-\alpha \right )}$
Neglecting

$\alpha$ due to common ion effect.

$\Rightarrow 10^{-5}=\displaystyle\frac{0.1 \alpha }{2}$

$\Rightarrow \alpha =2\times 10^{-4}$
Hence, it decreases from

$0.01 to 2\times 10^{-4}$ .

The percentage of one of the anion precipitated when another anion starts precipitation is:

(Given $\frac{1}{\sqrt{2}}=0.7$ )

0%
98.7%
0%
99.3%
0%
97.6%
0%
92.7%

The degree of dissociation

$(\alpha)$ of a weak electrolyte,

$\mathrm{A}_{\mathrm{x}}\mathrm{B}_{\mathrm{y}}$ is related to van't Hoff factor (i) by the expression:

0%
$\alpha =\displaystyle \frac{\mathrm{i}-1}{(\mathrm{x}+\mathrm{y}-1)}$
0%
$\alpha =\displaystyle \frac{\mathrm{i}-1}{\mathrm{x}+\mathrm{y}+1}$
0%
$\alpha =\displaystyle \frac{\mathrm{x}+\mathrm{y}-1}{\mathrm{i}-1}$
0%
$\alpha =\displaystyle \frac{\mathrm{x}+\mathrm{y}+1}{\mathrm{i}-1}$

Explanation

The dissociation of the weak electrolyte is as shown below.

$\displaystyle \underset {1-\alpha}{\mathrm{A}_{\mathrm{x}}\mathrm{B}_{\mathrm{y}}}\rightarrow \underset {x\alpha} {\mathrm{x}\mathrm{A}^{\mathrm{y}+}} + \underset {y\alpha} {\mathrm{y}\mathrm{B}^{\mathrm{x}-}}$

The vant hoff factor is

$\mathrm{i}=1-\alpha +\mathrm{x}\alpha +\mathrm{y}\alpha$

Hence,

$\alpha =\displaystyle \frac{\mathrm{i}-1}{(\mathrm{x}+\mathrm{y}-1)}$

Option A is correct. .

Given that

$K_w$ for water is

$10^{-13}M^2$ at

$2^oC$ , compute the sum of

$p{OH}$ and

$p{H}$ for a neutral aqueous solution at

$2^oC$ ?

0%
$7.0$
0%
$13.30$
0%
$14.0$
0%
$13.0$

Explanation

As we know,

$K_w = [H^+][OH^-];\quad pH+pOH = -logK_w = 13$ .

Which of the following is a buffer solution?

0%
$500\space mL$ of $0.1\space N\space CH_3COOH + 500\space mL$ of $0.1\space N\space NaOH$
0%
$500\space mL$ of $0.1\space N\space CH_3COOH + 500\space mL$ of $0.1\space N\space HCl$
0%
$500\space mL$ of $0.1\space N\space CH_3COOH + 500\space mL$ of $0.2\space N\space NaOH$
0%
$500\space mL$ of $0.2\space N\space CH_3COOH + 500\space mL$ of $0.1\space N\space NaOH$

The solubility of

$AgI$ in

$NaI$ solution is less than that in pure water because

0%
$AgI$ forms complex with $NaI$
0%
Of common ion effect
0%
Solubility product of $AgI$ is less than that of $NaI$
0%
The temperature of the solution decreases

The precipitate of

$CaF_2 (K_{sp} = 1.7 \times 10^{-10})$ is obtained when equal volumes of which of the following are mixed?

0%
$10^{-4} MCa^{+2} +10^{-4} MF^-$
0%
$10^{-2} M Ca^{+2} + 10^{-3} MF^-$
0%
$10^{-5} Ca^{+2} + 10^{-5} MF^-$
0%
$10^{-3} M Ca^{+2} + 10^{-5} MF^-$

Explanation

$10^{-2} M Ca^{2+} + 10^{-3} MF^-$

As equal volume are mixed so, V=2

Here, concetration product is

$\dfrac{10^{-2}}{2} \times\left(\dfrac {10^{-3}}{2}\right)^2 = {1.25}\times{10^{-9}} > K_{sp}$

So, here precipitation occurs.

$Ag_3 PO_4$ would be least soluble at 25

$^o$ C in

0%
0.1 M $AgNO_3$
0%
0.1 M $HNO_3$
0%
pure water
0%
0.1 M $Na_3PO_4$
0%
solubility in (a), (b), (c) or (d) is not different

Explanation

$Ag_3 PO_4$ is a weak electrolyte and

$AgNO_3$ is a strong electrolyte containing common ion

$(Ag^+ion)$ . Thus common ion effect is observed and the solubility of

$Ag_3 PO_4$ is suppressed.
Hence,

$Ag_3 PO_4$ is least soluble in 0.1M

$AgNO_3$ .

What is the percent ionization

$(\alpha)$ of a

$0.01M$

$HA$ solution?
(Given that

$K_a = 10^{-4}$ )

0%
$9.5\mbox{%}$
0%
$1\mbox{%}$
0%
$10.5\mbox{%}$
0%
$10\mbox{%}$

Explanation

As we know,

$\dfrac{c\alpha^2}{1-\alpha} = K_a$ , where

$\alpha$ is the degree of dissociation and

$K_a$ is the dissociation constant.

Given,

$c = 0.01M$ and

$K_a = 10^{-4}$ .

Therefore,

$k_a=\dfrac{0.01\times \alpha^2}{1-\alpha} = 10^{-4}$ ,

$0.01\alpha^2+10^{-4}\alpha-10^{-4}=0$

or,

$\alpha = 0.095$ .

$\mbox{%}$ dissocation is

$0.095\times100 = 9.5\mbox{%}$ .

What concentration of

$F-CH_2COOH, (K_a = 2.6\times10^{-3})$ is needed so that

$[H^+] = 2\times10^{-3}$ ?

0%
$1.53 \times10^{-3}\space M$
0%
$2.6\times10^{-3}\space M$
0%
$5.2\times10^{-3}\space M$
0%
$3.53\times10^{-3}\space M$

Explanation

As we know,

$\alpha = \sqrt{\dfrac{K_a}{C}}$

$[H^+] = C\alpha = \sqrt{C.K_a} = 2\times10^{-3}$
Putting the value of C

$=2\times10^{-3}\space M$ and

$K_a = 2.6 \times 10^{-3}$ ,

C =

$\dfrac {(2 \times 10^{-3})^2}{2.6 \times 10^{-3}} = 1.53 \times 10^{-3} M.$

The value of the ion product constant for water,

$(K_w)$ at

$60^{\small\circ}C$ is

$9.6\times10^{-14}\, M^2$ . What is the

$[H_3O^+]$ of a neutral aqueous solution at

$60^{\small\circ}C$ and the nature of an aqueous solution with a

$pH = 7.0$ at

$60^{\small\circ}C$ are respectively?

0%
$3.1\times10^{-8}, \space acidic$
0%
$3.1\times10^{-7}, \space neutral$
0%
$3.1\times10^{-8}, \space basic$
0%
$3.1\times10^{-7}, \space basic$

Explanation

As we know,

$K_w = [H^+][OH^-]; [H_3O^+] = \sqrt {K_w} = 3.1\times10^{-7}$ .

$pH= -log(3.1\times10^{-7})=6.5$

Neutral solution have pH < 7

So, the solution with pH= 7 will be basic in nature.

Which of the following mixtures can act as a buffer?

0%
$NaOH + HCOONa\ (1:1\space molar\space ratio)$
0%
$HCOOH + NaOH\ (2:1\space molar\space ratio)$
0%
$NH_4Cl+NaOH\ (2:1\space molar\space ratio)$
0%
$HCOOH + NaOH \ (1:1\space molar\space ratio)$

Explanation

An acidic buffer solution consists of solution of a weak acid and its salt with strong base. E.g.,

$HCOOH+NaOH\ (2:1\ molar \ ratio)$ .

A basic buffer solution consists of a mixture of a weak base and its salt with strong acid. E.g.,

$NH_4Cl+NaOH \: (2:1\ molar\ ratio)$ .

$HCOOH+NaOH\ (1:1\ molar\ ratio)$ is a weak acid-strong base salt and

$NaOH+HCOONa\ (1:1\ molar\ ratio)$ is a basic solution.

Which of the following expressions for

$\mbox{%}$ ionization of a monoacidic base

$(BOH)$ in aqueous solution is not correct at appreciable concentration?

0%
$100\times\sqrt{\displaystyle\frac{K_b}{c}}$
0%
$\displaystyle\frac{1}{1+10(pK_b - pOH)}$
0%
$\displaystyle\frac{K_w[H^+]}{K_b+K_w}$
0%
$\displaystyle\frac{K_b}{K_b+[OH^-]}$

Explanation

The expression for the dissociation of the base is

$BOH(aq) \rightleftharpoons B^+(aq) +OH^-(aq)$ .
Let

$h$ be the degree of dissociation.
At equilibrium, the concentrations of

$BOH, B^+ and OH^-$ are

$c(1-h), ch$ and

$ch$ respectively.

$K_b=\dfrac {[B^+][OH^-]}{[BOH]}=\dfrac {ch^2}{1-h}$
Since

$h$ is small,

$(1-h) \approx 1$ .
Hence,

$h= \sqrt{{\dfrac{k_b}{C}}}$ .
or,

$h= 100 \times \sqrt{{\dfrac{k_b}{C}}}\:\:\%$ .
Since total concentration of base is the sum of the concentration of

$BOH$ and

$B^+$ .
Hence,

$h=\dfrac {[B^+]}{[B^+]+[BOH]}=\dfrac {1}{1+\dfrac {[BOH]}{[B^+]}}=\dfrac {1}{1+\dfrac {[OH^-]}{K_b}}.$
or,

$\dfrac {K_b}{K_b+[OH^-]}=\dfrac {K_b[H^+]}{K_b[H^+]+K_w}$

Also,

$pOH=-log[OH^-]$ .

$[OH^-]=10^{-pOH}$

$K_b=10^{-pK_b}$

$h= \dfrac {1}{1+\dfrac {10^{-pOH}}{10^{-pK_b}}}=\dfrac {1}{1+10^{pK_b-pOH}}$

Which of the following mixtures constitute a buffer?

0%
$HCOOH + HCOONa$
0%
$Na_2CO_3+NaHCO_3$
0%
$NaCl+HCl$
0%
$NH_4Cl + (NH_4)_2SO_4$

Explanation

Acidic buffer solution : An acidic buffer solution consists of solution of a weak acid and its salt with strong base, for e.g.,

$HCOOH+HCOONa$

Basic buffer solution : A basic buffer solution consists of a mixture of a weak base and its salt with strong acid, for e.g.,

$NH_4OH+ NH_4Cl$

$NaCl+HCl$ is an acidic solution only.

$NH_4Cl + (NH_4)_2SO_4$ are both salts of weak base and strong acids.

Which of the following statement(s) is/are correct about the ionic product of water?

0%
$K_i$ (ionization constant of water) $< \space K_w$ (ionic product of water)
0%
$pK_i > pK_w$
0%
At $25^{\small\circ}C, \space K_i = 1.8\times10^{-14}$
0%
Ionic product of water at $10^{\small\circ}C$ is $10^{-14}$

Explanation

For water,

$K_i = \dfrac{[H^+][OH^-]}{[H_2O]}$

$K_w=[H^+][OH^-]$ , so

$K_i$ (ionization constant of water)

$< \space K_w$ (ionic product of water).

$(pK_i = -logK_i) > (pK_w = -logK_w)$ .

$25^oC,\ K_w = 10^{-14}$ . As temperature decreases, value of

$K_w$ also decreases from

$10^{-14}$ .

A solution consists of a mixture of

$0.01\ M\ KI$ and

$0.1\ M\ KCl$ . If solid

$AgNO_3$ is added to the solution, what is the concentration of

$I^-$ when

$AgCl$ begins to precipitate?

$[K_{sp\ (AgI)} = 1.5\times10^{-16}; \space K_{sp\ (AgCl)} = 1.8\times10^{-10}]$

0%
$3.5\times10^{-7}\ M$
0%
$6.1\times10^{-8}\ M$
0%
$2.2\times10^{-7}\ M$
0%
$8.3\times10^{-8}\ M$

Explanation

Given that a solution is

$0.01\ M\ KI$ and

$0.1\ M\ KCl$ .

When

$\text{AgCl}$ begins to precipitate,

$[Ag^+][Cl^-] = 1.8 \times 10^{-10}=K_{sp}$ of

$AgCl$

$[Ag^+] = 1.8 \times 10^{-9}$ as

$[Cl^-] =0.1$ which comes from

$KCl$ .

$K_{sp}$ of

$AgI$ is

$1.5\times10^{-16}$ .

So, at this time,

$[I^-] = \dfrac{K_{sp}}{[Ag^+]} = \dfrac{1.5 \times 10^{-16}}{1.8 \times 10^{-9}} = 8.3 \times 10^{-8}\ M$

Hence, the concentration of

$I^-$ when

$AgCl$ begins to precipitate is

$8.3\times 10^{-8}\ M$ .

Hence, the correct option is

$\text{D}$

A solution is

$0.10\space M\space Ba(NO_3)_2$ and

$0.10\space M\space Sr(NO_3)_2$ . If solid

$Na_2CrO_4$ , is added to the solution, what is

$[Ba^{2+}]$ when

$SrCrO_4$ begins to precipitate?

Given that :

$[K_{sp(BaCrO_4)} = 1.2\times10^{-10}; \space K_{sp(SrCrO_4)} = 3.5\times10^{-5}]$

0%
$7.4\times10^{-7}$
0%
$2.0\times10^{-7}$
0%
$6.1\times10^{-7}$
0%
$3.4\times10^{-7}$

Explanation

$K_{sp}(SrCrO_4)=[Sr^{2+}][CrO^{2-}_4]$

$[CrO^{2-}_4]=\dfrac{3.5 \times 10^{-5}}{0.1}=3.5 \times 10^{-4}$

$K_{sp}(BaCrO_4)=[Sr^{2+}][CrO^{2-}_4]$

$[CrO^{2-}_4]_{total}=[CrO^{2-}_4]$ from

$SrCrO_4$

$[Ba^{2+}]=\dfrac{1.2 \times 10^{-10}}{3.5 \times 10^{-4}}=3.4 \times 10^{-7}$

If the dissociation constant of

${ 5\times }10^{ -4\ }M$ aqueous solution of diethylamine is

${ 2.5\times }10^{ -5 }$ , its pH value is

0%
8.4
0%
3.95
0%
10.05
0%
2

$_{b_1}$ for X(OH)

$_{3}$ (a weak base) is 10

$^{-5}$ . What is the pH of its 0.1 M solution?

0%
3
0%
8
0%
11
0%
13

Which of the following statement(s) is/are correct?

0%
The conjugate acid of $NH_2^-$ is $NH_3$ .
0%
Solubility product increases with increase in concentration of ions.
0%
The change in $pH$ is negligible when a buffer solution is diluted.
0%
The concentration of $OH^-$ increases if some $HCl$ is added in an alkaline buffer solution.

Explanation

$NH_3\rightarrow NH_2^-+H^+$
So, the conjugate acid of

$NH_2^-$ is

$NH_3$ .

2) Solubility generally increases with an increase in temperature and so is the solubility product, but not always.

3) A solution that resists change in

$pH$ value upon addition of small amount of strong acid or base (less than

$1\%$ ) or when solution is diluted is called buffer solution.

Which of the following expressions is/are true?

0%
$[H^+] = [OH^-] = \sqrt{K_w}$ for a neutral solution
0%
$[OH^-]<\sqrt{K_w}$ for an acidic solution
0%
$pH+pOH = 14$ at all temperature
0%
$[OH^-] = 10^{-7}\space M$ at $25^{\small\circ}C$

Explanation

For neutral water,

$K_w = [H^+][OH^-]$ so

$[H^+] = [OH^-] = \sqrt{K_w}$ .

And for acidic solution as hydrogen ion conc. increases so hydroxide ion decreases and

$[OH^-]<\sqrt{K_w}$ .

$[OH^-] = 10^{-7}\space M$ at

$25^oC$ and

$pH+pOH=14$ at

$25^oC$ only.

The compound whose

$0.1\space M$ solution is acidic:

0%
Ammonium formate
0%
Ammonium sulphate
0%
Ammonium chloride
0%
Sodium formate

Explanation

Concept :-

A compound will be acidic if it form strong acidic solution after dissolving in water.

$HCOONH_4 + H_2 O \rightarrow \underset{Weak \, acid}{HCOOH} + \underset{weak \, base}{NH_4OH}$

$(NH_4)_2 SO_4 + 2H_2 O \rightarrow \rightarrow \underset{Strong \, acid}{H_2SO_4} + \underset{Weak\, base}{2NH_4OH}$

$NH_4 Cl + H_2O \rightarrow \underset{Weak \, base}{NH_4 OH} + \underset{Strong \, acid}{HCl}$

$HCO\overset{-}{O}N\overset{+}{a} + H_2O \rightarrow \underset{Strong \, base}{NaOH} + \underset{Weak \ acid}{HCOOH}$

Consider the following statements:
(a) Green crystals of ferrous sulphate become dirty white upon strong heating.
(b) The chemical formula of aluminium phyosphate is

$Al(PO_4)_3$ .
(c) Silver salts are mostly sensitive to light.
(d) The gas released in respiration is oxygen.
Which of the above statement is correct?

0%
(a)
0%
(b)
0%
(c)
0%
(d)

Simultaneous solubility of

$AgCNS\ (a)$ and

$AgBr\ (b)$ in a solution of water will be

${K}_{{sp}_{(AgBr)}}=5\times {10}^{-13}$ and

${K}_{{sp}_{(AgCNS)}}={10}^{-12}$

0%
$a=4.08\times {10}^{-7}mol$ ${litre}^{-1}$ ; $b=8.16\times {10}^{-7}$ $mol$ ${litre}^{-1}$
0%
$a=4.08\times {10}^{-7}mol$ ${litre}^{-1}$ ; $b=4.08\times {10}^{-7}$ $mol$ ${litre}^{-1}$
0%
$a=8.16\times {10}^{-7}mol$ ${litre}^{-1}$ ; $b=4.08\times {10}^{-7}$ $mol$ ${litre}^{-1}$
0%
None of these

Explanation

Suppose solubility of

$AgCNS$ and

$AgBr$ in a solution are

$a$ and

$b$ respectively.

$AgCNS(s)\rightleftharpoons \underset { a+b }{ { Ag }^{ + }(aq) } +\underset { a }{ { CNS }^{ - }(aq) }$

$AgBr(s)\rightleftharpoons \underset { a+b }{ { Ag }^{ + } } +\underset { b }{ { Br }^{ - } }$

$\therefore$

$[{Ag}^{2+}]=(a+b); [CN{S}^{-}]=a$ and

$[{Br}^{-}]=b$

For

$AgCNS:{ K }_{ { sp }_{ AgCNS }\quad }=\left[ { Ag }^{ + } \right] \left[ { CNS }^{ - } \right]$

$1\times {10}^{-12}=(a+b)(a)......(i)$

For

$AgBr: { K }_{ { sp }_{ AgBr }\quad }=\left[ { Ag }^{ + } \right] \left[ { Br }^{ - } \right]$

$5\times {10}^{-13}=(a+b)(b)........(ii)$

By Eqs

$(i)$ and

$(ii)$ , we get

$\cfrac{a}{b}=\cfrac{{10}^{-12}}{5\times {10}^{-13}}=2$ (

$a=2b$ )

$\therefore$ By Eq

$(i)$

$(2b+b)(2b)=1\times {10}^{-12}$

$\therefore$

$6{b}^{2}=1\times {10}^{-12}$

$b=4.08\times {10}^{-7}$

$mol$

${litre}^{-1}$

Similarly by Eq

$(ii)$

$[a+(a/2)]a=1\times {10}^{-12}$

$a=8.16\times {10}^{-7}$

$mol$

${litre}^{-1}$

Which can act as buffer?

0%
${NH}_{4}Cl+{NH}_{4}OH$
0%
${CH}_{3}COOH+{CH}_{3}COONa$
0%
$40$ mL of $0.1M$ $NaCN+10$ mL of $0.1M$ $HCl$
0%
All of the above

Explanation

Hint: Recollect the preparation of buffers.

Correct Answer: Option (D).

Explanation:

Buffer is a solution that resists the change in

$pH$ when an acid or a base is added to it.

Types of buffers:

$(i)$ Basic buffer is a mixture of a weak base and its salt with a strong acid.

$(ii)$ Acidic buffer is a mixture of a weak acid and its salt with a strong base.

Option (A):

$NH_4OH$ is a weak base and

$NH_4Cl$ is its salt with strong acid

$HCl$ .

Option (B):

$CH_3COOH$ is a weak acid and

$CH_3COONa$ is its salt with strong base

$NaOH$ .

Option (C): Let us first look at the reaction that takes place:

$NaCN+HCl\rightarrow NaCl+HCN$

Now, initial moles of

$HCl$ taken

$= 10 \times 0.1=1\ millimol$

Whereas, initial moles of

$NaCN$ taken

$=40 \times 0.1 = 4\ millimol$

Thus,

$NaCN$ is taken in far excess compared to

$HCl$ . So, after the reaction is over, much of

$NaCN$ remains unreacted. So, the resulting solution contains

$HCN$ which is a weak acid, and

$NaCN$ which is its salt with a strong base

$NaOH$ . So, it is also a buffer.

Which one is correct for

${H}_{2}O$ at

${25}^{o}C$

0%
Ionic product of water, ${K}_{w}={10}^{-14}$
0%
Equilibrium constant for dissociation of water ${K}_{c}=1.8\times {10}^{-16}$
0%
Aotoprotolysis constant of water, ${K}_{AP}=3.2\times {10}^{-18}$
0%
On heating ${K}_{w}$ increases with temperature

Explanation

Equilibrium constant for dissociation of water

${K}_{c}=1.8\times {10}^{-16}$

Ionic product of water,

${K}_{w} = [H^+][OH^-] = 10^{-7}\times10^{-7} ={10}^{-14}$

Also as we increase temperature, more no of ions will be produced and conc. of hydrogen ions as well as hydroxide ions will be increase and

$K_w$ increases.

Autoprotolysis constant of water,

${K}_{AP}=3.2\times {10}^{-18}$

The degree of dissociation of

$0.1M$

$HCN$ solution is:

0%
$6.4\times { 10 }^{ -5 }$
0%
$6.4\times { 10 }^{ - 3}$
0%
$6.4\times { 10 }^{ -2 }$
0%
$6.4\times { 10 }^{ -6 }$

Explanation

$\underset { c\\ 1-\alpha }{ HCN } \rightleftharpoons \underset {0\\ \alpha }{ { H }^{ + } } +\underset { 0\\ \alpha }{ { CN }^{ - } }$

${ \quad K }_{ a }=c\cdot { \alpha }^{ 2 }$

$4.1\times { 10 }^{ -10 }=0.1\times { \alpha }^{ 2 }$

$\alpha =6.4\times { 10 }^{ -5 }$

Hence, option A is correct.

Assertion: A is very dilute acidic solution of

$Cd^{2+}$ and

$Ni^{2+}$ gives a yellow precipitate of CdS on passing hydrogen sulphide.
Reason: Solubility product of CdS is more than that of NiS.

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct
0%
Both Assertion and Reason are incorrect

Explanation

A is very dilute acidic solution of

$Cd^{2+}$ and

$Ni^{2+}$ gives yellow precipitate of

$CdS$ on passing hydrogen sulphide because solubility product of

$CdS$ is less than that of

$NiS$ .

Nickel(II) sulfide

$NiS: 3\times 10^{–19}$

Cadmium sulfide

$CdS: 8\times 10^{–28}$

$[H^+]=\sqrt { \frac { { K }_{ w }{ K }_{ a } }{ C } }$ is suitable for

0%
$NaCl,$ $NH_4Cl$
0%
$CH_3COONa,$ $NaCN$
0%
$CH_3COONa,$ $(NH_4)_2SO_4$
0%
$CH_3COONH_4,$ $(NH_4)_2CO_3$

Aqueous solution sof

$H{NO}_{3},KOH,{CH}_{3}COOH,{CH}_{3}COONa$ of identical concentrations are provided. The pairs of solution swhich forms a buffer upon mixing is (are):

0%
$H{NO}_{3}$ and ${CH}_{3}COOH$
0%
$KOH+{CH}_{3}COONa$
0%
$H{NO}_{3}$ and ${CH}_{3}COONa$
0%
${CH}_{3}COOH+{CH}_{3}COONa$

Explanation

Following are the types of Buffer Solutions:
1. Acidic Buffer: Weak Acid + its Salt of Strong Base
2. Basic Buffer: Weak Base + its Salt of Strong Acid
3. Neutral Buffer: Salt of Weak Acid and Weak Base (rarely used)

Option A has a mixture of

$HNO_{3}$ (Strong Acid) and

$CH_{3}COOH$ (Weak Acid). Hence, it is not a Buffer solution.

Option B has a mixture of

$KOH$ (Strong Base) and

$CH_{3}COONa$ (Salt of Weak Acid and Strong Base). Hence, it is not a Buffer solution.

Option D has a mixture of

$CH_{3}COOH$ (Weak Acid) and

$CH_{3}COONa$ (its Salt with a Strong Base). Hence, it is a Buffer solution.

Option C has a mixture of

$HNO_{3}$ (Strong Acid) and

$CH_{3}COONa$ (its Salt with a Strong Base). Hence, it can act as a Buffer solution in the following case:

When Moles of

$CH_{3}COONa$ initially present is greater than the moles of

$HNO_{3}$ present in the mixture. This is because, following reaction would take place in the mixture:

$HNO_{3} + CH_{3}COONa \rightarrow CH_{3}COOH + NaNO_{3}$

We know that a mixture of

$CH_{3}COOH$ and

$CH_{3}COONa$ is a Buffer Solution which can only be formed when some of the

$CH_{3}COONa$ remains unreacted i.e. when

$CH_{3}COONa$ is taken in an excess.

Hence,option C and D are correct.

The given aqueous solution at

${25}^{o}C$ is:

0%
acidic if $[{H}^{+}]< \sqrt{{K}_{w}}$
0%
alkaline if $[{H}^{+}]< \sqrt{{K}_{w}}$
0%
acidic if $[{H}^{+}]> \sqrt{{K}_{w}}$
0%
neutral if $[{H}^{+}]= \sqrt{{K}_{w}}$

Explanation

$[H^+][OH^-] = K_w$

and

$[H^+] = [OH^-]$ in neutral solution

So, solution will be neutral if

$[{H}^{+}] = \sqrt{{K}_{w}}$

In alkaline solution,

$[H^+] < [OH^-]$

$[{H}^{+}]< \sqrt{{K}_{w}}$

In acidic solution,

$[H^+] > [OH^-]$

$[{H}^{+}] > \sqrt{{K}_{w}}$

The relation

$[{H}^{+}O]=\cfrac { { K }_{ w } }{ \left[ { H }_{ 3 }{ O }^{ + } \right] } +{ \left[ HCl \right] }_{ 0 }$ for an aqueous solution of

$HCl$ can be reduced to:

0%
$\left[ { H }_{ 3 }{ O }^{ + } \right] ={ \left[ HCl \right] }_{ 0 }\quad if\quad \left[ { H }_{ 3 }{ O }^{ + } \right] \ge { 10 }^{ -6 }\quad$
0%
$\left[ { H }_{ 3 }{ O }^{ + } \right] ={ \left[ HCl \right] }_{ 0 }\quad if\quad \cfrac { { K }_{ w } }{ \left[ { H }_{ 3 }{ O }^{ + } \right] } \ge { 10 }^{ -8 }\quad$
0%
$\left[ { H }_{ 3 }{ O }^{ + } \right] =\cfrac { { \left[ HCl \right] }_{ 0 }\pm \sqrt { { \left[ HCl \right] }_{ 0 }^{ 2 }+4{ K }_{ w } } }{ 2 } if\quad \left[ { H }_{ 3 }{ O }^{ + } \right] \le { 10 }^{ -6 }$
0%
$\left[ { H }_{ 3 }{ O }^{ + } \right] ={ \left[ HCl \right] }_{ 0 }\quad if\quad { \left[ HCl \right] }_{ 0 }\ge { 10 }^{ -6 }\quad$

Explanation

$HCl+{ H }_{ 2 }O\longrightarrow { H }_{ 3 }^{ + }O+{ Cl }^{ - }$

$\left[ { H }_{ 3 }{ O }^{ + } \right] =\cfrac { Kw }{ { \left[ { H }_{ 3 }{ O }^{ + } \right] } } +{ \left[ HCl \right] }_{ o }$

(A)When

$\left[ { H }_{ 3 }{ O }^{ + } \right] \ge { 10 }^{ -6 }$

$\cfrac { Kw }{ { \left[ { H }_{ 3 }{ O }^{ + } \right] } } \approx 0$

$\left[ { H }_{ 3 }{ O }^{ + } \right]={ \left[ HCl \right] }_{ o }$

(B) When

$\dfrac { Kw }{ \left[ { H }_{ 3 }{ O }^{ + } \right] } \ge { 10 }^{ -8 }\approx 0$

$\left[ { H }_{ 3 }{ O }^{ + } \right]={ \left[ HCl \right] }_{ o }$

(C)When

$\left[ { H }_{ 3 }{ O }^{ + } \right] \le { 10 }^{ -6 }$

Then

${ \left[ { H }_{ 3 }{ O }^{ + } \right] }^{ 2 }=Kw+{ \left[ HCl \right] }_{ o }\left[ { H }_{ 3 }{ O }^{ + } \right]$

${ \left[ { H }_{ 3 }{ O }^{ + } \right] }^{ 2 }-{ \left[ HCl \right] }_{ o }\left[ { H }_{ 3 }{ O }^{ + } \right] -Kw=0$

Compare with

$a{ x }^{ 2 }-bx+c=0$

$x=\cfrac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ { 2a } }$

$\left[ { H }_{ 3 }{ O }^{ + } \right] =\dfrac { { \left[ HCl \right] }_{ o }\pm \sqrt { { \left[ HCl \right] }_{ o }^{ 2 }+4Kw } }{ 2 }$

(D) When

${ \left[ HCl \right] }_{ o }\ge { 10 }^{ -6 }$

Here acid concentration is very high so automatically dissociation of water is negligible

$\left[ { H }_{ 3 }{ O }^{ + } \right] ={ \left[ HCl \right] }_{ o }$

Which are buffer mixtures?

0%
${H}_{3}{BO}_{3}$ and borax
0%
$NaOH$ and $Na{NO}_{3}$
0%
${CH}_{3}COONa$ and ${CH}_{3}COOH$
0%
${NH}_{4}OH$ and ${NH}_{4}Cl$

Assertion: On mixing equal volumes of 1 M

$HCl$ and of 2 M

$CH_3COONa$ , an acidic buffer solution is formed.
Reason: The resultant mixture contains

$CH_3COOH$ and

$CH_3COONa$ which are parts of acidic buffer.

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Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
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Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
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Assertion is correct but Reason is incorrect
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Assertion is incorrect but Reason is correct
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Both Assertion and Reason are incorrect

Explanation

A solution that resists change in pH value upon addition of small amount of strong acid or base (less than 1 %) or when solution is diluted is called buffer solution.

Acidic buffer solution : An acidic buffer solution consists of solution of a weak acid and its salt with strong base. The best-known example is a mixture of solution of acetic acid and its salt with strong base

$(CH_3COONa)$ .

Reaction:

$CH_3COONa + HCl \rightarrow CH_3COONa + CH_3COOH$

2 1 - -

1 - 1 1

which are parts of acidic buffer

The dissociation constant of

${NH}_{4}OH$ can be given as

0%
${ K }_{ b }=\cfrac { { \left( { c }_{ 3 } \right) }^{ 2 } }{ \left( { c }_{ 2 }-{ c }_{ 3 } \right) }$
0%
${ K }_{ b }=\cfrac { { \left( { c }_{ 3 } \right) }^{ 2 } }{ { c }_{ 2 } }$
0%
${ K }_{ b }=\cfrac { { c }_{ 3 } }{ \left( { c }_{ 2 }-{ c }_{ 3 } \right) }$
0%
${ K }_{ b }=\cfrac { { c }_{ 3 } }{ \left( { c }_{ 1 }-{ c }_{ 2 } \right) }$

Explanation

As given,

$\underset { { c }_{ 2 }\\ ({ c }_{ 2 }-{ c }_{ 3 }) }{ { NH }_{ 4 }OH } \rightleftharpoons \underset { 0\\ { c }_{ 3 } }{ { NH }_{ 4 }^{ + } } +\underset { 0\\ { c }_{ 3 } }{ { OH }^{ - } }$

$K_b = [NH_4^+][OH^-]/[NH_4OH]$

$\quad { K }_{ b }=\cfrac { { { c }_{ 3 } }^{ 2 } }{ \left( { c }_{ 2 }-{ c }_{ 3 } \right) }$

The

$pH$ of pure water at

${25}^{o}C$ and

${60}^{o}C$ are

$7$ and

$6.5$ respectively.

$HCl$ gas is passed through water at

${25}^{o}C$ till the resulting

$1$ litre solution which aquires a

$pH$ of

$3$ . Now

$4\times {10}^{-3}$ mole of

$NaCN$ are added into this solution. Also a fresh

$0.1M$

$HCN$ solution has

$pH$

$5.1936$ . Now in the one part of solution obtained after addition of

$NaCN$ , one milli mole of

$NaOH$ are added and in the second part of this solution

$0.5$ miili mole of

$HCl$ are added.

The dissociation constant of

$HCN$ is:

0%
$4.1\times {10}^{-10}$
0%
$4.1\times {10}^{-6}$
0%
$4.1\times {10}^{-3}$
0%
$4.1\times {10}^{-8}$

Explanation

$\underset { 1\\ 1-\alpha }{ HCN } \rightleftharpoons \underset { 0\\ \alpha }{ { H }^{ + } } +\underset { 0\\ \alpha }{ { CN }^{ - } }$
Also,

$\quad \left[ { H }^{ + } \right] =c\cdot \alpha =c\sqrt { \cfrac { { K }_{ a } }{ c } } =\sqrt { { K }_{ a }c }$

$-\log { { H }^{ + } } =-\cfrac { 1 }{ 2 } \left[ \log { { K }_{ a } } +\log { c } \right]$

$5.1936=-\cfrac { 1 }{ 2 } \left[ \log { { K }_{ a } } +\log { 0.1 } \right]$

$\therefore { \quad K }_{ a }=4.1\times {10}^{-10}$

Acetyl salicylic acid (aspirin) ionises in water as:

$HC_9H_7O_4+H_2O\rightarrow H_3O^+ + C_9H_7O_4^-;$

$(K_a = 2.75\times 10^{-9})$ . If two tablets of aspirin each of 0.32 g is dissolved in water to produce 250 mL solution, calculate

$[\overset{\circleddash}{O}H].$

0%
$1.61\times 10^{-9}M$
0%
$1.61\times 10^{-7}M$
0%
$1.61\times 10^{-3}M$
0%
$1.61\times 10^{-5}M$

Explanation

$[HC_9H_7O_4] = \displaystyle \frac{0.32\times 2\times 1000}{180\times 250} = 0.014\, M$

$\because \alpha = \displaystyle \sqrt{\frac{K_a}{C}} = \sqrt{\frac{2.75\times 10^{-9}}{0.014}} = 4.43\times 10^{-4}$

$\therefore [H^{\oplus}] = C.\alpha = 0.014 \times 4.43\times 10^{-4}=6.21\times 10^{-6}\, M$

$[C_9H_7O_4^{\ominus}]=C.\alpha = 6.21\times 10^{-6}\, M$

$[OH^{\circleddash}]=\displaystyle \frac{10^{-14}}{[H^{\oplus}]}=\frac{10^{-14}}{6.21\times 10^{-6}}= 1.61\times 10^{-9}M$

Calculate the equilibrium constants for the reactions with water of

$H_2PO_4^{\circleddash}, HPO_4^{2-}$ and

$PO_4^{3-}$ as base. Comparing the relative values of two equilibrium constants of

$H_2PO_4^{\circleddash}$ with water, deduce whether solutions of this ion in water are acidic or bases. Deduce whether solutions of

$HPO_4^{2-}$ are acidic or bases. Given

$K_1, K_2$ and

$K_3$ for

$H_3PO_4$ are

$7.1\times 10^{-3}, 6.3\times 10^{-8}$ and

$4.5\times 10^{-13}$ respectively .

0%
$1.4\times 10^{-12}, 1.6\times 10^{-7}, 2.2\times 10^{-2}, H_2PO_4^{\circleddash}$ is acidic and $HPO_4^{2-}$ is basic
0%
$2.2\times 10^{-12}, 1.6\times 10^{-7}, 1.4\times 10^{-2}, H_2PO_4^{\circleddash}$ is basic and $HPO_4^{2-}$ is basic
0%
$2.2\times 10^{-12}, 1.6\times 10^{-7}, 1.4\times 10^{-2}, H_2PO_4^{\circleddash}$ is acidic and $HPO_4^{2-}$ is basic
0%
$1.4\times 10^{-12}, 1.6\times 10^{-7}, 2.2\times 10^{-2}, H_2PO_4^{\circleddash}$ is basic and $HPO_4^{2-}$ is acidic

Explanation

$H_2PO_4^{\circleddash} + H_2O \rightleftharpoons H_3PO_4+\overset{\circleddash}{O}H$

$K_h = \displaystyle \frac{K_w}{K_a}=\frac{10^{-14}}{7.1\times 10^{-3}}=1.4\times 10^{-12}$
ii.

$HPO_4^{2-} + H_2O \rightleftharpoons H_2PO_4^{\circleddash}+\overset{\circleddash}{O}H$

$K_h = \displaystyle \frac{K_w}{K_2} = \frac{10^{-14}}{6.3\times 10^{-8}}=1.6\times 10^{-7}$
iii.

$PO_4^{3-}+H_2O \rightleftharpoons HPO_4^{2-}+\overset{\circleddash}{O}H$

$K_h = \displaystyle \frac{K_w}{K_3} = \frac{10^{-14}}{4.5\times 10^{-13}}=2.2\times 10^{-2}$

$H_2PO_4^{\circleddash}$ can react with water to yield

$HPO_4^{2-}$ and

$H_3O^{\oplus}$ with a constant

$6.3\times 10^{-8}$
Its reaction with water to yield

$H_3PO_4$ ,

$\overset{\circleddash}{O}H$ has a much smaller equilibrium constant,

$1.4 \times 10^{-12}$ , and so this ion in water is acidic comparision of the constants for

$HPO_4^{2-}(4.5\times 10^{-3}$ as an acid,

$1.6\times 10^{-7}$ as base

$)$ leads to the conclusion that this ion in solution is basic.

Which statement/relationship is correct?

0%
Upon hydrolysis of salt of a strong base and weak acid gives a solution with $pH < 7$
0%
$pH = -\log { \displaystyle\frac { 1 }{ \left[ { H }^{ + } \right] } }$
0%
Only at $25^{0}C$ the $pH$ of water is 7
0%
The value of $p{ K }_{ w }$ at $25^{0}C$ is 7

Explanation

As we know,

$K_w = [H^+][OH^-] = 10^{-14}$

So value of

$pK_w$ at

$25^{0}C$ is

$14$ .

At neutral point,

$[H^+] =\sqrt {K_w} = 10^{-7} M$

$pH = -log[H^+] = 7$

Also, upon hydrolysis of salt of a strong base and weak acid gives a solution with

$pH > 7$

The degree of dissociation of abscorbic acid solution is:

0%
$0.40$
0%
$0.33$
0%
$0.20$
0%
$0.15$

$K_{sp}\, of \, Mg(OH)_2$ is

$1\times 10^{-12}, 0.01\, M\, MgCl_2$ will be precipitating at the limiting

$pOH$

0%
$7$
0%
$8$
0%
$9$
0%
$5$

Explanation

According to given data,

$[OH^-]$ needed for precipitation of

$Mg(OH)_2=(\cfrac { { 10 }^{ -12 } }{ { 10 }^{ -2} })^{1/2} =({ 10 }^{ -10})^{1/2}M = 10^{-5} M$

$pOH = -log [OH^-] = 5$

Which of the following statements about a weak acid strong base titration is/are correct?

0%
The pH after the equivalence point of the weak acid strong base titration is determined by using the $K_b$ expression for the conjugate base.
0%
A buffer solution of weak acid and its conjugate base is formed before the equivalence is reached
0%
The pH at the equivalence point of a weak monoprotic acid strong base titration is equal to the pH at the equivalence point of a strong acid-strong base titration.
0%
The increase in pH in the region near the equivalence The increase in pH in the region near the equivalence point of a weak acid strong base titration is grater than the pH change in the same region of a strong acid strong base titration

Explanation

The titration curve is a graph of the volume of titrant, or in our case, the volume of a strong base plotted against the pH. There are several characteristics that are seen in all titration curves of a weak acid with a strong base.

The initial

$pH$ (before the addition of any strong base) is higher or less acidic than the titration of a strong acid.

There is a sharp increase in pH at the beginning of the titration. This is because the anion of the weak acid becomes a common ion that reduces the ionization of the acid.

After the sharp increase at the beginning of the titration, the curve only changes gradually. This is because the solution is acting as a buffer. This will continue until the base overcomes the capacity of the buffer.

In the middle of this gradually curve the half-neutralization occurs. At this point, the concentration of weak acid is equal to the concentration of its conjugate base. Therefore the

$pH=pKa$ . This point is called the half-neutralization because half of the acid has been neutralized.

At the equivalence point, the

$pH$ is greater than 7 because all of the acid (

$HA$ ) has been converted to its conjugate base (

$A-$ ) by the addition of

$NaOH$ and now the equilibrium moves backwards towards

$HA$ and produces hydroxide.

The steep portion of the curve prior to the equivalence point is short. It usually only occurs until a

$pH$ of around 10.

The solubility products of MA, MB, MC, and MD are

$1.8\times 10^{-10}, 4\times 10^{-3}, 4\times 10^{-8}$ and

$6\times 10^{-5}$ respectively. If a 0.01 M solution of MX is added dropwise to a mixture containing

$A^{\circleddash}, B^{\circleddash}, C^{\circleddash}$ , and

$D^{\circleddash}$ ions, then the one to be precipitated first will be:

0%
MA
0%
MB
0%
MC
0%
MD

Explanation

According to given data,

$[M^+]$ needed for precipitation of

$MA=\cfrac { 1.8\times{ 10 }^{ -10 } }{ { 10 }^{ -2} } =1.8\times{ 10 }^{ -8 }M$

$MB =\cfrac { 4\times{ 10 }^{ -3 } }{{ 10 }^{ -2 } } = 4\times{ 10 }^{ -1}M$

$MC=\cfrac { 4\times{ 10 }^{ -8 } }{ { 10 }^{ -2} } =4\times{ 10 }^{ -6 }M$

$MD=\cfrac { 6\times{ 10 }^{ -5 } }{ { 10 }^{ -2} } =6\times{ 10 }^{ -3 }M$

So, MA will first precipitate out as it requires lowest concentration of ion.

$EDTA,$ often abbreviated as

${H}_{4}Y$ , forms very stable complexes with almost all metal ions. Calculate the fraction of EDTA in the fully protonated form,

${H}_{4}Y$ in a solution obtained by dissolving

$0.1$ mol

${Na}_{4}Y$ in

$1$ litre.

Given, the acid dissociation constants of

${H}_{4}Y$ are as follows:

${k}_{1}=1.02\times {10}^{-2}$ ,

${k}_{2}=2.13\times {10}^{-3}$ ,

${k}_{3}=6.92\times {10}^{-7}$ ;

${k}_{4}=5.50\times {10}^{-11}$

0%
$3.82\times {10}^{-26}$
0%
$3.82\times {10}^{-6}$
0%
$3.82\times {10}^{-20}$
0%
$3.82\times {10}^{-30}$

Explanation

${Na}_{4}Y+{H}_{2}O\rightleftharpoons {Na}_{3}HY+NaOH\quad \cfrac{{k}_{w}}{{k}_{4}}=1.818\times {10}^{-4}$

$0.1$ Initial

$0.1-x$

$x$

$x$ Final

$\cfrac{{x}^{2}}{0.1-x}=1.818\times {10}^{-4}$

$\Rightarrow$

$5500.55{x}^{2}+x-0.1=0$

$x=4.17\times {10}^{-3}$

${Na}_{3}HY+{H}_{2}O\rightleftharpoons {Na}_{2}{H}_{2}Y+NaOH\cfrac{{k}_{w}}{{k}_{3}}=1.445\times {10}^{-8}$

$x$

$x-y$

$y$

$y+x$
Since

$y<<x$

$x-y\simeq x$

$x+y\simeq x$

$1.445\times {10}^{-8}=\cfrac{y\times x}{x}=y$

${Na}_{2}{H}_{2}Y+{H}_{2}O\rightleftharpoons Na{H}_{3}Y+NaOH\quad \cfrac {{k}_{w}}{{k}_{2}}=4.7\times {10}^{-12}$

$y$

$y-z$

$z$

$z+x$

$y-z\simeq y$ ,

$z+x\simeq x$

$4.7\times {10}^{-12}=\cfrac{z\times x}{y}$

$z=1.628\times {10}^{-17}$

$Na{H}_{3}Y+{H}_{2}O\rightleftharpoons {H}_{4}Y+NaOH$

$\cfrac {{k}_{w}}{{k}_{1}}=9.8\times {10}^{-13}$

$z$

$z-t$

$t$

$t+x$

$z-t\simeq z$

$t+x\simeq x$

$9.8\times {10}^{-13}=\cfrac{t.x}{z}$

$t=3.82\times {10}^{-26}$

If the equilibrium constant of

$BOH \rightleftharpoons B^{\oplus}+\overset{\circleddash}{O}H$ at

$25^oC$ is

$2.5\times 10^{-6}$ , then equilibrium constant for

$BOH + H^{\oplus} \rightleftharpoons B^{\oplus}+H_2O$ at the same temperature is

0%
$4.0\times 10^{-9}$
0%
$4.0\times 10^{5}$
0%
$2.5\times 10^{8}$
0%
$2.5\times 10^{-6}$

Explanation

If the equilibrium constant of

$BOH \rightleftharpoons B^{\oplus}+\overset{\circleddash}{O}H$ at

$25^oC$ is

$2.5\times 10^{-6}$ , then equilibrium constant for

$BOH + H^{\oplus} \rightleftharpoons B^{\oplus}+H_2O$ at the same temperature is

$2.5\times 10^{8}$

$BOH \rightleftharpoons B^{\oplus}+\overset{\circleddash}{O}H$ .

$K = 2.5\times 10^{-6}$ .....(1)

$\displaystyle H^+ + OH^- \rightleftharpoons H_2O$

$\dfrac {1} {K_w} = 10^{14}$ ......(2)
---------------------------------------------------------------------------------------------------

$BOH + H^{\oplus} \rightleftharpoons B^{\oplus}+H_2O$

$K' = ?$ ......(3)
The reaction (3) is obtained by adding reactions (1) and (2),

$\displaystyle K' = K \times \frac {1}{K_w} = 2.5 \times 10^{-6} \times 10^{14} = 2.5 \times 10^8$

$K_{sp}\,$ of

$\, Mg(OH)_2$ is

$1\times 10^{-12}, 0.01\, M\, MgCl_2$ will be precipitating at the limiting pH :

0%
8
0%
9
0%
10
0%
12

Explanation

According to given data,

$[OH^-]$ needed for precipitation of

$Mg(OH)_2=(\cfrac { { 10 }^{ -12 } }{ { 10 }^{ -2} })^{1/2} =({ 10 }^{ -10})^{1/2}M = 10^{-5} M$

$pOH = -log [OH^-] = 5$

So,

$pH = 14 - pOH = 9$

The value of

$[Mg^{2+}][OH^{\circleddash}]^2$ in a solution of 0.001 M

$Mg(OH)_2$ in

$Mg(NO_3)_2$ if the pH of solution is adjusted to 9 is:

$K_{sp}$ of

$Mg(OH)_2 = 8.9\times 10^{-12}$

0%
lesser than $K_{sp}$ and $Mg(OH)_2$ won't precipitate
0%
greater than $K_{sp}$ and $Mg(OH)_2$ won't precipitate
0%
lesser than $K_{sp}$ and $Mg(OH)_2$ will precipitate
0%
greater than $K_{sp}$ and $Mg(OH)_2$ will precipitate

Explanation

$pH = 9 \therefore [H^{\oplus}] = 10^{-9} M \, \, or \, \, [OH^{\circleddash}] = 10^{-5}M$

Now if

$Mg(NO_3)_2$ is present in a solution of

$[OH^{\circleddash}]=10^{-5}M$ , then

Product of ionic concentration

$=[Mg^{2+}][OH^{\circleddash}]^2$

$=[0.001][10^{-5}]^2$

$=10^{-13}$ less than

$K_{sp}$ of

$Mg(OH)_2$ (i.e.,

$8.9\times 10^{-12}$ )

$Mg(OH)_2$ will not precipitate.

Calculate

$[\overset{\circleddash}{O}H]$ which would be produced by each equilibrium concentration of

$NH_3$ in part (a). Predict whether

$Zn(OH)_2$ or

$Zn(OH)_4^{2-}$ would form in prefemce to

$Zn(NH_3)_4^{2+}$ upon addition of sufficient

$NH_3$ to produce the equilibrium concentration calculated in part (a).

0%
$3.33\times 10^{-4} M$
0%
$2.5\times 10^{-4} M$
0%
$4.5\times 10^{-4} M$
0%
$9.0\times 10^{-4} M$

Explanation

The concentration of

$\overset{\circleddash}{O}H$ in equilibrium with the

$NH_3$ concentration of part (a) is given by:

$NH_3 + H_2O \rightleftharpoons \overset{\oplus}{N}H_4+\overset{\circleddash}{O}H$ ,

(Let

$x = [\overset{\oplus}{N}H_4]=[\overset{\circleddash}{O}H])$

$K_b = \displaystyle \frac{ [\overset{\oplus}{N}H_4] [\overset{\circleddash}{O}H]}{[NH_3]}$

$1.8\times 10^{-5} = \displaystyle \frac{x^2}{4.5\times 10^{-2}}$

$\Longrightarrow x^2 = 8.1\times 10^{-7} \Longrightarrow x = [\overset{\circleddash}{O}H]= 9.0 \times 10^{-4}M$

$[\overset{\circleddash}{O}H] (9.0 \times 10^{-4} M)$ from the

$4.5\times 10^{-2} M \, NH_3$ is sufficient to precipitate

$Zn(OH)_2$ [part (b)] but not sufficient to form

$Zn(OH_4^{2-})$ [part (c)]

$0.00050$ mol

$NaH{CO}_{3}$ is added to

$1$ litre of a buffered solution of

$pH\ 8$ , then how much material will exist in each of the three forms

${H}_{2}{CO}_{3},H{CO}_{3}^{-}$ and

${CO}_{3}^{2-}$ ?

For

${H}_{2}{CO}_{3}$ ,

${K}_{1}=5\times {10}^{-7}$ ;

${K}_{2}=5\times {10}^{-13}$

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$[{H}_{2}{CO}_{3}]=9.85\times {10}^{-6}M$ , $[H{CO}_{3}^{-}]=4.9\times {10}^{-4}M$ , $[{CO}_{3}^{-2}]=2.45\times {10}^{-8}M$
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$[{H}_{2}{CO}_{3}]=4.9\times {10}^{-6}M$ , $[H{CO}_{3}^{-}]=9.85\times {10}^{-4}M$ , $[{CO}_{3}^{-2}]=2.45\times {10}^{-8}M$
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$[{H}_{2}{CO}_{3}]=9.85\times {10}^{-6}M$ , $[H{CO}_{3}^{-}]=2.45\times {10}^{-4}M$ , $[{CO}_{3}^{-2}]=4.9\times {10}^{-8}M$
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$[{H}_{2}{CO}_{3}]=92.45\times {10}^{-6}M$ , $[H{CO}_{3}^{-}]=4.9\times {10}^{-4}M$ , $[{CO}_{3}^{-2}]=9.85\times {10}^{-8}M$

Explanation

Given,

$pH=8, [{H}^{+}]={10}^{-8}, [{OH}^{-}={10}^{-6}$

$H{CO}_{3}^{-}\rightleftharpoons {H}^{+}+{CO}_{3}^{-2}$

$K=5\times {10}^{-13}$

$0.0005$

$0.0005-y-z$

${10}^{-8}$

$y$

$H{CO}_{3}^{-}+{H}_{2}O\rightleftharpoons {H}_{2}{CO}_{3}+{OH}^{-}$

$K=\cfrac{{K}_{w}}{{K}_{1}}=2\times {10}^{-8}$

$0.0005$

$0.0005-y-z$

$z$

${10}^{-6}$
Since equilibrium constant for first reaction is very less,

$y<<z$

$2\times {10}^{-8}=\cfrac{z({10}^{-6})}{0.0005-z}$

$51z=0.0005,\Rightarrow z=9.8\times {10}^{-6}$

$[{H}_{2}{CO}_{3}]=9.8\times {10}^{-6}M$

$[H{CO}_{3}^{-}]=0.0005-9.8\times {10}^{-6}=4.9\times {10}^{-4}M$

$5\times {10}^{-3}=\cfrac{{10}^{-8}\times y}{4.9\times {10}^{-4}}$

$[{CO}_{3}^{2-}]=y=2.45\times {10}^{-8}M$

CBSE Questions for Class 11 Medical Chemistry Equilibrium Quiz 11 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Equilibrium Quiz 11 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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