MCQ Questions for Class 11 Medical Chemistry Equilibrium Quiz 12

Compound A

$_{(l)}$ and B

$_{(l)}$ function as a weak acid and weak base respectively when dissolved in water. When pure A

$_{(l)}$ is dissolved in anhydrous B

$_{(l)}$ , A will function as:

0%
an acid
0%
a base
0%
a stronger acid than when dissolved in $H_2O$
0%
a stronger base than when it is dissolved in $H_2O$

Explanation

$A_{(l)}+H_2O\overset {K_a}{\rightleftharpoons}A'(aq)+H_2O^+$

$B_{(l)}+H_2O\overset {K_b}{\rightleftharpoons}BH^+{aq}+OH^-$

$A_{(l)}+B_{(l)}+2H_2O\overset {K_aK_b}{\rightleftharpoons} A'+BH^++H_3O^+ H_3O^++OH^-\overset {K_w}{\rightleftharpoons}2H_2O$

$A_{(l)}+B_{(l)}\underset {K_w}{\overset {K_aK_b}{\rightleftharpoons}}A'+BH^+$

Now, suppose

$\dfrac {K_aK_b}{K_w} > K_a$

$\Rightarrow$

$K_b > K_w$ , which is true because B is a stronger base than

$H_2O$ and hence, our assumption is correct.

$\Rightarrow A(l)$ is a stronger acid in

$B(l)$ compared to an aqueous medium.

Hence, the correct options are

$\text{A}$ and

$\text{C}$

Calculate the

${OH}^{-}$ concentration and the

${H}_{3}{PO}_{4}$ concentration of a solution prepared by dissolving

$0.1$ mol of

${Na}_{3}{PO}_{4}$ in sufficient water to make

$1L$ of solution.

${K}_{1}=7.1\times {10}^{-3}$ ,

${K}_{2}=6.3\times {10}^{-8}$ ;

${K}_{3}=4.5\times {10}^{-13}$

0%
$[{OH}^{-}]=1.36\times {10}^{-2}M$ , $[{H}_{3}{PO}_{4}]=2.96\times {10}^{-18}M$
0%
$[{OH}^{-}]=3.73\times {10}^{-2}M$ , $[{H}_{3}{PO}_{4}]=5.93\times {10}^{-18}M$
0%
$[{OH}^{-}]=5.9\times {10}^{-2}M$ , $[{H}_{3}{PO}_{4}]=6.45\times {10}^{-18}M$
0%
$[{OH}^{-}]=6.45\times {10}^{-2}M$ , $[{H}_{3}{PO}_{4}]=5.9\times {10}^{-18}M$

Explanation

${Na}_{3}{PO}_{4}+{H}_{2}O\longrightarrow {Na}_{2}H{PO}_{4}+NaOH$

$K=\cfrac{{K}_{w}}{{K}_{3}}=0.0222$

${Na}_{2}H{PO}_{4}+{H}_{2}O\longrightarrow Na{H}_{2}{PO}_{4}+NaOH$

$K=\cfrac{{K}_{w}}{{K}_{2}}=1.58\times {10}^{-7}$

$Na{H}_{2}{PO}_{4}+{H}_{2}O\longrightarrow {H}_{3}{PO}_{4}+NaOH$

$K=\cfrac{{K}_{w}}{{K}_{3}}=1.4\times {10}^{-12}$
Since equilibrium constant of 2nd and 3rd reaction is very less,

$[{OH}^{-}]$ will mainly come from 1st reaction.

${Na}_{3}{PO}_{4}+{H}_{2}O\longrightarrow {Na}_{2}H{PO}_{4}+NaOH$

$0.1$

$0.1-x$

$x$

$\cfrac{{x}^{2}}{0.1-x}=0.0222$

$\Rightarrow$

$45{x}^{2}+x-0.1=0$

$x=3.73\times {10}^{-2}$

$[{OH}^{-}]=x=3.73\times {10}^{-2}M$

${Na}_{2}H{PO}_{4}+NaOH\longrightarrow Na{H}_{2}{PO}_{4}+NaOH$

$x$

$x-y$

$y$

$y+x$

$x-y \simeq x$

$y+x\simeq x$

$1.58\times {10}^{-7}=\cfrac{(x+y)}{(x-y)}y=y$

$Na{H}_{2}{PO}_{4}+{H}_{2}O\longrightarrow {H}_{3}{PO}_{4}+NaOH$

$y$

$y-z$

$z$

$z+x$

$y-z\simeq y$ ,

$z+x\simeq x$

$1.4\times {10}^{-12}=\cfrac{z(x+z)}{(y-z)}=\cfrac{z\times x}{y}=\cfrac{z\times 3.73\times {10}^{2}}{1.58\times {10}^{-7}}$

$z=5..93\times {10}^{-18}$

$[{H}_{3}{PO}_{4}]=5.93\times {10}^{-18}M$

A solution contains

$HCl$ ,

${Cl}_{2}HCCOOH$ and

${CH}_{3}COOH$ at concentrations

$0.09M$ in

$HCl$ ,

$0.09M$ in

${Cl}_{2}HCCOOH$ and

$0.1M$ in

${CH}_{3}COOH$ ,

$pH$ for the solution is

$1$ . Ionization constant of

${CH}_{3}COOH={10}^{-5}$ . What is the magnitude of

$K$ for dichloroacetic acid?

0%
${k}{{a}_{2}}=1.25\times {10}^{-2}$
0%
${k}{{a}_{2}}=1.25\times {10}^{-1}$
0%
${k}{{a}_{2}}=1.25\times {10}^{-4}$
0%
${k}{{a}_{2}}=1.25\times {10}^{-3}$

Explanation

Given,

$HCl\longrightarrow 0.09M$

${CH}_{3}COOH\longrightarrow {C}_{1}=0.1M$ ,

${\alpha}_{1}, {K}_{{a}_{1}}={10}^{-5}$

${Cl}_{2}CHCOOH\longrightarrow {C}_{2}=0.09M$ ,

${\alpha}_{2}$ ,

${K}_{{a}_{2}}=?$

$pH=1, [{H}^{+}]=0.1$

$0.1=0.09+{C}_{1}{{\alpha}_{1}}+{C}_{2}{{\alpha}_{2}}$

${CH}_{3}COOH\rightleftharpoons {CH}_{3}CO{O}^{-}+{H}^{+}$

${C}_{1}$

${C}_{1}-{C}_{1}{\alpha}_{1}$

${C}_{1}{\alpha}_{1}$

$0.1$

${Cl}_{2}HCOOH\rightleftharpoons {Cl}_{2}HCCO{O}^{-}+{H}^{+}$

${C}_{2}$

${C}_{2}-{C}_{2}{\alpha}_{2}$

${C}_{2}{\alpha}_{2}$

$0.1$

${K}_{{a}_{1}}=\cfrac { \left( { C }_{ 1 }{ \alpha }_{ 1 } \right) \left( 0.1 \right) }{ { C }_{ 1 }\left( 1-{ \alpha }_{ 1 } \right) } \simeq { \alpha }_{ 1 }\times 0.1={ 10 }^{ -5 }$

${\alpha}_{1}={10}^{-4}$
Putting this in equation

$(i)$ , we get

${10}^{-4}\times 10.1+0.09{\alpha}_{2}=0.01$

${\alpha}_{2}=0.111$

${K}_{{a}_{2}}=\cfrac { \left( { C }_{ 2 }{ \alpha }_{ 2 } \right) \left( 0.1 \right) }{ { C }_{ 2 }\left( 1-{ \alpha }_{ 2 } \right) }$

$=\cfrac { \left( 0.111 \right) \left( 0.1 \right) }{ 1-0.111 }$

$=1.248\times { 10 }^{ -2 }$

${60}^{o}C$ , pure water has

$[{H}_{3}{O}^{+}]={10}^{-6.7}mol/lit$ .What is the value of

${K}_{W}$ at

${60}^{o}C$ :-

0%
${10}^{-6}$
0%
${10}^{-12}$
0%
${10}^{-67}$
0%
${10}^{-13.4}$

Explanation

In pure water,

$[H_3O^{+}]= [OH^{-}]$

$K_w = [H_3O^+][OH^-]$

As given,

$[{H}_{3}{O}^{+}]={10}^{-6.7}mol/lit$

So,

$K_w = {10}^{-6.7}\times{10}^{-6.7} = {10}^{-13.4}$

$90^oC$ , pure water

$[H_3O^{\oplus}]$ as

$10^{-6} mol\, L^{-1}$ what is the value of

$K_w$ at

$90^oC$ ?

0%
$10^{-6}$
0%
$10^{-12}$
0%
$10^{-14}$
0%
$10^{-8}$

Explanation

As we know,

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

$90^oC$ ,

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

$K_w = 10^{-12}$

Hence,option B is correct.

The

$pH$ of a solution of sodium hydroxide isWhat will be its

$pH$ when this solution is diluted?

0%
Less than 9
0%
More than 9
0%
Equal to 9
0%
Equal to 10

Explanation

The

$pH$ of a solution of sodium hydroxide is

$9$ .

Its

$pH$ will be less than

$9$ when this solution is diluted.

On dilution, the

$pH$ decreases as hydroxide ion concentration decreases and hydrogen ion concentration increases.

Option A is correct.

The solubility of

$Hg{I}_{2}$ in water decreases in presence of

$KI$ .

State whether the given statement is true or false.

0%
True
0%
False

Explanation

$Hg{I}_{2}$ forms soluble complex with

$KI$

$2KI+Hg{I}_{2}\longrightarrow {K}_{2}Hg{I}_{4}$

So solubility of

$Hg{I}_{2}$ in water increases in presence of

$KI$ .

The colour of a salt depends upon its

0%
density
0%
composition
0%
surface temperature
0%
radius

A buffer solution can be prepared from a mixture of:

(i) Sodium acetate and acetic acid in water
(ii) Sodium acetate and hydrochloric acid in water
(iii) Ammonia and ammonium chloride in water
(iv) Ammonia and sodium hydroxide in water

0%
(i), (ii)
0%
(ii), (iii)
0%
(iii), (iv)
0%
(i), (iii)

Explanation

An acidic buffer solution consists of a weak acid (such as acetic acid) and its salt (such as sodium acetate) with a strong base (such as

$NaOH$ ).

A buffer solution can be prepared from a mixture of:
(i) Sodium acetate and acetic acid in water
(ii) Sodium acetate and hydrochloric acid in water

Hydrochloric acid will partly neutralize sodium acetate to form acetic acid.

Hence, the option

$A$ is the correct answer.

Which one of the following statements is not correct?

0%
The pH of $1.0 \times {10}^{-8} M\ HCl$ is less than $7$
0%
The ionic product of water at $25^0C$ is $1.0 \times {10}^{-14}\ {mol}^{2} {L}^{-2}$
0%
${Cl}^{-}$ is a Lewis acid
0%
Bronsted-Lowry theory cannot explain the acidic character of $Al{Cl}_{3}$

Explanation

$Cl^-$ is not a Lewis acid. It is a Lewis base. Lewis acid is a chemical species that reacts with a Lewis base to form a Lewis adduct. A Lewis base, then, is any species that donates a pair of electrons to a Lewis acid to form a Lewis adduct.

For example,

$OH^−$ and

$NH_3$ are Lewis bases, because they can donate a lone pair of electrons.

$\displaystyle { BaCl }_{ 2 }$ dissociates in water to give one

$\displaystyle { Ba }^{ 2+ }$ ion and two

$\displaystyle { Cl }^{ - }$ ions. If concentrated

$\displaystyle HCl$ is added to this solution :

0%
$\displaystyle \left[ { Ba }^{ 2+ } \right]$ increases
0%
$\displaystyle \left[ { Ba }^{ 2+ } \right]$ remains constant
0%
$\displaystyle \left[ { OH }^{ - } \right]$ increases
0%
The number of moles of undissociated $\displaystyle { BaCl }_{ 2 }$ increases
0%
$\displaystyle \left[ { H }^{ + } \right]$ decreases

Explanation

The common ion effect describes the changes that occur with the introduction of ions to a solution containing that same ion.
The role that the common ion effect in solutions is mostly visible in the decrease of solubility of solids. Through the addition of common ions, the solubility of a compound generally decreases due to a shift in equilibrium.

$BaCl_2\rightarrow Ba^{2+} + 2Cl^{-}$

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

$Cl^{-}$ is the common ion so, the

$Ba^{2+}$ and

$2Cl^{-}$ combines ( associate ) to give the undissociated

$BaCl_2$ .
Hence , there is increase in concentration of undissociated

$BaCl_2$ .

Statement I: At equilibrium, the concentration of reactants and products remains constant.
Statement II: At equilibrium, the rates of the forward and reverse reactions are equal.

0%
Statement I and Statement II are true and Statement II is correct explanation of Statement I
0%
Statement I and Statement II are true and Statement II is not correct explanation of Statement I
0%
Statement I is true and Statement II is false
0%
Statement I is false and Statement I is true

Which of the following salts will have maximum cooling effect when

$0.5$ mole of the salt is dissolved in same amount of water. Integral heat of solution at

$298\ K$ is given for each salt.

0%
$KNO_{3} (\Delta =35.4\ kJ\ mol^{-1})$
0%
$NaCl (\Delta =5.35\ kJ\ mol^{-1})$
0%
$HBr (\Delta =83.3\ kJ\ mol^{-1})$
0%
$KOH (\Delta =55.6\ kJ\ mol^{-1})$

Explanation

More the heat absorbed, more will be the cooling effect. Hence more the positive value of

$\Delta H$ more the cooling effect.In the given question KBr has

$\Delta H =83.3 kJmol^-$ which is highest among all the gases.

At 298 K

$0.01 M\ { NH }_{ 4 }OH$ solution is

$4.3\%$ ionised.The ionization constant of

${ NH }_{ 4 }OH$ is :

0%
$1.84\times10^{-5}$
0%
$2.05\times 10^{-5}$
0%
$1.62\times 10^{-4}$
0%
$1.84\times 10^{-4}$

Which of the following symbol represents acid dissociation constant?

0%
$\displaystyle pH$
0%
$\displaystyle pOH$
0%
$\displaystyle { K }_{ a }$
0%
$\displaystyle { K }_{ b }$
0%
$\displaystyle { K }_{ w }$

Explanation

An acid dissociation constant,

$K_a$ , (also known as acidity constant, or acid-ionization constant) is a quantitative measure of the strength of an acid in solution. It is the equilibrium constant for a chemical reaction known as dissociation in the context of acid–base reactions. In aqueous solution, the equilibrium of acid dissociation can be written symbolically as:

\mathrm {HA+H_{2}O\rightleftharpoons A^{-}+H_{3}O^{+}}

where

$HA$ is a generic acid that dissociates into

$A−$ , known as the conjugate base of the acid and a hydrogen ion which combines with a water molecule to make an hydronium ion. In the example shown in the figure,

$HA$ represents acetic acid, and

$A−$ represents the acetate ion, the conjugate base.

The chemical species

$HA, A^−$ and

$H_3O^+$ are said to be in equilibrium when their concentrations do not change with the passing of time. The dissociation constant is usually written as a quotient of the equilibrium concentrations (in mol/L), denoted by

$[HA], [A−]$ and

$[H_3O^{+}]$

K_{\mathrm {a} }=\mathrm {\frac {[A^{-}][H_{3}O^{+}]}{[HA][H_{2}O]}}

Which of the following will form a buffer solution of 1L solution?

0%
0.2 mol of $\displaystyle HCl$ and 0.1 mole of $\displaystyle { K }_{ 2 }{ SO }_{ 3 }$
0%
0.2 mol of $\displaystyle NaOH$ and 0.4 mol of $\displaystyle HF$
0%
0.1 mol of $\displaystyle HBr$ and 0.1 mol of $\displaystyle { Ba\left( OH \right) }_{ 2 }$
0%
0.4 mol of $\displaystyle HCl$ and 0.2 mol of $\displaystyle { NH }_{ 3 }$
0%
0.2 mol of $\displaystyle NaOH$ and 0.4 mol of $\displaystyle HCl$

Explanation

The following will form a buffer solution of

$1\ L$ solution.

$0.2$ mol of

$NaOH$ and

$0.4$ mol of

$HF$ :
It is an example of acid buffer solution. An acid buffer solution contains a weak acid and its salt with a strong base.

$0.2$ mol of

$NaOH$ will react with

$0.2$ mol of

$HF$ to form

$0.2$ mol of

$NaF$ .

$0.2$ mol of

$HF$ will remain.
Thus, the mixture contains

$0.2$ mol of

$HF$ and

$0.2$ mol of

$NaF$ .

Assertion: AgCl will not dissolve in a concentrated solution.
Reason: The chloride ions from NaCl suppress the solubility of AgCl.

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

From the give options, which of the following compound will make best buffer solution?

0%
${ H }_{ 2 }O,\ 1\ M\ NaOH,\ 1\ M\ { H }_{ 2 }S{ O }_{ 4 }$
0%
${ H }_{ 2 }O,\ \ 1\ M\ C{ H }_{ 3 }COOH,\ 1\ M\ { Na }^{ + }C{ H }_{ 3 }CO{ O }^{ - }$
0%
${ H }_{ 2 }O,\ 1\ M\ C{ H }_{ 3 }COOH,\ 6\ M\ { Na }^{ + }C{ H }_{ 3 }CO{ O }^{ - }$
0%
${ H }_{ 2 }O,\ 1\ M\ C{ H }_{ 3 }COOH,\ 1\ M\ NaOH$

Explanation

Since the acidic buffer is a mixture of weak acid plus its salt with a strong base. So

$B$ is correct.

And there should be base as limiting reagent

$\left( { conc }^{ n }\quad of\quad base<{ conc }^{ n }\quad of\quad acid \right)$

A student performed an experiment to determine the solubility of a salt a various temperatures. The data from the experiment can be seen below:

Trial	Temp ( $\displaystyle ^{ \circ }{ C }$ )	Solubility in 100 g water
1	20	44
2	30	58
3	40	67
4	50	62
5	60	84

Which trial seems to be in error?

0%
1
0%
2
0%
3
0%
4
0%
5

The

$K_{sp}$ for

$Mg(OH)_2$ in water is

$1.2 \times 10^{-11}$ . In an acidic solution, if the

$Mg^{2+}$ concentration is

$1.2 \times 10^{-5} mol/L$ , what is the pH at which

$Mg(OH)_2$ just begins to precipitate?

0%
3
0%
4
0%
5
0%
11
0%
12

Explanation

$\displaystyle K_{sp} = [Mg^{2+}] [OH^-]$

$\displaystyle K_{sp} = 1.2 \times 10^{-11}$

$\displaystyle [Mg^{2+}] = 1.2 \times 10^{-5}$

$\displaystyle 1.2 \times 10^{-11} = 1.2 \times 10^{-5} \times [OH^-] ^2$

$\displaystyle [OH^-] ^2 = \dfrac {1.2 \times 10^{-11}}{1.2 \times 10^{-5}}$

$=1.0 \times 10^{-6}$

$\displaystyle [OH^-] = \sqrt {1.0 \times 10^{-6}}$

$=1.0 \times 10^{-3}$

$\displaystyle pOH = -log [OH^-]$

$= -log 1.0 \times 10^{-3}$

$=3$

$\displaystyle pH = 14 - POH$

$= 14 - 3$

$= 11$

At 298K,

$\displaystyle { K }_{ W }$ of water is :

0%
$\displaystyle 1\times { 10 }^{ -14 }$
0%
$\displaystyle 1\times { 10 }^{ -17 }$
0%
$\displaystyle 1\times { 10 }^{ -1 }$
0%
$\displaystyle 1\times { 10 }^{ -7 }$

Explanation

Ionic product of water

$(K_w)$ is the product of molar concentrations of hydrogen ions and hydroxide ions.

$\displaystyle K_w = [H^+][OH^-] = \displaystyle 1\times { 10 }^{ -14 }$

Assertion: Water makes a good buffer.
Reason: A good buffer will resist changes in

$\displaystyle pH$ .

0%

Both Assertion and Reason are true and Reason is the correct explanation of Assertion
0%

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

Assertion is true but Reason is false
0%

Assertion is false but Reason is true
0%

Both Assertion and Reason are false

Explanation

A buffer solution can be defined as a solution which resists a change in its pH when such a change is caused by the addition of a small amount of acid or base.
Buffer solutions can be obtained:
1: by mixing a weak acid with its salt with a strong base, eg; Acetic Acid + Soidum Acetate, Boric acid + Borax, Phthalic acid + Potassium acid phthalate.
2: by mixing a weak base with its salt with a strong acid, e.g; (a)

$NH_4OH + NH_4Cl$ , (b) Glycine + Glycine hydrochloride.
3: by a solution of ampholyte. The ampholytes or amphoteric electrolytes are the substances which show properties of both an acid and a base. Proteins and amino acids are the examples of such electrolytes.
4: by a mixture of an acid salt and a normal salt of a polybasic acid, e.g.,

$Na_2HPO_4 + Na_3PO_4$ , or a salt of weak acid and a weak base, such as

$CH_3COONH_4$ .

The following reaction occurs in a beaker:

$\displaystyle { Ag }^{ + }\left( aq \right) +{ Cl }^{ - }\left( aq \right) \rightarrow AgCl\left( s \right)$ . If a solution of sodium chloride were added to this beaker,

0%
The solubility of the sodium chloride would decrease
0%
The reaction would shift to the left
0%
The concentration of silver ions in solution would increase
0%
The solubility of the silver chloride would decrease
0%
The equilibrium would not shift at all

Explanation

$Ag^+ + Cl^- \rightarrow AgCl$

$NaCl \rightarrow Na^+ + Cl^-$
As

$Cl^-$ is the common ion in both the solution ( i.e.

$NaCl$ and

$AgCl$ ) , hence

$AgCl$ being weaker electrolyte is precipitated or solubility of the silver chloride would decrease.
Common ion Effect : The common ion effect is responsible for the reduction in the solubility of an ionic precipitate when a soluble compound containing one of the ions of the precipitate is added to the solution in equilibrium with the precipitate. It states that if the concentration of any one of the ions is increased, then, according to Le Chatelier's principle, some of the ions in excess should be removed from solution, by combining with the oppositely charged ions. Some of the salt will be precipitated until the ion product is equal to the solubility product. In short, the common ion effect is the suppression of the degree of dissociation of a weak electrolyte containing a common ion

An acidic buffer solution has

$[HA]=1.0\,M$ and

$[NaA]=1.0M$ . To

$(10+x)mL$ of this buffer solution

$9\,mL$ of

$1.0\,M\,HCl$ is added so that

$pH$ changes by one unit. The value of x is

0%
$0.1$
0%
$10$
0%
$1.5$
0%
$1.0$

$MgC{l}_{2}$ and

$NaCl$ are both salts that are used to treat roads. Which salt will have the greatest effect on the freezing point and why?

0%
NaCl because i = 2
0%
MgCl2 because i = 3
0%
MgCl2 because i = 2
0%
NaCl because i = 3
0%
Both salts would have the same effect

Explanation

Addition of salt on roads is used to melt the ice, as this causes the lowering of the freezing point of the ice. Hence, freezing point of the ice due to the addition of the salt decreases the freezing point lesser than the

$0^oC$ .
Which can not be obtained easily , hence ice melts.
Now , as per formula

$\Delta T_f = i k_f m$
Here ,

$i =$ Vant Hoff's Factor.
which depends upon the number of dissociated ions, more the ions are dissociated from a single formula unit, more will be value of the

$i$ .
and

$i \propto \Delta T_f$
hence , this means higher the value of

$i$ , more will be lowering of the freezing point.
So here the (B)

$MgCl_2$ is used , which has

$i = 3$ .

Which compound is used for bleaching cloths in laundry?

0%
Bleaching powder
0%
Washing powder
0%
Baking powder
0%
Plaster of Paris

The solubility of silver chloride ___________ in the presence of sodium chloride because of __________.

0%
increases; common ion effect
0%
increases; aldol condensation
0%
decreases; common ion effect
0%
decreases; aldol condensation

Explanation

The solubility of silver chloride decreases in the presence of sodium chloride because of common ion effect.

$\displaystyle AgCl \rightleftharpoons Ag^+ + Cl^-$

$\displaystyle NaCl \rightarrow Na^+ + Cl^-$

Sodium chloride is a strong electrolyte and completely dissociates to provide chloride ions that are common ions. The chloride ions shift the equilibrium for dissociation of

$AgCl$ towards left.

Uses of bleaching powder is/are:

0%
It is used as a disinfectant in sterilization of water
0%
It is used to bleach straw, ivory
0%
It is used as an oxidising agent in industry
0%
It is used for making wool unshrinkable

Assertion: When a non-volatile solute is added to pure water, the vapor pressure of the water will decrease.
Reason: All solutes dissociate into positive and negative ions.

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

Concentrated strong acid is added to a solid mixture of 0.015 mole samples of

$Fe(OH)_2$ and

$Ca(OH)_2$ placed in one litre of water. At what value of pH will the dissolution of each hydroxide be complete? (Assume negligible volume change)

$K_{sp}[Fe(OH)_2]=7.9\times 10^{-15}$ and

$K_{sp}[Cu(OH)_2]=1.6\times 10^{-19}$

0%
6.67
0%
7.86
0%
8.86
0%
7.26

Explanation

Let,

$\displaystyle [Fe^{2+}] = x \: M$ and

$\displaystyle [OH^-] = y \: M$ .

Therefore,

$\displaystyle [Ca^{2+} ] = 0.015-x \: M$ .

$\displaystyle K_{sp, Fe(OH)_2} = [Fe^{2+}] [OH^-]$

$\displaystyle 7.9 \times 10^{-15} = x y^2$ .....(1)

$\displaystyle K_{sp, Ca(OH)_2} = [Ca^{2+}] [OH^-]$

$\displaystyle 1.6 \times 10^{-19} = (0.015-x) y^2$ .....(2)

Dividing equation (2) by eqation (1) ,

$\displaystyle \dfrac { 1.6 \times 10^{-19} }{ 7.9 \times 10^{-15}} = \dfrac { (0.015-x) y^2}{ x y^2}$

$\displaystyle 2.025 \times 10^{-5} = \dfrac {0.015-x}{x}$

$\displaystyle 2.025 \times 10^{-5}x = 0.015-x$

$\displaystyle x = 0.014999$

Substitute value of x in equation (1),

$\displaystyle 7.9 \times 10^{-15} = x y^2$

$\displaystyle 7.9 \times 10^{-15} = 0.014999 \times y^2$

$\displaystyle y = 7.26 \times 10^{-7} \: M$

$\displaystyle pOH = -log [OH^-] = -log( 7.26 \times 10^{-7})= 6.14$

$\displaystyle pH = 14 - pOH = 14 - 6.14 = 7.86$

What is the net effect of the common ion?

0%
It reduces the solubility of the solute in the solution.
0%
It reduces the increases of the solute in the solution.
0%
It causes more product to be made.
0%
It causes more ions to form.
0%
There is no net effect overall.

Explanation

The net effect of the common ion is that it reduces the solubility of the solute in the solution.
The common ion effect is observed when the addition of an ion common to two solutes causes precipitation or reduces ionization.
Thus, if to a solution of weak electrolyte, a solution of strong electrolyte is added which furnishes an ion common to that furnished by the weak electrolyte, the ionization of the weak electrolyte is suppressed.
For example, ammonium hydroxide is a weak electrolyte and ionizes to a small extent to give ammonium ions and hydroxide ions.

$\displaystyle NH_4OH \rightleftharpoons NH_4^+ + OH^-$
A strong electrolyte NaOH is added which furnishes hydroxide ions that are common ions. This suppresses the equilibrium for the dissociation of ammonium hydroxide.

$\displaystyle NaOH \rightarrow Na^+ + OH^-$

$pK_{b}$ of

$NH_{3}$ is

$4.74$ and

$pK_{b}$ of

$A^{-}, B^{-}$ and

$C^{-}$ are

$4, 5$ and

$6$ respectively. Aqueous solution of

$0.01\ M$ has

$pH$ in the increasing order.

0%
$NH_{4}A < NH_{4}B < NH_{4}C$
0%
$NH_{4}C < NH_{4}B < NH_{4}A$
0%
$NH_{4}C < NH_{4}A < NH_{4}B$
0%
All have equal $pH$

Explanation

$\because Salts\ NH_{4}A, NH_{4}B, NH_{4}C$ are of weak acid and weak base.

$\therefore pH = 7 + \dfrac {pK_{a}}{2} - \dfrac {pK_{b}}{2}$

Thus, greater the value of

$pK_{a}$ of

$HA$ , greater the

$pH$ .

$pK_{b} (A^{-}, B^{-}, C^{-}) | \underset {4}{A^{-}} < \underset {5}{B^{-}} < \underset {6}{C^{-}}$

$\therefore pK_{a} (HA, HB, HC)|\underset {8}{HC} < \underset {9}{HB} < \underset {10}{HA}$

Thus, order of

$pH$ values is

$(NH_{4}C < NH_{4}B < NH_{4}A)$

A litre of solution is saturated with

$AgCl$ . To this solution if

$1.0\times { 10 }^{ -4 }$ mole of solid

$NaCl$ is added, what will be the

$\left[ { Ag }^{ + } \right]$ assuming no volume change?

0%
More
0%
Less
0%
Equal
0%
Zero

Explanation

A litre of solution is saturated with

$AgCl$ .

To this solution if

$1.0 \times 10^{-4}$ mole of solid

$NaCl$ is added,

$\displaystyle [Ag^+]$ will be less.

$\displaystyle AgCl \rightleftharpoons Ag^+ + Cl^-$

$\displaystyle NaCl \rightarrow Na^+ + Cl^-$

Thus, the dissociation of

$NaCl$ provides chloride ions that are common to

$AgCl$ . Due to common ion effect, the dissociation of

$AgCl$ is suppressed and

$\displaystyle [Ag^+]$ decreases.

Hence, option B is correct.

Which mixture forms a buffer when dissolved in 1 L of water?

0%
0.2mol NaOH + 0.2 mol HBr
0%
0.2mol NaCl + 0.3mol HCl
0%
0.4 mol $HNO_2$ + 0.2mol NaOH
0%
0.5 mol $NH_3$ + 0.5mol HCl

Equilibrium constant for the reaction,

$NH_{4}OH + H^{+}\rightleftharpoons NH_{4}^{+} + H_{2}O$ is

$1.8\times 10^{9}$ . Hence, equilibrium constant for

$NH_{3}(aq) + H_{2}O \rightleftharpoons NH_{4}^{+} + OH^{-}$ is:

0%
$1.8\times 10^{-5}$
0%
$1.8\times 10^{5}$
0%
$1.8\times 10^{-9}$
0%
$5.59\times 10^{-10}$

Explanation

$NH_{4}OH + H^{+}\rightleftharpoons NH_{4}^{+} + H_{2}O$

$\therefore \dfrac {[NH_{4}^{+}][H_{2}O]}{[NH_{4}OH][H^{+}]} = 1.8\times 10^{9}$

Again,

$NH_{3} + H_{2}O\rightarrow NH_{4}OH \rightleftharpoons NH_{4}^{+} + OH^{-}$

$\therefore K' =\dfrac {[NH_{4}^{+}][OH^{-}]}{[NH_{4}OH]}$

Now

$\dfrac {K'}{K} = \dfrac {[NH_{4}^{+}][OH^{-}]}{[NH_{4}OH]}\times \dfrac {[NH_{4}OH][H^{+}]}{[NH_{4}^{+}][H_{2}O]}$

$= [OH^{-}][H^{+}] (\because H_{2}O$ is in excess)

$= K_{w} = 1\times 10^{-14}$

$\therefore K' = K\times 1\times 10^{-14}$

$= 1.8\times 10^{9} \times 10^{-14} = 1.8\times 10^{-5}$

The solubility of

$AgI$ in

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

0%
$Agl$ forms complex with $Nal$
0%
of common ion effect
0%
solubility product of $Agl$ is less
0%
the temperature of the solution decreases

Explanation

The solubility of sparingly soluble salt is reduced in a solution that contains an ion in common with the salt. Since

$AgI$ and

$NaI$ have common ion

$I^-$ the solubility of

$AgI$ in

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

1.0 g of a weak monobasic acid (

$HA$ ), when dissolved in 150 mL water, lowers the freezing point by

$0.186^o C$ . Also 1.0 g of the same acid required 125 mL of a 0.10N Nate solution for complete neutralisation. Determine dissociation constant

$(k_a)$ of the weak acid.

$k_f$ of water is 1.86 K kg

$mol^{-1}$ .

0%
$4.16 \times 10^{-3}$
0%
$2.38 \times 10^{-6}$
0%
$6.43 \times 10^{-3}$
0%
$3.78 \times 10^{-6}$

Explanation

From equation reaction,

$meq.$ of acid

$= 125 \times 0.1 =12.5$

$Eq. wt.$ of acid

$=\dfrac{1000}{12.5}=80$ = mol. wt.
Molarity

$=\dfrac{1}{8} \times \dfrac{100}{15}=0.083$

$-\Delta T_t=ik_tm=(1+\alpha )\times 1.86 \times 0.083$

$\alpha =0.2$

$K_a=\dfrac{C\alpha ^2}{1-\alpha }=\dfrac{0.083\times (0.2)^2}{0.8}=4.16 \times 10^{-3}$

Hence, the correct option is

$\text{A}$

When common salt is added to a saturated solution of soap, soap is precipitated. This is based on the principle of

0%
Common ion effect
0%
Principle of solubility product
0%
Adsorption from solution
0%
Peptisation

Which among the following groups is acidic in nature?

0%
Grapes and distilled water
0%
Soap solution and vinegar
0%
Milk of magnesia and milk
0%
Ant sting and vinegar

Explanation

A. Grapes contain tartaric acid. Distilled water is neutral.

B. Soap solution is basic in nature. Vinegar contains acetic acid (

${ CH }_{ 3 }COOH$ )

C. Milk of magnesia (

${ Mg }{ (OH) }_{ 2 }$ ) is basic in nature. Milk has lactic acid.

D. Ant sting contains formic acid and vinegar contains acetic acid (

${ CH }_{ 3 }COOH$ )

Hence, D is the answer.

During electrolysis of molten NaCl, some water was added. What will happen?

0%
Electrolysis will stop
0%
Hydrogen will be evolved
0%
Some amount of caustic soda will be formed
0%
A fire is likely

Explanation

$Na +H_2O \rightarrow \underset{\text{(caustic soda)}}{NaOH} + \dfrac{1}{2} H_2+ Q$
A fire is likely to take place due to vigorous reaction of sodium with water.

For the following equilibrium,

$N_2O_4\rightleftharpoons 2NO_2$ in gaseous phase,

$NO_2$ is 50% of the total volume when equilibrium is set up. Hence, percent of dissociation of

$N_2O_4$ is:

0%
50%
0%
25%
0%
66.66%
0%
33.33%

Explanation

$\displaystyle N_2O_4\rightleftharpoons 2NO_2$

$\displaystyle NO_2$ is 50% of the total volume when equilibrium is set up.

Thus, the volume fraction (at equilibrium) of

$\displaystyle NO_2 = \dfrac {50}{100} = 0.5$

Volume fraction (at equilibrium) is proportional to mole fraction. The mole fraction (at equilibrium) of

$\displaystyle NO_2 = 0.5$

Let initially n moles of

$\displaystyle N_2O_4$ are present and a is the degree of dissociation.

The equilibrium number of moles of

$\displaystyle N_2O_4 = n (1-\alpha)$

The equilibrium number of moles of

$\displaystyle NO_2 = 2n\alpha$

The mole fraction of

$\displaystyle NO_2 =\dfrac {2n\alpha}{n(1-\alpha) + 2n\alpha}=\dfrac {2n\alpha}{n(1+\alpha)}$ . But it is equal to 0.5

$\displaystyle \dfrac {2n\alpha}{n(1+\alpha)}=0.5$

$\displaystyle 4n\alpha = n+n\alpha$

$\displaystyle 3n\alpha = n$

$\displaystyle \alpha = \dfrac {1}{3}$

The percent dissociation

$\displaystyle =100\alpha = \dfrac {100}{3}=33.33$ %.

Thus, the option (D) is the answer.

Very strong acids, such as

$HNO_3$ and

$HCl$ , appear to be equally strong in water. This "leveling effect" of water is because

0%
$OH^-$ is a stronger base than the conjugate bases of $HNO_3$ and $HCl$
0%
$H_3O^+$ is a stronger acid than $HNO_3$ and $HCl$
0%
$H_2O$ is a stronger base than the conjugate bases of $HNO_3$ and $HCl$
0%
$H_2O$ is a weaker base than the conjugate bases of $HNO_3$ and $HCl$

Explanation

Very strong acids, such as

$\displaystyle HNO_3$ and

$\displaystyle HCl$ , appear to be equally strong in the water. This "leveling effect" of water is because

$\displaystyle H_2O$ is a stronger base than the conjugate bases of

$\displaystyle HNO_3$ and

$\displaystyle HCl$ .

Hence, option (C) is the correct answer.

Note: When a strong acid is dissolved in water, it reacts with water to form hydronium ion

$\displaystyle (H_3O^+)$ .

$\displaystyle HCl + H_2O \rightarrow H_3O^+ + Cl^-$

Which of the following pairs will show common ion effect?

0%
Barium chloride + barium sulphate
0%
Silver cyanide + potassium nitrite
0%
Ammonium hydroxide + ammonium chloride
0%
Sodium chloride + hydrogen chloride

Explanation

The suppression of the dissociation of weak acid/base on the addition of its own ion is called the common ion effect.

Ammonium hydroxide is a weak base and it can be dissociated as,

$N{H_4}OH + HCl \rightleftarrows N{H_4}^ + + O{H^ - }$

Ammonium chloride is a salt and it is obtained from a neutralization of weak base and strong acid
( $N{H_4}OH + HCl$ ) . Ammonium chloride can be dissociated as,

$N{H_4}Cl \rightleftarrows N{H_4}^ + + C{l^ - }$

On $N{H_4}OH$ and $N{H_4}Cl$ , $N{H_4}^ +$ $N{H_4}^ +$ (Ammonium ion) is common and thus, it suppresses the dissociation of ammonium hydroxide as per Le Chatelier's principle.

Thus, ammonium hydroxide and ammonium chloride show a common ion effect.

The hydrated salt

$Na_{2}SO_{4}, 10H_{2}O$ undergoes

$X\%$ loss in weight on heating and becomes anhydrous. The value of

$X$ will be:

0%
$10$
0%
$45$
0%
$56$
0%
$70$

Explanation

Molecular mass of

$Na_2SO_410H_2O=2\times23+32+16\times14+1\times20=322$

On heating the

$Na_2SO_410H_2O$ it becomes anhydrous that means 10

$H_2O$ molecules separated

Molecular mass of

$10H_2O=20\times1+10\times16=180$

% loss

$=\dfrac{180}{322}\times100=55.9$ %

$=56$ %

(A) pH of

$10^{-7}$ M NaOH solution exists between 7 to 7.3 at

$25^o C$ .
(R) Due to common ion effect ionization of water is suppressed.

0%
Both (R) and (A) are true and reason is the correct explanation of assertion
0%
Both (R) and (A) are true but reason is not correct explanation of assertion
0%
Assertion (A) is true but reason (R) is false
0%
Assertion (A) and reason (R) both are false
0%
Assertion (A) is false but reason (R) is true

Explanation

$pH$ of

${ 10 }^{ -7 }$

$M$

$NaOH$ will be slightly greater than

$7$ . It is between

$7$ to

$7.3$ at

${ 25 }^{ 0 }C$ due to presence of very small quantity of excess

${ OH }^{ - }$ .

R) Due to common ion effect ionisation of water is suppressed.

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

So in presence of common ion

${ H }^{ + }$ or

${ OH }^{ - }$ the ionisation of

${ H }_{ 2 }O$ will be decreased.

So both

$(R)$ and

$(A)$ are true and reason is the correct explanation of assertion.

Which of the following mixtures will be buffer?

0%
${ CH }_{ 3 }COOH+{ CH }_{ 3 }COO{ NH }_{ 4 }$
0%
$HCl+NaCl$
0%
Borax+boric acid
0%
${ CH }_{ 3 }COOH+{ CH }_{ 3 }COONa$

Explanation

A buffer solution is one which resists changes in pH when small quantities of an acid or an alkali are added to it.

An acidic buffer solution is simply one which has a pH less than 7. Acidic buffer solutions are commonly made from a weak acid and one of its salts - often a sodium salt.EX: $CH_3COOH+CH_3COONa$

An alkaline buffer solution has a pH greater than 7. Alkaline buffer solutions are commonly made from a weak base and one of its salts.EX: Borax and boric acid.

Hence option C, D are correct.

(A) If water is heated to 350 K, then pOH will increase to
(R)

$K_w$ increases with increase in temperature.

0%
Both (R) and (A) are true and reason is the. correct explanation of assertion
0%
Both (R) and (A) are true but reason is not correct explanation of assertion
0%
Assertion (A) is true but reason (R) is false
0%
Assertion (A) and reason (R) both are false
0%
Assertion (A) is false but reason (R) is true

Explanation

If water is heated to 350 K, then pOH will increase to 8.

$K_w$ increases with increase in temperature. As

$K_w$ increases pOH increases, pH decreases.

So both (R) and (A) are true and reason is the. correct explanation of assertion.

Hence option A is correct.

An excess of

$Ag_2CrO_4(s)$ is added to a

$5\times 10^{-3}M$

$K_2CrO_4$ solution. The concentration of

$Ag^+$ in the solution is closest to.[Solubility product for

$Ag_2CrO_4=1.1\times 10^{-12}$ ]

0%
$2.2\times 10^{-10}$ M
0%
$1.5\times 10^{-5}$ M
0%
$1.0\times 10^{-6}$ M
0%
$5.0\times 10^{-3}$ M

Explanation

Since,

$K_2CrO_4$ is a strong electrolyte so the concentration of chromate ion will be almost equal to the concentration of

$K_2CrO_4$ .

$Ag_2CrO_4\leftrightharpoons 2Ag^+ + CrO_4^{2-}$

$S$

$2S$

$5\times 10^{-3}\ M$

Now,

$K_{sp}=[Ag^+]^2[CrO_4^{2-}]$

$1.1\times 10^{-12}=[Ag^+]^2[5\times 10^{-3}]$

$[Ag^+]=1.5\times 10^{-5}$

Which of the following would dissolve

$Pb(OH)_{2}$ more than pure water?

0%
Buffer solution having $pH = 6$
0%
$0.01\ M\ PbCl_{2}$ solution
0%
$0.01\ M\ NH_{4}OH$ solution
0%
All of the above

The solubility product constant

$Ksp$ of

$Mg(OH)_{2}$ is

$9.0\times 10^{-12}$ . If a solution is

$0.010\ M$ with respect to

$Mg^{2+}$ ion. What is the maximum hydroxide ion concentration which could be present without causing the precipitation of

$Mg(OH)_{2}$ ?

0%
$1.5\times 10^{-7}M$
0%
$3.0\times 10^{-7}M$
0%
$1.5\times 10^{-5}M$
0%
$3.0\times 10^{-5}M$

Explanation

$Ksp(Mg(OH)_2)=9.0\times 10^{-12}$

$(Mg(OH)_2 \leftrightharpoons Mg^{2+}+2[OH]^-$

$Ksp=[Mg^{2+}][OH^-]^2$

$9\times 10^{-12}=(10^{-12})(OH^-)^2$

$[OH^-]^2=3^2\times (10^{-5})^2$

$[OH^-]=3.0\times 10^{-5}M$

Maximum Hydroxide-ion concentration.

CBSE Questions for Class 11 Medical Chemistry Equilibrium Quiz 12 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Equilibrium Quiz 12 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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