MCQ Questions for Class 11 Medical Chemistry Equilibrium Quiz 10

Calculate ionisation constant for pyridinium chloride,
Given that

$H^+$ ion concent ration is

$3.6 \times 10^{-4} M$ and its concentration is

$0.02 M$ .

0%
$6.4 \times 10^2$
0%
$6.4 \times 10^{-6}$
0%
$6.4 \times 10^{-8}$
0%
$12 \times 10^{-8}$

$pH$ of water is

$7.0$ at

$25^{\circ}C$ . If water is heated to

$70^{\circ}C$ , the

0%
$pH$ will decrease and the sample becomes acidic.
0%
$pH$ will increase but the sample will remain neutral.
0%
$pH$ will remain constant as $7$ .
0%
$pH$ will decrease but the sample will remain neutral.

Explanation

We know that at

$25^{\circ}C$ dissociation constant

$K_w$ of water is,

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

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

So,

$[H^+]=[OH^-]=10^{-7}M$

Thus

$pH=7$ for

$H_2O$ at

$25^{\circ}C$

Dissociation of water is an endothermic process. With an increase in temperature dissociation constant of water increases. So, the dissociation constant is more i.e.,

$K_w>10^{-14}.$

Therefore,

$[H^+]>10^{-7}M$ and

$pH$ is less than 7 but the solution will remain neutral because nothing is added to water to change the concentration of

$H^+$ or

$OH^-$ ion.

Hence, the correct option is (D).

In comparison of ferrous salts, ferric salts are:

0%
more stable
0%
less stable
0%
equal stable
0%
none of these

Explanation

Ferric salts are more stable than Ferrous salts.

$Fe^{3+}$ ion are more stable compared to

$Fe^{2+}$ due to its electronic configuration. The electronic configuration of

$Fe^{3+}$ is [Ar]

$3d^{5}4s^{0}$ i.e half-filled which makes it more stable.

Hence, the correct option is (A).

Which of the following statements is not true for Mohr's salt?

0%
It decolourises $KMnO_4$ solution
0%
It is a double salt
0%
Oxidation state of iron is $+3$
0%
It is a primary standard

Explanation

Mohr's salt is a complex with the formula

$(NH_{4})_{2}Fe(SO_{4})_{2}.6H_{2}O$ .

It acts as a reducing agent which reduces Manganese (Mn) present in

$KMnO_{4}$ from +7 state to +2 state which results in decolourisation. It is a primary standard and also a double salt as it contains two different cations

$Fe^{2+}$ and

$NH^{+}_{4}$ . In this complex, iron is present in +1 state.

Hence, the correct option is (C).

$pH$ of

$0.01 M - (NH_{4})_{2}SO_{4})$ and

$0.02 M - NH_{4}OH$ buffer (

$pK_{a}$ of

$NH_{4}^{+} = 9.26$ ) is

0%
$9.26 + log 2$
0%
$9.26 - log 2$
0%
$4.74 +log 2$
0%
$9.26$

The conjugate base of sulphuric acid is _________

0%
Sodium hydroxide
0%
Hydrochloric acid
0%
Bisulphate ion
0%
Barium hydroxide

Explanation

Conjugate base of

$H_2SO_4$ is

$HSO_4^-$ .

In which of the following solutions, ions are present

0%
Sucrose in water
0%
Sulphur in $CS_{2}$
0%
Caesium nitrate in water
0%
Ethanol in water

Explanation

being an ionic salt Caesium nitrate

$CsNO_3$ in water it contain ions.

$CsNO_3\rightarrow Cs^-+NO^-_3$

Pure

$NaCl$ is prepared by saturating a cold saturated solution of common salt in water with

$HCl$ gas. The principle used is

0%
Le Chatelier principle
0%
Displacement law
0%
Common ion effect
0%
Fractional distillation

Explanation

Pure NaCl is prepared by saturating a cold saturated solution of common salt in water with HCl gas and is based on the

$\text{common ion effect.}$

HCl is a strong electrolyte and provides an ion

$(Cl^-)$ that is common to that provided by the weak electrolyte. Thus, the ionization of weak electrolytes is suppressed.

Sodium chloride is purified by passing hydrogen chloride gas in an impure solution of sodium chloride. It is base on

0%
Buffer action
0%
Common ion effect
0%
Association of salt
0%
Hydrolysis of salt

Explanation

$\text{Purification of NaCl by the passage of HCl through brine is based on the common ion effect.}$

$\text{HCl is a strong electrolyte and provides an ion }(Cl^−)\text{ that is common to that provided by the weak electrolyte.}$

$\text{Thus, the ionization of weak electrolytes is suppressed.}$

The degree of dissociation in a weak electrolyte increases________

0%
on increasing dilution
0%
on increasing pressure
0%
on decreasing dilution
0%
none of these

To obtain a buffer which be suitable for maintaining a

$pH$ of about

$4-5$ , we need to have in solution a mixture of

0%
A strong base + its salt with a weak acid
0%
A weak base + its with a strong acid
0%
A strong acid+ its salt with a weak base
0%
A weak acid+its salt with a strong base

Explanation

the pH of buffer solution is about 4−5 it means we have to prepare acidic buffer as Acid buffer has acidic pH and is prepared by

$\text{mixing a weak acid and it's salt with a strong base.}$ For example An aqueous solution of an equal concentration of acetic acid and sodium acetate has a pH of 4.7

Which of the following is a buffer?

0%
$NaOH+CH_{3}COONa$
0%
$NaOH+Na_{2}SO_{4}$
0%
$K_{2}SO_{4}+H_{2}SO_{4}$
0%
$NH_{4}OH+NH_{4}Cl$

Amongst the following the buffer solution is______

0%
$NH_{4}Cl+NH_{4}OH$ solution
0%
$NH_{4}Cl+NaOH$ solution
0%
$NH_{4}OH+HCl$ solution
0%
$NaOH+HCl$ solution

Explanation

$NH_{4}Cl$ and

$NH_{4}OH$ is a buffer solution (weak base and salt of strong acid).

$25^{o}C$ the

$pH$ value of a solution is

$6$ . The solution is

0%
Basic
0%
Acidic
0%
Neutral
0%
Both (b) and (c)

A buffer solution is a mixture of

0%
Strong acid and strong base
0%
Weak acid and weak base
0%
Weak acid and conjugate acid
0%
Weak acid and conjugate base

Explanation

A buffer solution (more precisely, pH buffer or hydrogen ion buffer) is an aqueous solution consisting of a

$\text{mixture of}$

$\text{a weak acid and its conjugate base, or vice versa.}$ Buffer solutions are used as a means of keeping pH at a nearly constant value in a wide variety of chemical applications.

Le-chatelier principle is applicable

0%
Both for physical and chemical equilibrium
0%
Only for chemical equilibrium
0%
Only for physical equilibrium
0%
Neither for (b) nor for (c)

The ionisation constant of phenol is higher than that of ethanol because___________

0%
Phenoxide ion is bulkier than ethanoxide
0%
Phenoxide ion is stronger base than ethanoxide
0%
Phenoxide ion is stabilised through delocalisation
0%
Phenoxide ion is less stable than ethoxide

A solution that resists changes in pH upon the addition of a small amount of strong acid or strong base known as_____

0%
A colloid
0%
A crystalloid
0%
A buffer
0%
A indicator

Explanation

Because buffer solution have a constant

$pH$ .

Which one is buffer solution

0%
$[PO_{4}^{-}][HPO^{-}_{4}]$
0%
$[PO_{3}^{3-}][H_{2}PO_{4}^{-}]$
0%
$[HPO_{4}^{-}][H_{2}PO_{4}^{-}]$
0%
All of these

Explanation

[Normal salt+acidic salt] is a buffer solution.

$\underset {normal\ salt}{[PO_{3}^{3-}]} \underset {acidic\ salt}{[H_{2}PO_{4}^{-}]}$

Option B is correct.

Which one of the following is not a buffer solution?

0%
$0.8\ M\ H_2S +0.8\ M\ KHS$
0%
$2\ M\ C_6H_5 NH_2 +2 M\ C_6H_5\ NH_3^+ Br$
0%
$3\ M\ H_2CO_3+ 3\ M\ KHCO_3$
0%
$0.05\ M\ KClO_4 + 0.05\ M\ HClO_4$

Explanation

Buffer solutions can be obtained by mixing a weak acid with its salt formed with a strong base or by mixing a weak base with its salt formed with a strong acid.

Because

$HClO_4$ is a strong acid. While buffer is a mixture of weak acid and their salt.

Which of the following will produce a buffer solution when mixed in equal volumes?

0%
$0.1 mol\ dm^{-3}$ $NH_{4}OH$ and $0.1 mol \ dm^{-3}$ $HCl$ .
0%
$0.05 mol\ dm^{-3}$ $NH_{4}OH$ and $0.1 mol\ dm^{-3}$ $HCl$ .
0%
$0.1 mol\ dm^{-3}$ $NH_{4}OH$ and $0.05 mol\ dm^{-3}$ $HCl$ .
0%
$0.1 mol\ dm^{-3}$ $CH_{4}COONa$ and $0.1 mol\ dm^{-3}$ $NaOH$ .

Explanation

$0.1 mol\ dm^{-3}$

$NH_{4}OH$ and

$0.05 mol\ dm^{-3}$

$HCl$ .

As a mixture of ammonium chloride and ammonium hydroxide acts as a buffer around pH

$9.25$ .

The main salt soluble in sea water is

0%
$MgCl_2$
0%
$NaCl$
0%
$MgSO_4$
0%
$CaSO_4$

State true or false
Salts are neutral.

0%
True
0%
False

Why curd and other sour substances should not be stored in brass and copper vessels?

0%
Curd and sour substances contain bases
0%
Curd and sour substances contain acids
0%
Both A and B
0%
None of A or B

Explanation

Curd and sour substances contain acids. These acids react with the metals in brass and copper vessels to produce salt and hydrogen gas. This will corrode the metal and result in the spoilage of food.

Therefore, option

$B$ is correct.

If first dissociation of

$X(OH)_3$ is

$100\mbox{%}$ whereas second dissociation is

$50\mbox{%}$ and third dissociation is negligible then the

$pH$ of

$4\times10^{-3}\space M\space X(OH)_3$ is :

0%
$11.78$
0%
$10.78$
0%
$2.5$
0%
$2.22$

Explanation

For polyprotic base:

$[OH^-] = C\alpha_1+C\alpha_2 = C(\alpha_1+\alpha_2)=4\times10^{-3} \times 1.5 = 6\times 10^{-3}$

So,

$[H^+] = \dfrac{10^{-14}}{6\times 10^{-3}}={1.67\times 10^{-12}}$

$pH = -log({1.67\times 10^{-12}})=11.78$

Solubility product of

$Al(OH)_{3}$ and

$Zn(OH)_{2}$ are

$8.5\times 10^{-23}$ and

$1.8\times 10^{-14}$ respectively. If both

$Al^{3+}$ and

$Zn^{2+}$ ions are present in a solution, which one will precipitate first on addition of

$NH_{4}OH$ ?

0%
$Al(OH)_{3}$
0%
$Zn(OH)_{2}$
0%
$Mg (OH)_{ 2}$
0%
$Ca (OH)_{ 2}$

Explanation

The value of solubility product for

$Al(OH)_{3}$ is

$8.5\times 10^{-23}$ .

It is much lower than the value of the solubility product for

$Zn(OH)_{2}$ which is

$1.8\times 10^{-14}$ .

Hence, when both

$Al^{3+}$ and

$Zn^{2+}$ ions are present in the solution and as

$NH_{4}OH$ is added,

$Al(OH)_{3}$ will precipitate first as the solubility product of

$Al(OH)_3$ is quite low.

Hence, its ionic product will exceed its solubility product at a very low concentration of

$OH^-$ ions.

Option A is correct.

The dissociation of water at

$25^{\circ}$ C is

$1.9 \times 10^{-7}\%$ and the density of water is

$1.0$ g/cc. The ionisation constant of water is :

0%
$3.42\times 10^{-6}$
0%
$3.42\times 10^{-8}$
0%
$1.00\times 10^{-14}$
0%
$2.00 \times 10^{-16}$

Explanation

The degree of dissociation

$(\alpha)$ for water is

$\dfrac {1.9 \times 10^{-7}} {100}=1.9 \times 10^{-9}$ .

$1$ L of water (molecular mass

$= 18$ g/mol) weighs

$1000$ g.

Hence, concentration of water

$=\dfrac {1000}{18}$ M.

The expression for the ionization constant of water is

$K=\dfrac{[H^+][OH^-]}{[H_2O]}=C (\alpha)^2$ .

Substituting values in the above expression, we get:

$K=\dfrac {1000}{18} \times (1.9 \times 10^{-9})^2=2.0 \times 10^{-16}$ .

Which of the following statements is correct?

0%
The $pH$ of $1.0 \times 10^{-8}M$ solution of $HCl$ is $8$
0%
The conjugate base of $H_{2}PO_{4}^{-2}$ is $H PO_{4}^{-2}$
0%
Autoprotolysis constant of water increases with temperature
0%
When a solution of weak monoprotic acid is titrated against a strong base, at half neutralization point $pH = \frac{1}{2}p{Ka}$

Explanation

$HCl$ in water always creates an acidic solution. A solution with pH 8 is basic but the pH of the solution with

$10^{-8}$ molarity

$HCl$ must be less than 7.

Species formed after protonation of is called a conjugate base. The conjugate base of

$H_2PO_4^{-2}$ is

$HPO_4^{-3}$ .

Autoprotolysis is the dissociation of water which increases with increase in temperature.

Half neutralization point occurs when half of the acid moles are consumed by strong base. Therefore,

$pH = 1/2 \left(pK_a - log (c/2) \right)$ .

Total number of moles for the reaction

$2HI$

$\rightleftharpoons$

$H_{2}+I_{2}$ , if

$\alpha$ is degree of dissociation is :

0%
$2$
0%
$2-\alpha$
0%
$1$
0%
$1-\alpha$

Explanation

	$HI$	$\rightleftharpoons$ $H_2$ +	$I_2$
Initial moles	2	0	0
Change	$-\alpha$	$\dfrac {\alpha} {2}$	$\dfrac {\alpha} {2}$
moles at equilibrium	$2-\alpha$	$\dfrac {\alpha} {2}$	$\dfrac {\alpha} {2}$

The reaction is

$2HI \rightleftharpoons H_{2}+I_{2}$ .

$\alpha$ is the degree of dissociation.

Total number of moles

$=2-\alpha+\frac {\alpha} {2}+\frac {\alpha} {2}=2$ .

If the dissociation constant of a weak acid is

$1.0 \times 10^{-5}$ , then the equilibrium constant for the reaction of the acid with a strong base is:

0%
$1.0 \times 10^{-5}$
0%
$1.0 \times 10^{-9}$
0%
$1.0 \times 10^{9}$
0%
$1.0 \times 10^{14}$

Explanation

The equilibrium reaction for the dissociation of the weak acid is as follows:

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

The expression for the dissociation constant

$(K_a)$ is as given below:

$K_a= \dfrac {[H^+][A^-]} {[HA]}$

$.....(i)$

The reaction for the neutralization of weak acid

$HA$ with strong base

$OH^-$ is as follows:

$HA+OH^- \rightleftharpoons A^- + H_2O$

The expression for the equilibrium constant is as given below:

$K=\dfrac {[A^-]} {[HA][OH^-]}$

$......(ii)$

On dividing equation

$(i)$ by equation

$(ii)$ , we get

$\dfrac {K_a} {K}=\dfrac {\dfrac {[H^+][A^-]} {[HA]}} {\dfrac {[A^-]} {[HA][OH^-]}}$

Hence,

$\dfrac {K_a} {K}=[H^+][OH^-]=K_w=10^{-14}$

Thus,

$K=\dfrac {K_a} {K_w}=\dfrac {10^{-5}} {10^{-14}}=10^9$

Hence, the equilibrium constant for the reaction of the acid with a strong base is

$1.0\times 10^9$ .

The solubility product of

$BaSO_{4}$ is

$1.5\times 10^{-9}$ . The precipitation in a

$0.01M$

$Ba^{2+}$ solution will start on adding

$H_{2}SO_{4}$ of concentration:

0%
$1\times 10^{-9}M$
0%
$1.5\times 10^{-7}M$
0%
$2\times 10^{-7} M$
0%
$1\times 10^{-6} M$

Explanation

The solubility product of

$BaSO_{4}$ is

$1.5\times 10^{-9}$ .

$K_{sp}=[Ba^{2+}][SO_4^{2-}]$

Substitute values in the above expression.

$1.5\times 10^{-9}=0.01 \times [SO_4^{2-}]$

Hence,

$[SO_4^{2-}]=1.5\times 10^{-7}\ M$ .

Thus, the sulfuric acid solution of the concentration

$1.5\times 10^{-7}\ M$ should be added to start precipitation of barium sulphate.

Option B is correct.

A precipitate of calcium oxalate will not dissolve in:

0%
$HCl$
0%
$HNO_{3}$
0%
$aqua\: regia$
0%
$CH_{3}COOH$

Blood

$pH$ is controlled by the concentrations of

$H_{2}CO_{3}$ and

$HCO_{3}{^{-}}$ . In the presence of

$NaHCO_{3}$ ,

$pH$ of blood:

0%
increases
0%
decreases
0%
does not change
0%
first decreases and then increases

Explanation

Acidic buffer solution is prepared by mixing a weak acid and its salt with a strong base.
Thus, a mixture of

$H_{2}CO_{3}$ and

$HCO_{3}^{-}$ acts as an acidic buffer solution.
The expression for the pH of the acidic buffer solution is as given below:

$pH=pK_a+log \frac {[salt]} {[acid]}$
In the presence of

$NaHCO_{3}$ , the concentration of salt increases. Hence, the term

$log \frac {[salt]} {[acid]}$ increases. This increases the

$pH$ of blood.

pH of

$0.01M (NH_4)_2SO_4$ and

$0.02M$

$NH_4OH$ buffer

$(pK_{a\ (NH_4^+)}=9.26)$ is

0%
$9.26+ log1$
0%
$9.26 +log2$
0%
$4.74$
0%
$4.74 +log2$

$pK_{ b }$ For

$CN^-$ at

$25^{ o } C$ is 4.The pH of 0.5M aqueous NaCN solution is:-

0%
12
0%
10
0%
11.5
0%
11

Which of the following mixture is not a buffer solution?

0%
100 ml of 0.5 $NH_4Cl$ and 100 ml of 0.6 $N\ NH_{4}OH$
0%
100 ml of 0.6 $N$ $HCN$ and 100 ml of 0.4 $N$ $NaOH$
0%
100 ml of 0.2 $N$ $NH_{4}OH$ and 100 ml of 0.2 $N \ CH_{3}COOH$
0%
100 ml of 0.4 $N$ $HCl$ and 100 ml of 0.4 $N$ $NaOH$

Explanation

An acidic buffer solution is prepared by mixing a weak acid and its salt with a strong base. A basic buffer solution is prepared by mixing a weak base and its salt with strong acid. A buffer solution cannot be obtained by mixing a strong acid with a strong base.

$HCl$ is a strong acid and

$NaOH$ is a strong base, therefore it forms salt and not a buffer solution.

$NH_4OH$ and

$CH_3COOH$ is a mixture of a weak acid and weak base, therefore it forms salt and not buffer solution.

Hence option C & D is correct.

A solution contains

$0.09\ M\ HCl,\ 0.09\ M\ CHCl_{2}COOH$ , and

$0.1\ M\ CH_{3}COOH$ . If total

$[H^{+}] = 0.1$ and

$K_{a}$ for

$CH_{3}COOH =10^{-5},\ K_{a}$ for

$CCl_{2}HCOOH$ is:

0%
$1.35 \times10^{-4}$
0%
$0.18 \times10^{-2}$
0%
$0.18\times10^{-5}$
0%
$1.25\times10^{-2}$

Explanation

For a weak acid, the expression for the hydrogen ion concentration is

$[H^+]=\sqrt {c \times K_a}$ .
Hence the total hydrogen ion concentraton of he solution will be,

$[H^+]_{total}=[H^+]_{HCl}+[H^+]_{CHCl_2COOH}[H^+]_{CH_3COOH}$ .

$[H^+]_{total}=0.1=0.09+\sqrt {c_{CHCl_2COOH} \times K_{a,CHCl_2COOH}}+\sqrt {c_{CH_3COOH} \times K_{a,CH_3COOH}}$

$0.01=\sqrt {0.09 \times K_{a,CHCl_2COOH}}+\sqrt {0.1 \times 10^{-5}}$

Hence,

$K_{a,CHCl_2COOH}=9 \times 10^{-4}$ .

The degree of dissociation of 0.1 M weak acid HA is 0.5%. If 2 ml of 1.0 M HA solution is diluted to 32 ml, the degree of dissociation of acid and

$H_{3}O^{+}$ ion concentration in the resulting solution will be respectively:

0%
$0.02$ and $3.125 \times10^{-4}$
0%
$1.25 \times10^{-3}$ and $0.02$
0%
$0.632$ and $3.95 \times 10^{-4}$
0%
$0.02$ and $8.0\times10^{-12}$

Explanation

Given,

$\alpha =0.5 \%= 0.005$ and

$C= 0.1\ M$

The expression for the degree of dissociation is

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

$0.005= \sqrt {\dfrac{K_a}{0.1}}$

$K_a=2.5 \times 10^{-6}$ .

When the solution is diluted, the molarity of the solution is given by the following expression.

$M_1V_1=M_2V_2$

$2 \times 1 = 32 \times M_2$

Hence,

$M_2=\dfrac {1} {16}$ .

For this diluted solution,

$\alpha = \sqrt {\dfrac{K_a}{C}}=0.632\%$ .
The hydrogen ion concentration is

$[H^+]=c \alpha =\dfrac {1} {16} \times 0.00632 =3.955 \times 10^{-4}$ .

The first and second dissociation constants of an acid

$H_{2}A$ are

$1.0\times10^{-5}$ and

$5.0\times 10^{-10}$ respectively. The overall dissociation constant of the acid will be:

0%
$5.0\times10^{-5}$
0%
$5.0\times10^{15}$
0%
$5.0\times10^{-15}$
0%
$2.0\times10^{5}$

Explanation

For the dissociation of a dibasic acid, the overall dissociation constant is equal to the product of the first and second dissociation constants.
Therefore,

$K=K_1 \times K_2=1.0 \times 10^{-5} \times 5.0 \times 10^{-10}= 5.0 \times 10^{-15}$

Hence, the overall dissociation constant is

$5.0\times 10^{-15}$ .

The degree of dissociation of an electrolyte is :

0%
directly proportional to its concentration
0%
directly proportional to the square of its concentration
0%
inversely proportional to its concentration
0%
inversely proportional to the square root of its concentration

Explanation

For a weak electrolyte, the dissociation constant can be written as:

$K_d=\dfrac{\alpha^2}{1-\alpha} C$ , where

$\alpha$ (degree of dissociation)

$<< 1$ and C is the concentration of electrolyte.

$\therefore \dfrac{K_d}{C} =\dfrac{\alpha^2}{1}$ .

$\therefore \alpha=\sqrt{\dfrac{K_d}{C}}$ .

$\therefore \alpha \propto \dfrac{1}{\sqrt{C}}$ .

The degree of dissociation is inversely proportional to the square root of concentration.

What is the

$K_{b}$ of a weak base that produces an

$OH^{-}$ per molecule if a 0.05 M solution is 2.5% ionized?

0%
$7.8 \times10^{-8}$
0%
$1.6 \times10^{-6}$
0%
$3.125 \times10^{-5}$
0%
$1.2 \times10^{-3}$

Explanation

When a solution is

$2.5\%$ ionized, its degree of dissociation

$\alpha$ is

$\dfrac {2.5} {100}=0.025$ .

The concentration C is 0.05 M.

The relationship between the dissociation constant

$K_b$ and the degree of dissociation is

$K_b=C \times \alpha ^2$ .

Substitute values in the above expression.

$K_b=0.05 \times (0.025)^2=3.125 \times 10^{-5}$ .

If the degree of dissociation of water at

$90^{\circ}C$ is

$1.28\times10^{-8}$ , then the ionization constant of water at

$90^{\circ}C$ is:

0%
$1.28 \times10^{-14}$
0%
$12.81 \times10^{-15}$
0%
$9.07 \times10^{-15}$
0%
$7.52 \times10^{-12}$

The dissociation constants of a weak acid HA and weak base BOH are

$2\times 10^{-5}$ and

$5\times 10^{-6}$ respectively. The equilibrium constant for the neutralisation reaction of the two is :

0%
$1.0\times 10^{4}$
0%
$1.0\times 10^{-4}$
0%
$1.0\times 10^{-10}$
0%
$2.5\times 10^{-1}$

Explanation

For the neutralization reaction of a weak acid and weak base, the expression for the equilibrium constant K is as given below.

$K= \displaystyle\frac {K_a \times K_b} {K_w}$

By substituting values in the above expression, we get

$K= \displaystyle\frac {2 \times 10^{-5} \times 5 \times 10^{-6}} {1 \times 10^{-14}}=1.0 \times 10^{4}$ .

Aqueous solutions of

$HNO_{3}, KOH, CH_{3}COOH$ and

$CH_{3}COONa$ of identical concentrations are provided. The pair(s) of solutions which form a buffer upon mixing are:

0%
$\mathrm{H}\mathrm{N}\mathrm{O}_{3}$ and $\mathrm{C}\mathrm{H}_{3}\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H}$
0%
KOH and $\mathrm{C}\mathrm{H}_{3}\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{N}\mathrm{a}$
0%
$\mathrm{H}\mathrm{N}\mathrm{O}_{3}$ and $\mathrm{C}\mathrm{H}_{3}\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{N}\mathrm{a}$
0%
$\mathrm{C}\mathrm{H}_{3}\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H}$ and $\mathrm{C}\mathrm{H}_{3}\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{N}\mathrm{a}$

Explanation

A buffer solution consists of weak acid and its salt with strong base or weak base and its salt with strong acid.

$CH_3COONa + HNO_3 \rightarrow CH_3COOH + NaNO_3$

Now the solution contains acetic acid (weak acid) and sodium acetate (salt with strong base NaOH). Hence, it is a buffer solution.

(D) Acetic acid (weak acid) and sodium acetate (salt with strong base NaOH) form a buffer solution.

Thus, the options (C) and (D) form buffer solutions.

Silver ions are added to the solution with

$[Br^{-}]=[Cl^{-}]=[CO_{3}^{2-}]=[AsO_{4}^{3-}] = 0.1M$
Which compound will precipitate at the lowest [

$Ag^{+}$ ] ?

0%
$AgBr(K_{sp}= 5\times 10^{-13})$
0%
$AgCl(K_{sp}= 1.8\times 10^{-10})$
0%
$Ag_{2}CO_{3}(K_{sp}= 8.1\times 10^{-12})$
0%
$Ag_{3}AsO_{4}(K_{sp}= 10^{-22})$

Explanation

Let S moles per litre be the solubility of the salt

$A_xB_y$ .
The expression for the solubility product will be

$K_{sp}=[A^{y+}]^x[B^{x-}]^y=(x \times S)^x(y \times y)^y$ .
Also

$[A^{y+}]=(x \times S$ and

$B^{x-})=y \times y$ .
When values are substituted in the above expressions for

$AgBr, AgCl$ ,

$Ag_{2}CO_{3}$ and

$Ag_{3}AsO_{4}$ , lowest

$[Ag^{+}]$ (that will precipitate the compound) is obtained for

$AgBr.$

The initial rate of hydrolysis methyl acetate (

$1\ M$ ) by a weak acid (

$HA,\ 1M$ ) is

$1/100^{th}$ of that of a strong acid (

$HX,\ 1M$ ), at

$25^{o}C$ . The

$K_{a}$ of

$HA$ is:

0%
1 x 10 $^{-4}$
0%
1 x 10 $^{-5}$
0%
1 x 10 $^{-6}$
0%
1 x 10 $^{-3}$

Explanation

The rate of the hydrolysis of methyl acetate is directly proportional to the hydrogen ion concentration.
In case of weak acid, the equilibrium constant is given by the expression

$K_a=\dfrac {[H^+][A^-]}{[HA]}=\dfrac {[H^+]^2}{[HA]}$
Hence,

$[H^+]=\sqrt {K_a[HA]}$

Hence, the expression for the ratio of the rate of the hydrolysis of methyl acetate in presence of weak acid and strong acid becomes

$\dfrac {r_{weak \ acid}} {r_{strong \ acid}}=\dfrac {\sqrt {K_a[HA]}}{[HX]}=\dfrac {1} {100}$

Substitute

$[HA]=1 M$

Hence,

$\dfrac {\sqrt {K_a \times 1}}{1}=\dfrac {1} {100}$

$K_a=1 \times 10^{-4}$

From application of thermodynamics on chemical reaction,
we get

$\Delta G = \Delta G^{0} + RT$ ln Q
Also

$\Delta G = \Delta H -T\Delta S$ .
If

$\Delta G = 0$ , reaction is at equilibrium.
If

$\Delta G > 0$ reaction is non-spontaneous under given condition.

$If \Delta G < 0$ , reaction is spontaneous under given condition.

Consider the reaction given below.

$A(s) \rightleftharpoons 2B(g) \ \ \ \ \ \ \ \ \ \Delta H^{0}= 160$ KJ/mol.

$\Delta S^{0} = 400$ J/mol-K at

$400$ K
Which of the following is correct at

$400$ K?

0%
On adding more $A(s)$ , more $B(g)$ is produced, when $A(s)$ and $B(g)$ were in equilibrium
0%
The equilibrium constant at $400$ K can't be found
0%
The reaction is at equilibrium at $400$ K and standard condition
0%
The $\Delta G$ of the reaction is greater than zero, at $400$ K and standard condition

Explanation

Th relationship between the standard free energy change, standard enthalpy change and entropy change is

$\Delta G^o=\Delta H^o-T\Delta S^o$ .

$\Delta H^o=160 \ kJ/mol=160000 \ J/mol$ .
Substitute values in the above expression.

$\Delta G^o=160000 \ J/mol-400K \times 400 \ J/mol-K=0 \ J/mol$ .
Under standard conditions,

$K_p=1$ , so

$lnK_p=0$ .
Thus

$\Delta G = \Delta G^{0} + RT \ ln\ K_p=0 + RT(0)=0$ .
Hence, the system is at equilibrium when a temperature is 400 K and standard conditions are used.

Find the concentration of

$H^+$ ions in an aqueous solution, which is saturated with

$H_2S$ (

$0.1M$ ) as well as

$H_2CO_3$ (

$0.2 M$ ).

[Use data:

$K_{1}=10^{-7},K_{2}=10^{-14}$ for

$H_{2}S, K_{1}=4\times10^{-7}, K_{2}=4\times10^{-11}$ for

$H_{2}CO_{3} ]$

0%
$3\times10^{-4}M$
0%
$3.83\times10^{-4}M$
0%
$2.83\times10^{-4}M$
0%
None of these

Explanation

Let the contribution of

$[H^+]$ from

$H_2S$ be

$x$ and from

$H_2CO_3$ be

$y$ . Since the second dissociation constants of those acids are very small, we consider only the first dissociation.

$H_2S\leftrightharpoons H^++HS^-$

$x+y$

$x$

$H_2CO_3\leftrightharpoons H^++HCO_3^-$

$x+y$

$y$

$K_{1(H_2S)}=\cfrac{[H^+][HS^-]}{[H_2S]}=\cfrac{x(x+y)}{0.1}=10^{-7}$

Given

$K_1=10^{-7}$

$\therefore x(x+y)=10^{-8}\quad$ ---------

$(1)$

Similarly,

$K_{1(H_2CO_3)}=\cfrac{y(x+y)}{0.2}=4\times 10^{-7}$

$\therefore y(x+y)=8\times 10^{-8}\quad$ -------

$(2)$

From

$(1)$ and

$(2)$ ,

$y=8x$

Subsitituting this in

$(1)$ ,

$x(9x)=10^{-8}$

$\implies x^2=\cfrac {10^{-8}}{9}$

$\therefore x=\cfrac {1}{3}\times 10^{-4}$ and

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

Total

$H^+$ concentration

$=x+y=\left(\cfrac{1+8}{3}\right)\times 10^{-4}$

$M=3\times 10^{-4}$

$M$

Note: Hydrogen ion from 2nd dissociation is negligible in both the cases.

White salt is readily soluble in water and gives a colourless solution with

$pH$ of about

$9$ . The salt would be:

0%
$NH_4NO_3$
0%
$CH_3COONa$
0%
$CH_3COONH_4$
0%
$CaCO_3$

Explanation

After dissolution in water, the salt gives basic

$pH$ , it is evident that the salt is formed by a strong base and a weak acid.

For the following given salts,

$HNO_3 + NH_4OH \rightarrow NH_4NO_3+H_2O$ (Strong Acid + Weak Base)

$CH_3COOH + NaOH \rightarrow CH_3COONa+H_2O$ (Weak acid + Strong Base)

$CH_3COOH + NH_4OH \rightarrow CH_3COONH_4+H_2O$ (Weak Acid + Weak Base)

$Ca(OH)_2 + H_2CO_3 \rightarrow CaCO_3+2H_2O$ (Weak acid + Strong Base)

But

$CaCO_3$ is insoluble in water, hence our answer is

$CH_3COONa$

Option B is correct.

Concentration of

$H^+$ ions in 0.1 M

$H_2CO_3$ is

$(K_{1}=4\times10^{-7},K_{2}=4\times10^{-11}):$

0%
$2\times10^{-4}M$
0%
$4\times10^{-9}M$
0%
$2\times10^{-3}M$
0%
None of these

Explanation

$[H^{+}]=\sqrt{K_{1}C_{o}}=\sqrt{4\times10^{-7}\times0.1} =2\times10^{-4}M$ .

CBSE Questions for Class 11 Medical Chemistry Equilibrium Quiz 10 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Equilibrium Quiz 10 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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