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CHXI07:EQUILIBRIUM

314733 The dissociation constants of aniline, acetic acid and water at $25^{\circ} C$ are respectively, $3.83 \times 10^{- 10}$ , $1.75 \times 10^{- 15}$ and $1.008 \times 10^{- 14}$ . What is the percentage hydrolysis of aniline acetate in a decimolar solution?

1 50

2 55

3 60

4 12

Explanation:

The degree of hydrolysis ' $h$ ' of a salt of a weak base is given by $h = \sqrt{\frac{K_{w}}{K_{a} \times K_{b}}}$
$= \sqrt{\frac{1 .008 \times 1 0^{- 4}}{1 .75 \times 1 0^{- 5} \times 3 .83 \times 1 0^{- 10}}} = 1 .219$
But, ' $h$ ' cannot be greater than 1. This indicates that the degree of hydrolysis in this case is very large and therefore, the above formula, which is based on the assumption that ' $h$ ' is small, is not applicable.
In the present case, therefore, we have to calculate ' $h$ ' as follows:
$K_{h} = \frac{K_{w}}{K_{a} \times K_{b}} K_{h} = \frac{h^{2}}{{(1 - h)}^{2}} \frac{h}{1 - h} = \sqrt{K_{h}}$
$= \sqrt{\frac{K_{w}}{K_{a} \times K_{b}}} = \sqrt{1 .219} = 0 .5495$
Percentage hydrolysis $= 55$

CHXI07:EQUILIBRIUM

314734 What will be the $pH$ of an aqueous solution of 1.0 $M$ ammonium formate $({HCOONH}_{4})$ ?
(Given : ${pK}_{a} = 3.8$ and ${pK}_{b} = 4.8$ )

1 7.5

2 3.4

3 6.5

4 10.2

Explanation:

$pH = \frac{1}{2} ({pK}_{w} + {pK}_{a} - {pK}_{b}) = 6.5$

CHXI07:EQUILIBRIUM

314735 The $pH$ of aqueous solution of $N H_{4} CN$
$K_{a} of HCN is 9 .2 \times 1 0^{-10}$
$K_{b} of N H_{4} OH is 1 .8 \times 1 0^{-5}$ is

1

> 7

2

< 7

3 7

4 14

Explanation:

$K_{a}$ of $HCN = 9.2 \times 10^{- 10}, {pK}_{a} = 9.04$
$K_{b}$ of ${NH}_{4} OH = 1.8 \times 10^{- 5}, {pK}_{b} = 4.74$
$p H = 7 + \frac{1}{2} [p K_{a} - p K_{b}] =$
$7 + \frac{1}{2} [9.04 - 4.74] = 9.15$

CHXI07:EQUILIBRIUM

314736 The extent of hydrolysis of $0 .0050 M K_{2} Cr O_{4}$ is (The ionisation constants of $H_{2} {CrO}_{4}$ are $K a_{1} = 1 .6, K a_{2} = 3 .1 \times 1 0^{- 7})$

1

0.5 %

2

0.026 %

3

0.26 %

4

0.15 %

Explanation:

Just as in the ionisation of polyprotic acids, so in the hydrolysis of their salts, the reaction proceeds in successive stages. The extent of the second stage is generally very small compared with the first. This is particularly true in this case, where $H_{2} {CrO}_{4}$ is essentially a strong acid with respect to its first ionisation. The equation of interest is
${CrO}_{4}^{2 -} + H_{2} O ⇌ {HCrO}_{4}^{-} + {OH}^{-}$
Which indicates that the conjugate acid of hydrolysing ${CrO}_{4}^{2 -}$ is ${HCrO}_{4}^{-}$ . As the ionisation constant for ${HCrO}_{4}^{-}$ is ${Ka}_{2}$ , the hydrolysis constant for the reaction is $K_{W} / {Ka}_{2}$ .
$\frac{[{OH}^{-}] [{HCrO}_{4}^{-}]}{[{CrO}_{4}^{2 -}]} = K_{h} = \frac{K_{w}}{Ka} = \frac{10^{- 14}}{3.1 \times 10^{- 7}} = 3.2 \times 10^{- 8}$
Let $x = [{OH}^{-}] = [{HCrO}_{4}^{-}]$ .
Then $[{CrO}_{4}^{2 -}] = 0.005 - x \approx 0.005$ and
$\frac{x^{2}}{0.005} = 3.2 \times 10^{- 8} x = 1.3 \times 10^{- 5}$
$%$ hydrolysis $= (1.3 \times 10^{- 5} \times 100) / 0.005 = 0.26 %$

CHXI07:EQUILIBRIUM

314733 The dissociation constants of aniline, acetic acid and water at $25^{\circ} C$ are respectively, $3.83 \times 10^{- 10}$ , $1.75 \times 10^{- 15}$ and $1.008 \times 10^{- 14}$ . What is the percentage hydrolysis of aniline acetate in a decimolar solution?

1 50

2 55

3 60

4 12

Explanation:

The degree of hydrolysis ' $h$ ' of a salt of a weak base is given by $h = \sqrt{\frac{K_{w}}{K_{a} \times K_{b}}}$
$= \sqrt{\frac{1 .008 \times 1 0^{- 4}}{1 .75 \times 1 0^{- 5} \times 3 .83 \times 1 0^{- 10}}} = 1 .219$
But, ' $h$ ' cannot be greater than 1. This indicates that the degree of hydrolysis in this case is very large and therefore, the above formula, which is based on the assumption that ' $h$ ' is small, is not applicable.
In the present case, therefore, we have to calculate ' $h$ ' as follows:
$K_{h} = \frac{K_{w}}{K_{a} \times K_{b}} K_{h} = \frac{h^{2}}{{(1 - h)}^{2}} \frac{h}{1 - h} = \sqrt{K_{h}}$
$= \sqrt{\frac{K_{w}}{K_{a} \times K_{b}}} = \sqrt{1 .219} = 0 .5495$
Percentage hydrolysis $= 55$

CHXI07:EQUILIBRIUM

314734 What will be the $pH$ of an aqueous solution of 1.0 $M$ ammonium formate $({HCOONH}_{4})$ ?
(Given : ${pK}_{a} = 3.8$ and ${pK}_{b} = 4.8$ )

1 7.5

2 3.4

3 6.5

4 10.2

Explanation:

$pH = \frac{1}{2} ({pK}_{w} + {pK}_{a} - {pK}_{b}) = 6.5$

CHXI07:EQUILIBRIUM

314735 The $pH$ of aqueous solution of $N H_{4} CN$
$K_{a} of HCN is 9 .2 \times 1 0^{-10}$
$K_{b} of N H_{4} OH is 1 .8 \times 1 0^{-5}$ is

1

> 7

2

< 7

3 7

4 14

Explanation:

$K_{a}$ of $HCN = 9.2 \times 10^{- 10}, {pK}_{a} = 9.04$
$K_{b}$ of ${NH}_{4} OH = 1.8 \times 10^{- 5}, {pK}_{b} = 4.74$
$p H = 7 + \frac{1}{2} [p K_{a} - p K_{b}] =$
$7 + \frac{1}{2} [9.04 - 4.74] = 9.15$

CHXI07:EQUILIBRIUM

314736 The extent of hydrolysis of $0 .0050 M K_{2} Cr O_{4}$ is (The ionisation constants of $H_{2} {CrO}_{4}$ are $K a_{1} = 1 .6, K a_{2} = 3 .1 \times 1 0^{- 7})$

1

0.5 %

2

0.026 %

3

0.26 %

4

0.15 %

Explanation:

Just as in the ionisation of polyprotic acids, so in the hydrolysis of their salts, the reaction proceeds in successive stages. The extent of the second stage is generally very small compared with the first. This is particularly true in this case, where $H_{2} {CrO}_{4}$ is essentially a strong acid with respect to its first ionisation. The equation of interest is
${CrO}_{4}^{2 -} + H_{2} O ⇌ {HCrO}_{4}^{-} + {OH}^{-}$
Which indicates that the conjugate acid of hydrolysing ${CrO}_{4}^{2 -}$ is ${HCrO}_{4}^{-}$ . As the ionisation constant for ${HCrO}_{4}^{-}$ is ${Ka}_{2}$ , the hydrolysis constant for the reaction is $K_{W} / {Ka}_{2}$ .
$\frac{[{OH}^{-}] [{HCrO}_{4}^{-}]}{[{CrO}_{4}^{2 -}]} = K_{h} = \frac{K_{w}}{Ka} = \frac{10^{- 14}}{3.1 \times 10^{- 7}} = 3.2 \times 10^{- 8}$
Let $x = [{OH}^{-}] = [{HCrO}_{4}^{-}]$ .
Then $[{CrO}_{4}^{2 -}] = 0.005 - x \approx 0.005$ and
$\frac{x^{2}}{0.005} = 3.2 \times 10^{- 8} x = 1.3 \times 10^{- 5}$
$%$ hydrolysis $= (1.3 \times 10^{- 5} \times 100) / 0.005 = 0.26 %$

CHXI07:EQUILIBRIUM

314733 The dissociation constants of aniline, acetic acid and water at $25^{\circ} C$ are respectively, $3.83 \times 10^{- 10}$ , $1.75 \times 10^{- 15}$ and $1.008 \times 10^{- 14}$ . What is the percentage hydrolysis of aniline acetate in a decimolar solution?

1 50

2 55

3 60

4 12

Explanation:

The degree of hydrolysis ' $h$ ' of a salt of a weak base is given by $h = \sqrt{\frac{K_{w}}{K_{a} \times K_{b}}}$
$= \sqrt{\frac{1 .008 \times 1 0^{- 4}}{1 .75 \times 1 0^{- 5} \times 3 .83 \times 1 0^{- 10}}} = 1 .219$
But, ' $h$ ' cannot be greater than 1. This indicates that the degree of hydrolysis in this case is very large and therefore, the above formula, which is based on the assumption that ' $h$ ' is small, is not applicable.
In the present case, therefore, we have to calculate ' $h$ ' as follows:
$K_{h} = \frac{K_{w}}{K_{a} \times K_{b}} K_{h} = \frac{h^{2}}{{(1 - h)}^{2}} \frac{h}{1 - h} = \sqrt{K_{h}}$
$= \sqrt{\frac{K_{w}}{K_{a} \times K_{b}}} = \sqrt{1 .219} = 0 .5495$
Percentage hydrolysis $= 55$

CHXI07:EQUILIBRIUM

314734 What will be the $pH$ of an aqueous solution of 1.0 $M$ ammonium formate $({HCOONH}_{4})$ ?
(Given : ${pK}_{a} = 3.8$ and ${pK}_{b} = 4.8$ )

1 7.5

2 3.4

3 6.5

4 10.2

Explanation:

$pH = \frac{1}{2} ({pK}_{w} + {pK}_{a} - {pK}_{b}) = 6.5$

CHXI07:EQUILIBRIUM

314735 The $pH$ of aqueous solution of $N H_{4} CN$
$K_{a} of HCN is 9 .2 \times 1 0^{-10}$
$K_{b} of N H_{4} OH is 1 .8 \times 1 0^{-5}$ is

1

> 7

2

< 7

3 7

4 14

Explanation:

$K_{a}$ of $HCN = 9.2 \times 10^{- 10}, {pK}_{a} = 9.04$
$K_{b}$ of ${NH}_{4} OH = 1.8 \times 10^{- 5}, {pK}_{b} = 4.74$
$p H = 7 + \frac{1}{2} [p K_{a} - p K_{b}] =$
$7 + \frac{1}{2} [9.04 - 4.74] = 9.15$

CHXI07:EQUILIBRIUM

314736 The extent of hydrolysis of $0 .0050 M K_{2} Cr O_{4}$ is (The ionisation constants of $H_{2} {CrO}_{4}$ are $K a_{1} = 1 .6, K a_{2} = 3 .1 \times 1 0^{- 7})$

1

0.5 %

2

0.026 %

3

0.26 %

4

0.15 %

Explanation:

Just as in the ionisation of polyprotic acids, so in the hydrolysis of their salts, the reaction proceeds in successive stages. The extent of the second stage is generally very small compared with the first. This is particularly true in this case, where $H_{2} {CrO}_{4}$ is essentially a strong acid with respect to its first ionisation. The equation of interest is
${CrO}_{4}^{2 -} + H_{2} O ⇌ {HCrO}_{4}^{-} + {OH}^{-}$
Which indicates that the conjugate acid of hydrolysing ${CrO}_{4}^{2 -}$ is ${HCrO}_{4}^{-}$ . As the ionisation constant for ${HCrO}_{4}^{-}$ is ${Ka}_{2}$ , the hydrolysis constant for the reaction is $K_{W} / {Ka}_{2}$ .
$\frac{[{OH}^{-}] [{HCrO}_{4}^{-}]}{[{CrO}_{4}^{2 -}]} = K_{h} = \frac{K_{w}}{Ka} = \frac{10^{- 14}}{3.1 \times 10^{- 7}} = 3.2 \times 10^{- 8}$
Let $x = [{OH}^{-}] = [{HCrO}_{4}^{-}]$ .
Then $[{CrO}_{4}^{2 -}] = 0.005 - x \approx 0.005$ and
$\frac{x^{2}}{0.005} = 3.2 \times 10^{- 8} x = 1.3 \times 10^{- 5}$
$%$ hydrolysis $= (1.3 \times 10^{- 5} \times 100) / 0.005 = 0.26 %$

CHXI07:EQUILIBRIUM

314733 The dissociation constants of aniline, acetic acid and water at $25^{\circ} C$ are respectively, $3.83 \times 10^{- 10}$ , $1.75 \times 10^{- 15}$ and $1.008 \times 10^{- 14}$ . What is the percentage hydrolysis of aniline acetate in a decimolar solution?

1 50

2 55

3 60

4 12

Explanation:

The degree of hydrolysis ' $h$ ' of a salt of a weak base is given by $h = \sqrt{\frac{K_{w}}{K_{a} \times K_{b}}}$
$= \sqrt{\frac{1 .008 \times 1 0^{- 4}}{1 .75 \times 1 0^{- 5} \times 3 .83 \times 1 0^{- 10}}} = 1 .219$
But, ' $h$ ' cannot be greater than 1. This indicates that the degree of hydrolysis in this case is very large and therefore, the above formula, which is based on the assumption that ' $h$ ' is small, is not applicable.
In the present case, therefore, we have to calculate ' $h$ ' as follows:
$K_{h} = \frac{K_{w}}{K_{a} \times K_{b}} K_{h} = \frac{h^{2}}{{(1 - h)}^{2}} \frac{h}{1 - h} = \sqrt{K_{h}}$
$= \sqrt{\frac{K_{w}}{K_{a} \times K_{b}}} = \sqrt{1 .219} = 0 .5495$
Percentage hydrolysis $= 55$

CHXI07:EQUILIBRIUM

314734 What will be the $pH$ of an aqueous solution of 1.0 $M$ ammonium formate $({HCOONH}_{4})$ ?
(Given : ${pK}_{a} = 3.8$ and ${pK}_{b} = 4.8$ )

1 7.5

2 3.4

3 6.5

4 10.2

Explanation:

$pH = \frac{1}{2} ({pK}_{w} + {pK}_{a} - {pK}_{b}) = 6.5$

CHXI07:EQUILIBRIUM

314735 The $pH$ of aqueous solution of $N H_{4} CN$
$K_{a} of HCN is 9 .2 \times 1 0^{-10}$
$K_{b} of N H_{4} OH is 1 .8 \times 1 0^{-5}$ is

1

> 7

2

< 7

3 7

4 14

Explanation:

$K_{a}$ of $HCN = 9.2 \times 10^{- 10}, {pK}_{a} = 9.04$
$K_{b}$ of ${NH}_{4} OH = 1.8 \times 10^{- 5}, {pK}_{b} = 4.74$
$p H = 7 + \frac{1}{2} [p K_{a} - p K_{b}] =$
$7 + \frac{1}{2} [9.04 - 4.74] = 9.15$

CHXI07:EQUILIBRIUM

314736 The extent of hydrolysis of $0 .0050 M K_{2} Cr O_{4}$ is (The ionisation constants of $H_{2} {CrO}_{4}$ are $K a_{1} = 1 .6, K a_{2} = 3 .1 \times 1 0^{- 7})$

1

0.5 %

2

0.026 %

3

0.26 %

4

0.15 %

Explanation:

Just as in the ionisation of polyprotic acids, so in the hydrolysis of their salts, the reaction proceeds in successive stages. The extent of the second stage is generally very small compared with the first. This is particularly true in this case, where $H_{2} {CrO}_{4}$ is essentially a strong acid with respect to its first ionisation. The equation of interest is
${CrO}_{4}^{2 -} + H_{2} O ⇌ {HCrO}_{4}^{-} + {OH}^{-}$
Which indicates that the conjugate acid of hydrolysing ${CrO}_{4}^{2 -}$ is ${HCrO}_{4}^{-}$ . As the ionisation constant for ${HCrO}_{4}^{-}$ is ${Ka}_{2}$ , the hydrolysis constant for the reaction is $K_{W} / {Ka}_{2}$ .
$\frac{[{OH}^{-}] [{HCrO}_{4}^{-}]}{[{CrO}_{4}^{2 -}]} = K_{h} = \frac{K_{w}}{Ka} = \frac{10^{- 14}}{3.1 \times 10^{- 7}} = 3.2 \times 10^{- 8}$
Let $x = [{OH}^{-}] = [{HCrO}_{4}^{-}]$ .
Then $[{CrO}_{4}^{2 -}] = 0.005 - x \approx 0.005$ and
$\frac{x^{2}}{0.005} = 3.2 \times 10^{- 8} x = 1.3 \times 10^{- 5}$
$%$ hydrolysis $= (1.3 \times 10^{- 5} \times 100) / 0.005 = 0.26 %$

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