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CHXI07:EQUILIBRIUM

314755 The ionisation constant of nitrous acid is $4.5 \times$ $10^{- 4}$ . The $pH$ and degree of hydrolysis of 0.04 $M$ sodium nitrite solution are, respectively.

1

7.98, 2.36 \times 10^{- 5}

2

6.98, 2.36 \times 10^{- 5}

3

7.98, 2.36 \times 10^{- 6}

4

6.98, 2.36 \times 10^{- 6}

Explanation:

$G i v e n, [NaN O_{2}] = 0 . 04 M$
$K_{a}$ of ${HNO}_{2} = 4.5 \times 10^{- 4}$
$p K_{a} = - \log K_{a} = - \log (4.5 \times 10^{- 4})$
$p K_{a} = - 4 - 0.6532 = 3.3468 \approx 3.35$
$pH = 7 + \frac{p K_{a} + \log C}{2}$
$pH = 7 + \frac{3.35 + \log 0.04}{2} = 7.98$
$pH = 7.98$ .
Degree of hydrolysis,
$h = \sqrt{\frac{K_{w}}{K_{a} C}} = \sqrt{\frac{10^{- 14}}{4.5 \times 10^{- 4} \times 0.04}}$

CHXI07:EQUILIBRIUM

314756 $pH$ of a salt solution of weak acid $({pK}_{a} = 4)$ and weak base $({pK}_{b} = 5)$ at $25^{\circ}$ is

1 6.5

2 6

3 7

4 7.5

Explanation:

$pH$ of buffer solution of weak acid and weak base is calculated, using the following formula.
$pH = \frac{1}{2} [{pK}_{w} + {pK}_{a} - {pK}_{b}]$
Given: ${pK}_{a} = 4, {pK}_{b} = 5, {pK}_{w} = 14$
$∴ pH = \frac{1}{2} [14 + 4 - 5] = 6.5$

AIIMS - 2019

CHXI07:EQUILIBRIUM

314757 The $pH$ of $0 .1 M N H_{4} Cl$ solution is 5.13. What will be the dissociation constant of ${NH}_{4} OH$ ?

1

1.8 \times 10^{- 9}

2

1.8 \times 10^{- 5}

3 None

4

1.8 \times 10^{- 7}

Explanation:

${NH}_{4} Cl$ is salt of weak base and strong acid, formula for $pH$ of its hydrolysis may be calculated as,
$pH = \frac{1}{2} [{pK}_{w} - {pK}_{b} - \log C]$
where $pH = 5.13$
${pK}_{W} = - \log K_{W} = - \log 10^{- 14} = 14$
$\log C = \log 0.1 = - 1$
Substituting the values in eq. (1) we get
$5.13 = \frac{1}{2} [14 - {pK}_{b} + 1] 10.26 = 15 - {pK}_{b} {pK}_{a} = 4.74 ∴ - \log K_{b} = {pK}_{b} = 4.74 K_{b} = anti \log (- 4.74) K_{b} = 1.8 \times 10^{- 5}$

CHXI07:EQUILIBRIUM

314758 Match the column I with column II and mark the appropriate choice.
Column I
Column II
A
$C H_{3} C O O N a$
P
Almost neutral, pH>7 or <7
B
$N H_{4} C l$
Q
Acidic, pH <7
C
$N a N O_{3}$
R
Alkaline, pH>7
D
$C H_{3} C O O N H_{4}$
S
Neutral, pH=7

1

A - P,    B - Q,    C - R,    D - S

2

A - Q,    B - R,    C - S,    D - P

3

A - R,    B - Q,    C - S,    D - P

4

A - S,    B - P,    C - R,    D - Q

Explanation:

(A)
${CH}_{3} COONa ⇌ \underset{weak acid}{{CH}_{3} {COO}^{-}} + \underset{strong base}{{Na}^{+}} ($ Basic $)$
(B) ${NH}_{4} Cl ⇌ \underset{Weak base}{{NH}_{4}^{+}} + \underset{Strong acid}{{Cl}^{-}} ($ Acidic)
(C) ${NaNO}_{3} ⇌ \underset{Strong base}{{Na}^{+}} + \underset{Strong acis}{{NO}_{3}^{-}} ($ Neutral $)$
(D)
${CH}_{3} {COONH}_{4} ⇌ \underset{Weak acid}{{CH}_{3} {COO}^{-}} + \underset{Weak base}{{NH}_{4}^{+}}$ (Almost neutral)
So, the option (3) is correct.

CHXI07:EQUILIBRIUM

314755 The ionisation constant of nitrous acid is $4.5 \times$ $10^{- 4}$ . The $pH$ and degree of hydrolysis of 0.04 $M$ sodium nitrite solution are, respectively.

1

7.98, 2.36 \times 10^{- 5}

2

6.98, 2.36 \times 10^{- 5}

3

7.98, 2.36 \times 10^{- 6}

4

6.98, 2.36 \times 10^{- 6}

Explanation:

$G i v e n, [NaN O_{2}] = 0 . 04 M$
$K_{a}$ of ${HNO}_{2} = 4.5 \times 10^{- 4}$
$p K_{a} = - \log K_{a} = - \log (4.5 \times 10^{- 4})$
$p K_{a} = - 4 - 0.6532 = 3.3468 \approx 3.35$
$pH = 7 + \frac{p K_{a} + \log C}{2}$
$pH = 7 + \frac{3.35 + \log 0.04}{2} = 7.98$
$pH = 7.98$ .
Degree of hydrolysis,
$h = \sqrt{\frac{K_{w}}{K_{a} C}} = \sqrt{\frac{10^{- 14}}{4.5 \times 10^{- 4} \times 0.04}}$

CHXI07:EQUILIBRIUM

314756 $pH$ of a salt solution of weak acid $({pK}_{a} = 4)$ and weak base $({pK}_{b} = 5)$ at $25^{\circ}$ is

1 6.5

2 6

3 7

4 7.5

Explanation:

$pH$ of buffer solution of weak acid and weak base is calculated, using the following formula.
$pH = \frac{1}{2} [{pK}_{w} + {pK}_{a} - {pK}_{b}]$
Given: ${pK}_{a} = 4, {pK}_{b} = 5, {pK}_{w} = 14$
$∴ pH = \frac{1}{2} [14 + 4 - 5] = 6.5$

AIIMS - 2019

CHXI07:EQUILIBRIUM

314757 The $pH$ of $0 .1 M N H_{4} Cl$ solution is 5.13. What will be the dissociation constant of ${NH}_{4} OH$ ?

1

1.8 \times 10^{- 9}

2

1.8 \times 10^{- 5}

3 None

4

1.8 \times 10^{- 7}

Explanation:

${NH}_{4} Cl$ is salt of weak base and strong acid, formula for $pH$ of its hydrolysis may be calculated as,
$pH = \frac{1}{2} [{pK}_{w} - {pK}_{b} - \log C]$
where $pH = 5.13$
${pK}_{W} = - \log K_{W} = - \log 10^{- 14} = 14$
$\log C = \log 0.1 = - 1$
Substituting the values in eq. (1) we get
$5.13 = \frac{1}{2} [14 - {pK}_{b} + 1] 10.26 = 15 - {pK}_{b} {pK}_{a} = 4.74 ∴ - \log K_{b} = {pK}_{b} = 4.74 K_{b} = anti \log (- 4.74) K_{b} = 1.8 \times 10^{- 5}$

CHXI07:EQUILIBRIUM

314758 Match the column I with column II and mark the appropriate choice.
Column I
Column II
A
$C H_{3} C O O N a$
P
Almost neutral, pH>7 or <7
B
$N H_{4} C l$
Q
Acidic, pH <7
C
$N a N O_{3}$
R
Alkaline, pH>7
D
$C H_{3} C O O N H_{4}$
S
Neutral, pH=7

1

A - P,    B - Q,    C - R,    D - S

2

A - Q,    B - R,    C - S,    D - P

3

A - R,    B - Q,    C - S,    D - P

4

A - S,    B - P,    C - R,    D - Q

Explanation:

(A)
${CH}_{3} COONa ⇌ \underset{weak acid}{{CH}_{3} {COO}^{-}} + \underset{strong base}{{Na}^{+}} ($ Basic $)$
(B) ${NH}_{4} Cl ⇌ \underset{Weak base}{{NH}_{4}^{+}} + \underset{Strong acid}{{Cl}^{-}} ($ Acidic)
(C) ${NaNO}_{3} ⇌ \underset{Strong base}{{Na}^{+}} + \underset{Strong acis}{{NO}_{3}^{-}} ($ Neutral $)$
(D)
${CH}_{3} {COONH}_{4} ⇌ \underset{Weak acid}{{CH}_{3} {COO}^{-}} + \underset{Weak base}{{NH}_{4}^{+}}$ (Almost neutral)
So, the option (3) is correct.

CHXI07:EQUILIBRIUM

314755 The ionisation constant of nitrous acid is $4.5 \times$ $10^{- 4}$ . The $pH$ and degree of hydrolysis of 0.04 $M$ sodium nitrite solution are, respectively.

1

7.98, 2.36 \times 10^{- 5}

2

6.98, 2.36 \times 10^{- 5}

3

7.98, 2.36 \times 10^{- 6}

4

6.98, 2.36 \times 10^{- 6}

Explanation:

$G i v e n, [NaN O_{2}] = 0 . 04 M$
$K_{a}$ of ${HNO}_{2} = 4.5 \times 10^{- 4}$
$p K_{a} = - \log K_{a} = - \log (4.5 \times 10^{- 4})$
$p K_{a} = - 4 - 0.6532 = 3.3468 \approx 3.35$
$pH = 7 + \frac{p K_{a} + \log C}{2}$
$pH = 7 + \frac{3.35 + \log 0.04}{2} = 7.98$
$pH = 7.98$ .
Degree of hydrolysis,
$h = \sqrt{\frac{K_{w}}{K_{a} C}} = \sqrt{\frac{10^{- 14}}{4.5 \times 10^{- 4} \times 0.04}}$

CHXI07:EQUILIBRIUM

314756 $pH$ of a salt solution of weak acid $({pK}_{a} = 4)$ and weak base $({pK}_{b} = 5)$ at $25^{\circ}$ is

1 6.5

2 6

3 7

4 7.5

Explanation:

$pH$ of buffer solution of weak acid and weak base is calculated, using the following formula.
$pH = \frac{1}{2} [{pK}_{w} + {pK}_{a} - {pK}_{b}]$
Given: ${pK}_{a} = 4, {pK}_{b} = 5, {pK}_{w} = 14$
$∴ pH = \frac{1}{2} [14 + 4 - 5] = 6.5$

AIIMS - 2019

CHXI07:EQUILIBRIUM

314757 The $pH$ of $0 .1 M N H_{4} Cl$ solution is 5.13. What will be the dissociation constant of ${NH}_{4} OH$ ?

1

1.8 \times 10^{- 9}

2

1.8 \times 10^{- 5}

3 None

4

1.8 \times 10^{- 7}

Explanation:

${NH}_{4} Cl$ is salt of weak base and strong acid, formula for $pH$ of its hydrolysis may be calculated as,
$pH = \frac{1}{2} [{pK}_{w} - {pK}_{b} - \log C]$
where $pH = 5.13$
${pK}_{W} = - \log K_{W} = - \log 10^{- 14} = 14$
$\log C = \log 0.1 = - 1$
Substituting the values in eq. (1) we get
$5.13 = \frac{1}{2} [14 - {pK}_{b} + 1] 10.26 = 15 - {pK}_{b} {pK}_{a} = 4.74 ∴ - \log K_{b} = {pK}_{b} = 4.74 K_{b} = anti \log (- 4.74) K_{b} = 1.8 \times 10^{- 5}$

CHXI07:EQUILIBRIUM

314758 Match the column I with column II and mark the appropriate choice.
Column I
Column II
A
$C H_{3} C O O N a$
P
Almost neutral, pH>7 or <7
B
$N H_{4} C l$
Q
Acidic, pH <7
C
$N a N O_{3}$
R
Alkaline, pH>7
D
$C H_{3} C O O N H_{4}$
S
Neutral, pH=7

1

A - P,    B - Q,    C - R,    D - S

2

A - Q,    B - R,    C - S,    D - P

3

A - R,    B - Q,    C - S,    D - P

4

A - S,    B - P,    C - R,    D - Q

Explanation:

(A)
${CH}_{3} COONa ⇌ \underset{weak acid}{{CH}_{3} {COO}^{-}} + \underset{strong base}{{Na}^{+}} ($ Basic $)$
(B) ${NH}_{4} Cl ⇌ \underset{Weak base}{{NH}_{4}^{+}} + \underset{Strong acid}{{Cl}^{-}} ($ Acidic)
(C) ${NaNO}_{3} ⇌ \underset{Strong base}{{Na}^{+}} + \underset{Strong acis}{{NO}_{3}^{-}} ($ Neutral $)$
(D)
${CH}_{3} {COONH}_{4} ⇌ \underset{Weak acid}{{CH}_{3} {COO}^{-}} + \underset{Weak base}{{NH}_{4}^{+}}$ (Almost neutral)
So, the option (3) is correct.

CHXI07:EQUILIBRIUM

314755 The ionisation constant of nitrous acid is $4.5 \times$ $10^{- 4}$ . The $pH$ and degree of hydrolysis of 0.04 $M$ sodium nitrite solution are, respectively.

1

7.98, 2.36 \times 10^{- 5}

2

6.98, 2.36 \times 10^{- 5}

3

7.98, 2.36 \times 10^{- 6}

4

6.98, 2.36 \times 10^{- 6}

Explanation:

$G i v e n, [NaN O_{2}] = 0 . 04 M$
$K_{a}$ of ${HNO}_{2} = 4.5 \times 10^{- 4}$
$p K_{a} = - \log K_{a} = - \log (4.5 \times 10^{- 4})$
$p K_{a} = - 4 - 0.6532 = 3.3468 \approx 3.35$
$pH = 7 + \frac{p K_{a} + \log C}{2}$
$pH = 7 + \frac{3.35 + \log 0.04}{2} = 7.98$
$pH = 7.98$ .
Degree of hydrolysis,
$h = \sqrt{\frac{K_{w}}{K_{a} C}} = \sqrt{\frac{10^{- 14}}{4.5 \times 10^{- 4} \times 0.04}}$

CHXI07:EQUILIBRIUM

314756 $pH$ of a salt solution of weak acid $({pK}_{a} = 4)$ and weak base $({pK}_{b} = 5)$ at $25^{\circ}$ is

1 6.5

2 6

3 7

4 7.5

Explanation:

$pH$ of buffer solution of weak acid and weak base is calculated, using the following formula.
$pH = \frac{1}{2} [{pK}_{w} + {pK}_{a} - {pK}_{b}]$
Given: ${pK}_{a} = 4, {pK}_{b} = 5, {pK}_{w} = 14$
$∴ pH = \frac{1}{2} [14 + 4 - 5] = 6.5$

AIIMS - 2019

CHXI07:EQUILIBRIUM

314757 The $pH$ of $0 .1 M N H_{4} Cl$ solution is 5.13. What will be the dissociation constant of ${NH}_{4} OH$ ?

1

1.8 \times 10^{- 9}

2

1.8 \times 10^{- 5}

3 None

4

1.8 \times 10^{- 7}

Explanation:

${NH}_{4} Cl$ is salt of weak base and strong acid, formula for $pH$ of its hydrolysis may be calculated as,
$pH = \frac{1}{2} [{pK}_{w} - {pK}_{b} - \log C]$
where $pH = 5.13$
${pK}_{W} = - \log K_{W} = - \log 10^{- 14} = 14$
$\log C = \log 0.1 = - 1$
Substituting the values in eq. (1) we get
$5.13 = \frac{1}{2} [14 - {pK}_{b} + 1] 10.26 = 15 - {pK}_{b} {pK}_{a} = 4.74 ∴ - \log K_{b} = {pK}_{b} = 4.74 K_{b} = anti \log (- 4.74) K_{b} = 1.8 \times 10^{- 5}$

CHXI07:EQUILIBRIUM

314758 Match the column I with column II and mark the appropriate choice.
Column I
Column II
A
$C H_{3} C O O N a$
P
Almost neutral, pH>7 or <7
B
$N H_{4} C l$
Q
Acidic, pH <7
C
$N a N O_{3}$
R
Alkaline, pH>7
D
$C H_{3} C O O N H_{4}$
S
Neutral, pH=7

1

A - P,    B - Q,    C - R,    D - S

2

A - Q,    B - R,    C - S,    D - P

3

A - R,    B - Q,    C - S,    D - P

4

A - S,    B - P,    C - R,    D - Q

Explanation:

(A)
${CH}_{3} COONa ⇌ \underset{weak acid}{{CH}_{3} {COO}^{-}} + \underset{strong base}{{Na}^{+}} ($ Basic $)$
(B) ${NH}_{4} Cl ⇌ \underset{Weak base}{{NH}_{4}^{+}} + \underset{Strong acid}{{Cl}^{-}} ($ Acidic)
(C) ${NaNO}_{3} ⇌ \underset{Strong base}{{Na}^{+}} + \underset{Strong acis}{{NO}_{3}^{-}} ($ Neutral $)$
(D)
${CH}_{3} {COONH}_{4} ⇌ \underset{Weak acid}{{CH}_{3} {COO}^{-}} + \underset{Weak base}{{NH}_{4}^{+}}$ (Almost neutral)
So, the option (3) is correct.

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