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CHXI07:EQUILIBRIUM

314522 pH of an aqueous solution of $0.6 M {NH}_{3}$ and $0.4 M {NH}_{4} Cl$ is 9.4 $(p K_{b} = 4.74)$ . The new pH when $0.1 M Ca (OH)_{2}$ solution is added to it is

1 9.86

2 3.75

3 7.25

4 10.5

Explanation:

Since $Ca (OH)_{2}$ is completely ionised.
supporting img

BBB rule: On adding base,to hte basic buffer,cocentration of base increases and salt decreases.
$∴$ New concentartaion of base and salt are:
$[$ Base $] = [{NH}_{3}] = 0.6 + 0.2 = 0.8 M$
$[$ Salt $] = [{NH}_{4} Cl] = 0.4 - 0.2 = 0.2 M$
$∴ pOH = p K_{b} + \log [\frac{Salt}{Base}]$
$\begin{array}{l} p O H = 4.74 + \log (\frac{0.2}{0.8}) \\ = 4.74 - 2 \log 2 = 4.74 - 2 \times 0.30 \\ = 4.14 \\ p H = 14 - 4.14 = 9.86 \end{array}$

CHXI07:EQUILIBRIUM

314523 Which one of the following mixture does not act as a buffer solution?

1 Boric acid an borax

2 Sodium acetate and sodium propionate

3 Sodium phosphate and disodium hydrogen phosphate

4 Sodium propionate and propionic acid

CHXI07:EQUILIBRIUM

314524 Read the Statement -A and Statement -B carefully to mark the correct options given below:
Statement A :
Mixture of ${CH}_{3} COOH$ and ${CH}_{3} {COONH}_{4}$ is an example of acidic buffer.
Statement B :
Acidic buffer contains equimolar mixture of a weak acid and its salt with strong base.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

Acidic buffer contains a mixture of a weak acid and a salt of this weak acid with a strong base. Also, ${CH}_{3} COOH$ is a weak acid and
${CH}_{3} {COONH}_{4}$ is a salt of ${CH}_{3} COOH$ with weak base. So, the mixture of ${CH}_{3} COOH$ and ${CH}_{3} {COONH}_{4}$ is not an example of acidic buffer. Correct Option is 2.

CHXI07:EQUILIBRIUM

314525 The $pH$ of an acid buffer can be raised by 2 units by

1 Increasing the concentration of both weak acid and salt by two moles

2 Increasing the concentration of both the acid and salt by 10 times

3 Diluting the solution by 10 times

4 Increasing the concentration of the salt by 10 times and decreasing concentration of the acid by 10 times.

Explanation:

$p H_{1} = p K_{a} + log (\frac{Salt}{Acid})$
$p H_{1} = p K_{a} + log (\frac{x_{1}}{y_{1}})$ , $p H_{2} = p K_{a} + log (\frac{x_{2}}{y_{2}})$
$p H_{2} - p H_{1} = log \frac{x_{2}}{y_{2}} - log \frac{x_{1}}{y_{1}}$ $2 = log (\frac{x_{2} / y_{2}}{x_{1} / y_{1}})$ $∴ \frac{x_{2} / y_{2}}{x_{1} / y_{1}} = 1 0^{2} = 100$
This is only possible, if the concentration of salt is increased by 10 times and the concentration of acid is decreased by 10 times.

CHXI07:EQUILIBRIUM

314526 To a buffer solution of ${CH}_{3} COOH$ and ${CH}_{3} COONa$ , some $HCl$ is added. The excess of $[H^{+}]$ ions are removed by the following equation

1

{CH}_{3} COOH + {OH}^{-} \to {CH}_{3} {COO}^{-} + H_{2} O

2

{CH}_{3} {COO}^{-} + H^{+} \to {CH}_{3} COOH

3

{Na}^{+} + {OH}^{-} \to NaOH

4

{CH}_{3} {COO}^{-} + {Na}^{+} \to {CH}_{3} COONa

Explanation:

$\underset{(from buffer)}{{CH}_{3} {COO}^{-}} + \underset{(from acid)}{H^{+}} \to {CH}_{3} COOH$

CHXI07:EQUILIBRIUM

314522 pH of an aqueous solution of $0.6 M {NH}_{3}$ and $0.4 M {NH}_{4} Cl$ is 9.4 $(p K_{b} = 4.74)$ . The new pH when $0.1 M Ca (OH)_{2}$ solution is added to it is

1 9.86

2 3.75

3 7.25

4 10.5

Explanation:

Since $Ca (OH)_{2}$ is completely ionised.
supporting img

BBB rule: On adding base,to hte basic buffer,cocentration of base increases and salt decreases.
$∴$ New concentartaion of base and salt are:
$[$ Base $] = [{NH}_{3}] = 0.6 + 0.2 = 0.8 M$
$[$ Salt $] = [{NH}_{4} Cl] = 0.4 - 0.2 = 0.2 M$
$∴ pOH = p K_{b} + \log [\frac{Salt}{Base}]$
$\begin{array}{l} p O H = 4.74 + \log (\frac{0.2}{0.8}) \\ = 4.74 - 2 \log 2 = 4.74 - 2 \times 0.30 \\ = 4.14 \\ p H = 14 - 4.14 = 9.86 \end{array}$

CHXI07:EQUILIBRIUM

314523 Which one of the following mixture does not act as a buffer solution?

1 Boric acid an borax

2 Sodium acetate and sodium propionate

3 Sodium phosphate and disodium hydrogen phosphate

4 Sodium propionate and propionic acid

CHXI07:EQUILIBRIUM

314524 Read the Statement -A and Statement -B carefully to mark the correct options given below:
Statement A :
Mixture of ${CH}_{3} COOH$ and ${CH}_{3} {COONH}_{4}$ is an example of acidic buffer.
Statement B :
Acidic buffer contains equimolar mixture of a weak acid and its salt with strong base.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

Acidic buffer contains a mixture of a weak acid and a salt of this weak acid with a strong base. Also, ${CH}_{3} COOH$ is a weak acid and
${CH}_{3} {COONH}_{4}$ is a salt of ${CH}_{3} COOH$ with weak base. So, the mixture of ${CH}_{3} COOH$ and ${CH}_{3} {COONH}_{4}$ is not an example of acidic buffer. Correct Option is 2.

CHXI07:EQUILIBRIUM

314525 The $pH$ of an acid buffer can be raised by 2 units by

1 Increasing the concentration of both weak acid and salt by two moles

2 Increasing the concentration of both the acid and salt by 10 times

3 Diluting the solution by 10 times

4 Increasing the concentration of the salt by 10 times and decreasing concentration of the acid by 10 times.

Explanation:

$p H_{1} = p K_{a} + log (\frac{Salt}{Acid})$
$p H_{1} = p K_{a} + log (\frac{x_{1}}{y_{1}})$ , $p H_{2} = p K_{a} + log (\frac{x_{2}}{y_{2}})$
$p H_{2} - p H_{1} = log \frac{x_{2}}{y_{2}} - log \frac{x_{1}}{y_{1}}$ $2 = log (\frac{x_{2} / y_{2}}{x_{1} / y_{1}})$ $∴ \frac{x_{2} / y_{2}}{x_{1} / y_{1}} = 1 0^{2} = 100$
This is only possible, if the concentration of salt is increased by 10 times and the concentration of acid is decreased by 10 times.

CHXI07:EQUILIBRIUM

314526 To a buffer solution of ${CH}_{3} COOH$ and ${CH}_{3} COONa$ , some $HCl$ is added. The excess of $[H^{+}]$ ions are removed by the following equation

1

{CH}_{3} COOH + {OH}^{-} \to {CH}_{3} {COO}^{-} + H_{2} O

2

{CH}_{3} {COO}^{-} + H^{+} \to {CH}_{3} COOH

3

{Na}^{+} + {OH}^{-} \to NaOH

4

{CH}_{3} {COO}^{-} + {Na}^{+} \to {CH}_{3} COONa

Explanation:

$\underset{(from buffer)}{{CH}_{3} {COO}^{-}} + \underset{(from acid)}{H^{+}} \to {CH}_{3} COOH$

CHXI07:EQUILIBRIUM

314522 pH of an aqueous solution of $0.6 M {NH}_{3}$ and $0.4 M {NH}_{4} Cl$ is 9.4 $(p K_{b} = 4.74)$ . The new pH when $0.1 M Ca (OH)_{2}$ solution is added to it is

1 9.86

2 3.75

3 7.25

4 10.5

Explanation:

Since $Ca (OH)_{2}$ is completely ionised.
supporting img

BBB rule: On adding base,to hte basic buffer,cocentration of base increases and salt decreases.
$∴$ New concentartaion of base and salt are:
$[$ Base $] = [{NH}_{3}] = 0.6 + 0.2 = 0.8 M$
$[$ Salt $] = [{NH}_{4} Cl] = 0.4 - 0.2 = 0.2 M$
$∴ pOH = p K_{b} + \log [\frac{Salt}{Base}]$
$\begin{array}{l} p O H = 4.74 + \log (\frac{0.2}{0.8}) \\ = 4.74 - 2 \log 2 = 4.74 - 2 \times 0.30 \\ = 4.14 \\ p H = 14 - 4.14 = 9.86 \end{array}$

CHXI07:EQUILIBRIUM

314523 Which one of the following mixture does not act as a buffer solution?

1 Boric acid an borax

2 Sodium acetate and sodium propionate

3 Sodium phosphate and disodium hydrogen phosphate

4 Sodium propionate and propionic acid

CHXI07:EQUILIBRIUM

314524 Read the Statement -A and Statement -B carefully to mark the correct options given below:
Statement A :
Mixture of ${CH}_{3} COOH$ and ${CH}_{3} {COONH}_{4}$ is an example of acidic buffer.
Statement B :
Acidic buffer contains equimolar mixture of a weak acid and its salt with strong base.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

Acidic buffer contains a mixture of a weak acid and a salt of this weak acid with a strong base. Also, ${CH}_{3} COOH$ is a weak acid and
${CH}_{3} {COONH}_{4}$ is a salt of ${CH}_{3} COOH$ with weak base. So, the mixture of ${CH}_{3} COOH$ and ${CH}_{3} {COONH}_{4}$ is not an example of acidic buffer. Correct Option is 2.

CHXI07:EQUILIBRIUM

314525 The $pH$ of an acid buffer can be raised by 2 units by

1 Increasing the concentration of both weak acid and salt by two moles

2 Increasing the concentration of both the acid and salt by 10 times

3 Diluting the solution by 10 times

4 Increasing the concentration of the salt by 10 times and decreasing concentration of the acid by 10 times.

Explanation:

$p H_{1} = p K_{a} + log (\frac{Salt}{Acid})$
$p H_{1} = p K_{a} + log (\frac{x_{1}}{y_{1}})$ , $p H_{2} = p K_{a} + log (\frac{x_{2}}{y_{2}})$
$p H_{2} - p H_{1} = log \frac{x_{2}}{y_{2}} - log \frac{x_{1}}{y_{1}}$ $2 = log (\frac{x_{2} / y_{2}}{x_{1} / y_{1}})$ $∴ \frac{x_{2} / y_{2}}{x_{1} / y_{1}} = 1 0^{2} = 100$
This is only possible, if the concentration of salt is increased by 10 times and the concentration of acid is decreased by 10 times.

CHXI07:EQUILIBRIUM

314526 To a buffer solution of ${CH}_{3} COOH$ and ${CH}_{3} COONa$ , some $HCl$ is added. The excess of $[H^{+}]$ ions are removed by the following equation

1

{CH}_{3} COOH + {OH}^{-} \to {CH}_{3} {COO}^{-} + H_{2} O

2

{CH}_{3} {COO}^{-} + H^{+} \to {CH}_{3} COOH

3

{Na}^{+} + {OH}^{-} \to NaOH

4

{CH}_{3} {COO}^{-} + {Na}^{+} \to {CH}_{3} COONa

Explanation:

$\underset{(from buffer)}{{CH}_{3} {COO}^{-}} + \underset{(from acid)}{H^{+}} \to {CH}_{3} COOH$

CHXI07:EQUILIBRIUM

314522 pH of an aqueous solution of $0.6 M {NH}_{3}$ and $0.4 M {NH}_{4} Cl$ is 9.4 $(p K_{b} = 4.74)$ . The new pH when $0.1 M Ca (OH)_{2}$ solution is added to it is

1 9.86

2 3.75

3 7.25

4 10.5

Explanation:

Since $Ca (OH)_{2}$ is completely ionised.
supporting img

BBB rule: On adding base,to hte basic buffer,cocentration of base increases and salt decreases.
$∴$ New concentartaion of base and salt are:
$[$ Base $] = [{NH}_{3}] = 0.6 + 0.2 = 0.8 M$
$[$ Salt $] = [{NH}_{4} Cl] = 0.4 - 0.2 = 0.2 M$
$∴ pOH = p K_{b} + \log [\frac{Salt}{Base}]$
$\begin{array}{l} p O H = 4.74 + \log (\frac{0.2}{0.8}) \\ = 4.74 - 2 \log 2 = 4.74 - 2 \times 0.30 \\ = 4.14 \\ p H = 14 - 4.14 = 9.86 \end{array}$

CHXI07:EQUILIBRIUM

314523 Which one of the following mixture does not act as a buffer solution?

1 Boric acid an borax

2 Sodium acetate and sodium propionate

3 Sodium phosphate and disodium hydrogen phosphate

4 Sodium propionate and propionic acid

CHXI07:EQUILIBRIUM

314524 Read the Statement -A and Statement -B carefully to mark the correct options given below:
Statement A :
Mixture of ${CH}_{3} COOH$ and ${CH}_{3} {COONH}_{4}$ is an example of acidic buffer.
Statement B :
Acidic buffer contains equimolar mixture of a weak acid and its salt with strong base.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

Acidic buffer contains a mixture of a weak acid and a salt of this weak acid with a strong base. Also, ${CH}_{3} COOH$ is a weak acid and
${CH}_{3} {COONH}_{4}$ is a salt of ${CH}_{3} COOH$ with weak base. So, the mixture of ${CH}_{3} COOH$ and ${CH}_{3} {COONH}_{4}$ is not an example of acidic buffer. Correct Option is 2.

CHXI07:EQUILIBRIUM

314525 The $pH$ of an acid buffer can be raised by 2 units by

1 Increasing the concentration of both weak acid and salt by two moles

2 Increasing the concentration of both the acid and salt by 10 times

3 Diluting the solution by 10 times

4 Increasing the concentration of the salt by 10 times and decreasing concentration of the acid by 10 times.

Explanation:

$p H_{1} = p K_{a} + log (\frac{Salt}{Acid})$
$p H_{1} = p K_{a} + log (\frac{x_{1}}{y_{1}})$ , $p H_{2} = p K_{a} + log (\frac{x_{2}}{y_{2}})$
$p H_{2} - p H_{1} = log \frac{x_{2}}{y_{2}} - log \frac{x_{1}}{y_{1}}$ $2 = log (\frac{x_{2} / y_{2}}{x_{1} / y_{1}})$ $∴ \frac{x_{2} / y_{2}}{x_{1} / y_{1}} = 1 0^{2} = 100$
This is only possible, if the concentration of salt is increased by 10 times and the concentration of acid is decreased by 10 times.

CHXI07:EQUILIBRIUM

314526 To a buffer solution of ${CH}_{3} COOH$ and ${CH}_{3} COONa$ , some $HCl$ is added. The excess of $[H^{+}]$ ions are removed by the following equation

1

{CH}_{3} COOH + {OH}^{-} \to {CH}_{3} {COO}^{-} + H_{2} O

2

{CH}_{3} {COO}^{-} + H^{+} \to {CH}_{3} COOH

3

{Na}^{+} + {OH}^{-} \to NaOH

4

{CH}_{3} {COO}^{-} + {Na}^{+} \to {CH}_{3} COONa

Explanation:

$\underset{(from buffer)}{{CH}_{3} {COO}^{-}} + \underset{(from acid)}{H^{+}} \to {CH}_{3} COOH$

CHXI07:EQUILIBRIUM

314522 pH of an aqueous solution of $0.6 M {NH}_{3}$ and $0.4 M {NH}_{4} Cl$ is 9.4 $(p K_{b} = 4.74)$ . The new pH when $0.1 M Ca (OH)_{2}$ solution is added to it is

1 9.86

2 3.75

3 7.25

4 10.5

Explanation:

Since $Ca (OH)_{2}$ is completely ionised.
supporting img

BBB rule: On adding base,to hte basic buffer,cocentration of base increases and salt decreases.
$∴$ New concentartaion of base and salt are:
$[$ Base $] = [{NH}_{3}] = 0.6 + 0.2 = 0.8 M$
$[$ Salt $] = [{NH}_{4} Cl] = 0.4 - 0.2 = 0.2 M$
$∴ pOH = p K_{b} + \log [\frac{Salt}{Base}]$
$\begin{array}{l} p O H = 4.74 + \log (\frac{0.2}{0.8}) \\ = 4.74 - 2 \log 2 = 4.74 - 2 \times 0.30 \\ = 4.14 \\ p H = 14 - 4.14 = 9.86 \end{array}$

CHXI07:EQUILIBRIUM

314523 Which one of the following mixture does not act as a buffer solution?

1 Boric acid an borax

2 Sodium acetate and sodium propionate

3 Sodium phosphate and disodium hydrogen phosphate

4 Sodium propionate and propionic acid

CHXI07:EQUILIBRIUM

314524 Read the Statement -A and Statement -B carefully to mark the correct options given below:
Statement A :
Mixture of ${CH}_{3} COOH$ and ${CH}_{3} {COONH}_{4}$ is an example of acidic buffer.
Statement B :
Acidic buffer contains equimolar mixture of a weak acid and its salt with strong base.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

Acidic buffer contains a mixture of a weak acid and a salt of this weak acid with a strong base. Also, ${CH}_{3} COOH$ is a weak acid and
${CH}_{3} {COONH}_{4}$ is a salt of ${CH}_{3} COOH$ with weak base. So, the mixture of ${CH}_{3} COOH$ and ${CH}_{3} {COONH}_{4}$ is not an example of acidic buffer. Correct Option is 2.

CHXI07:EQUILIBRIUM

314525 The $pH$ of an acid buffer can be raised by 2 units by

1 Increasing the concentration of both weak acid and salt by two moles

2 Increasing the concentration of both the acid and salt by 10 times

3 Diluting the solution by 10 times

4 Increasing the concentration of the salt by 10 times and decreasing concentration of the acid by 10 times.

Explanation:

$p H_{1} = p K_{a} + log (\frac{Salt}{Acid})$
$p H_{1} = p K_{a} + log (\frac{x_{1}}{y_{1}})$ , $p H_{2} = p K_{a} + log (\frac{x_{2}}{y_{2}})$
$p H_{2} - p H_{1} = log \frac{x_{2}}{y_{2}} - log \frac{x_{1}}{y_{1}}$ $2 = log (\frac{x_{2} / y_{2}}{x_{1} / y_{1}})$ $∴ \frac{x_{2} / y_{2}}{x_{1} / y_{1}} = 1 0^{2} = 100$
This is only possible, if the concentration of salt is increased by 10 times and the concentration of acid is decreased by 10 times.

CHXI07:EQUILIBRIUM

314526 To a buffer solution of ${CH}_{3} COOH$ and ${CH}_{3} COONa$ , some $HCl$ is added. The excess of $[H^{+}]$ ions are removed by the following equation

1

{CH}_{3} COOH + {OH}^{-} \to {CH}_{3} {COO}^{-} + H_{2} O

2

{CH}_{3} {COO}^{-} + H^{+} \to {CH}_{3} COOH

3

{Na}^{+} + {OH}^{-} \to NaOH

4

{CH}_{3} {COO}^{-} + {Na}^{+} \to {CH}_{3} COONa

Explanation:

$\underset{(from buffer)}{{CH}_{3} {COO}^{-}} + \underset{(from acid)}{H^{+}} \to {CH}_{3} COOH$

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