QBManager

Ionic Equilibrium

229723 A buffer solution has equal volumes of $0.2 M$ ${NH}_{4} OH$ and $0.02 M NH {NC}_{4}$ . The ${pK}_{b}$ of the base is 5 . The $pH$ is

1 10

2 9

3 4

4 7

Explanation:

Given thatvolume of ${NH}_{4} OH = 0.2 M$ and volume of ${NH}_{4} Cl = 0.02 M$ ${pK}_{b} = 5$
According to the Henderson's equation.
$\begin{aligned} pOH = {pK}_{b} + \log \frac{[salt]}{[base]} \\ = 5 + \log \frac{0.02}{0.2} = 5 + \log \frac{1}{10} = 5 + (- 1) = 4 \\ pH = 14 - pOH = 14 - 4 = 10 \end{aligned}$

COMEDK-2017**

Ionic Equilibrium

229724 When 0.1 mole of an acid is added to $2 L$ of a buffer solution, the $pH$ of the buffer decreases by 0.5 . The buffer capacity of the solution is

1 0.6

2 0.4

3 0.2

4 0.1

Explanation:

Given data:
Number of moles of acid added $= \frac{0.1}{2}$ per litre Change of $pH = 0.5$
Now,
Buffer capacity $= \frac{number of moles of acid added/L}{change in pH}$
Buffer capacity $= \frac{0.1}{0.5 \times 2}$
Buffer capacity $= 0.1$

2008

Ionic Equilibrium

229726 Assertion : Mixture of ${CH}_{3} COOH$ and
**Reason :** Acidic buffer contains equimolar mixture of a weak acid and its salt with weak base.

1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

2 If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.

3 In the Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

5 If the assertion is incorrect but the Reason is correct.

Explanation:

Acidic buffer contains equimolar mixture of weak acid and its salt with strong base.
${CH}_{3} COOH / {CH}_{3} {COONH}_{4}$ is not an example of acidic buffer.

AIIMS-(2007)

Ionic Equilibrium

229730 For a buffer of a mixture of $0.12 mol L^{- 1}$ ${CH}_{3} COOH$ and $0.12 mol L^{- 1} {CH}_{3} COONa$ , the buffer capacity is

1 1.38

2 0.130

3 0.06

4 0.60

Explanation:

Buffer capacity $(β) = \frac{Δ C_{a}}{Δ pH}$
Where $C_{a} =$ concentration of acid for acidic buffer
Concentration of acid = concentration of salt $∴ Δ C_{a} = 0.12 {molL}^{- 1}$
Hence, we use the following formula for calculating its $pH$ . ${pH}_{pH}^{pH} - \log [H^{+}] = - \log (0.12) = 0.92$
Buffer capacity $(β) = \frac{0.12}{0.92} = \frac{12}{92} = \frac{6}{46} = \frac{3}{23} = 0.130$

AMU-2017

Ionic Equilibrium

229723 A buffer solution has equal volumes of $0.2 M$ ${NH}_{4} OH$ and $0.02 M NH {NC}_{4}$ . The ${pK}_{b}$ of the base is 5 . The $pH$ is

1 10

2 9

3 4

4 7

Explanation:

Given thatvolume of ${NH}_{4} OH = 0.2 M$ and volume of ${NH}_{4} Cl = 0.02 M$ ${pK}_{b} = 5$
According to the Henderson's equation.
$\begin{aligned} pOH = {pK}_{b} + \log \frac{[salt]}{[base]} \\ = 5 + \log \frac{0.02}{0.2} = 5 + \log \frac{1}{10} = 5 + (- 1) = 4 \\ pH = 14 - pOH = 14 - 4 = 10 \end{aligned}$

COMEDK-2017**

Ionic Equilibrium

229724 When 0.1 mole of an acid is added to $2 L$ of a buffer solution, the $pH$ of the buffer decreases by 0.5 . The buffer capacity of the solution is

1 0.6

2 0.4

3 0.2

4 0.1

Explanation:

Given data:
Number of moles of acid added $= \frac{0.1}{2}$ per litre Change of $pH = 0.5$
Now,
Buffer capacity $= \frac{number of moles of acid added/L}{change in pH}$
Buffer capacity $= \frac{0.1}{0.5 \times 2}$
Buffer capacity $= 0.1$

2008

Ionic Equilibrium

229726 Assertion : Mixture of ${CH}_{3} COOH$ and
**Reason :** Acidic buffer contains equimolar mixture of a weak acid and its salt with weak base.

1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

2 If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.

3 In the Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

5 If the assertion is incorrect but the Reason is correct.

Explanation:

Acidic buffer contains equimolar mixture of weak acid and its salt with strong base.
${CH}_{3} COOH / {CH}_{3} {COONH}_{4}$ is not an example of acidic buffer.

AIIMS-(2007)

Ionic Equilibrium

229730 For a buffer of a mixture of $0.12 mol L^{- 1}$ ${CH}_{3} COOH$ and $0.12 mol L^{- 1} {CH}_{3} COONa$ , the buffer capacity is

1 1.38

2 0.130

3 0.06

4 0.60

Explanation:

Buffer capacity $(β) = \frac{Δ C_{a}}{Δ pH}$
Where $C_{a} =$ concentration of acid for acidic buffer
Concentration of acid = concentration of salt $∴ Δ C_{a} = 0.12 {molL}^{- 1}$
Hence, we use the following formula for calculating its $pH$ . ${pH}_{pH}^{pH} - \log [H^{+}] = - \log (0.12) = 0.92$
Buffer capacity $(β) = \frac{0.12}{0.92} = \frac{12}{92} = \frac{6}{46} = \frac{3}{23} = 0.130$

AMU-2017

Ionic Equilibrium

229723 A buffer solution has equal volumes of $0.2 M$ ${NH}_{4} OH$ and $0.02 M NH {NC}_{4}$ . The ${pK}_{b}$ of the base is 5 . The $pH$ is

1 10

2 9

3 4

4 7

Explanation:

Given thatvolume of ${NH}_{4} OH = 0.2 M$ and volume of ${NH}_{4} Cl = 0.02 M$ ${pK}_{b} = 5$
According to the Henderson's equation.
$\begin{aligned} pOH = {pK}_{b} + \log \frac{[salt]}{[base]} \\ = 5 + \log \frac{0.02}{0.2} = 5 + \log \frac{1}{10} = 5 + (- 1) = 4 \\ pH = 14 - pOH = 14 - 4 = 10 \end{aligned}$

COMEDK-2017**

Ionic Equilibrium

229724 When 0.1 mole of an acid is added to $2 L$ of a buffer solution, the $pH$ of the buffer decreases by 0.5 . The buffer capacity of the solution is

1 0.6

2 0.4

3 0.2

4 0.1

Explanation:

Given data:
Number of moles of acid added $= \frac{0.1}{2}$ per litre Change of $pH = 0.5$
Now,
Buffer capacity $= \frac{number of moles of acid added/L}{change in pH}$
Buffer capacity $= \frac{0.1}{0.5 \times 2}$
Buffer capacity $= 0.1$

2008

Ionic Equilibrium

229726 Assertion : Mixture of ${CH}_{3} COOH$ and
**Reason :** Acidic buffer contains equimolar mixture of a weak acid and its salt with weak base.

1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

2 If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.

3 In the Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

5 If the assertion is incorrect but the Reason is correct.

Explanation:

Acidic buffer contains equimolar mixture of weak acid and its salt with strong base.
${CH}_{3} COOH / {CH}_{3} {COONH}_{4}$ is not an example of acidic buffer.

AIIMS-(2007)

Ionic Equilibrium

229730 For a buffer of a mixture of $0.12 mol L^{- 1}$ ${CH}_{3} COOH$ and $0.12 mol L^{- 1} {CH}_{3} COONa$ , the buffer capacity is

1 1.38

2 0.130

3 0.06

4 0.60

Explanation:

Buffer capacity $(β) = \frac{Δ C_{a}}{Δ pH}$
Where $C_{a} =$ concentration of acid for acidic buffer
Concentration of acid = concentration of salt $∴ Δ C_{a} = 0.12 {molL}^{- 1}$
Hence, we use the following formula for calculating its $pH$ . ${pH}_{pH}^{pH} - \log [H^{+}] = - \log (0.12) = 0.92$
Buffer capacity $(β) = \frac{0.12}{0.92} = \frac{12}{92} = \frac{6}{46} = \frac{3}{23} = 0.130$

AMU-2017

Ionic Equilibrium

229723 A buffer solution has equal volumes of $0.2 M$ ${NH}_{4} OH$ and $0.02 M NH {NC}_{4}$ . The ${pK}_{b}$ of the base is 5 . The $pH$ is

1 10

2 9

3 4

4 7

Explanation:

Given thatvolume of ${NH}_{4} OH = 0.2 M$ and volume of ${NH}_{4} Cl = 0.02 M$ ${pK}_{b} = 5$
According to the Henderson's equation.
$\begin{aligned} pOH = {pK}_{b} + \log \frac{[salt]}{[base]} \\ = 5 + \log \frac{0.02}{0.2} = 5 + \log \frac{1}{10} = 5 + (- 1) = 4 \\ pH = 14 - pOH = 14 - 4 = 10 \end{aligned}$

COMEDK-2017**

Ionic Equilibrium

229724 When 0.1 mole of an acid is added to $2 L$ of a buffer solution, the $pH$ of the buffer decreases by 0.5 . The buffer capacity of the solution is

1 0.6

2 0.4

3 0.2

4 0.1

Explanation:

Given data:
Number of moles of acid added $= \frac{0.1}{2}$ per litre Change of $pH = 0.5$
Now,
Buffer capacity $= \frac{number of moles of acid added/L}{change in pH}$
Buffer capacity $= \frac{0.1}{0.5 \times 2}$
Buffer capacity $= 0.1$

2008

Ionic Equilibrium

229726 Assertion : Mixture of ${CH}_{3} COOH$ and
**Reason :** Acidic buffer contains equimolar mixture of a weak acid and its salt with weak base.

1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

2 If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.

3 In the Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

5 If the assertion is incorrect but the Reason is correct.

Explanation:

Acidic buffer contains equimolar mixture of weak acid and its salt with strong base.
${CH}_{3} COOH / {CH}_{3} {COONH}_{4}$ is not an example of acidic buffer.

AIIMS-(2007)

Ionic Equilibrium

229730 For a buffer of a mixture of $0.12 mol L^{- 1}$ ${CH}_{3} COOH$ and $0.12 mol L^{- 1} {CH}_{3} COONa$ , the buffer capacity is

1 1.38

2 0.130

3 0.06

4 0.60

Explanation:

Buffer capacity $(β) = \frac{Δ C_{a}}{Δ pH}$
Where $C_{a} =$ concentration of acid for acidic buffer
Concentration of acid = concentration of salt $∴ Δ C_{a} = 0.12 {molL}^{- 1}$
Hence, we use the following formula for calculating its $pH$ . ${pH}_{pH}^{pH} - \log [H^{+}] = - \log (0.12) = 0.92$
Buffer capacity $(β) = \frac{0.12}{0.92} = \frac{12}{92} = \frac{6}{46} = \frac{3}{23} = 0.130$

AMU-2017