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Buffer Solutions and Titrations Practice Questions & Answers

25 Questions·Free Practice·MCQ with Explanations

Module 12: Aqueous Equilibria

Advanced acid-base chemistry focusing on resistance to pH change and analytical techniques.

Key Concepts:

Buffers: Common ion effect and preparation of buffers.
Henderson–Hasselbalch Equation: Calculating buffer pH.
Titrations: Strong Acid-Strong Base, Weak-Strong curves.
Indicators: Choosing the correct indicator for endpoints.

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Which of the following mixtures would result in a buffer solution?

$\text{HCl} \text{ and } \text{NaCl}$
$\text{NaOH} \text{ and } \text{H}_2\text{O}$
$\text{CH}_3\text{COOH} \text{ and } \text{CH}_3\text{COONa}$
$\text{NH}_3 \text{ and } \text{NaOH}$

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Correct Answer: Option C -

$\text{CH}_3\text{COOH} \text{ and } \text{CH}_3\text{COONa}$

Explanation:

A buffer solution consists of a weak acid and its conjugate base (or a weak base and its conjugate acid). Option (c) contains acetic acid (weak acid) and sodium acetate (source of conjugate base), forming a valid buffer. Option (a) involves a strong acid, and Option (d) involves two bases.

How does an acidic buffer resist changes in pH when a small amount of strong base ( $\text{OH}^-$ ) is added?

The $\text{OH}^-$ ions react with the water to form $\text{H}_3\text{O}^+$ .
The $\text{OH}^-$ ions react with the weak acid component ( $\text{HA}$ ) to produce water and the conjugate base ( $\text{A}^-$ ).
The $\text{OH}^-$ ions react with the conjugate base ( $\text{A}^-$ ) to reform the weak acid.
The buffer precipitates the $\text{OH}^-$ ions out of the solution.

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Correct Answer: Option B -

The $\text{OH}^-$ ions react with the weak acid component ( $\text{HA}$ ) to produce water and the conjugate base ( $\text{A}^-$ ).

Explanation:

When a strong base is added, the hydroxide ions ( $\text{OH}^-$ ) are neutralized by the weak acid component of the buffer: $\text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O}$ . This converts the strong base into the much weaker conjugate base, preventing a drastic pH change.

Which of the following statements best describes buffer capacity?

The ability of a solution to change color at the equivalence point.
The amount of acid or base that can be added to a buffer before the pH changes significantly.
The pH range in which an indicator works effectively.
The volume of titrant required to reach the end point.

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Correct Answer: Option B -

The amount of acid or base that can be added to a buffer before the pH changes significantly.

Explanation:

Buffer capacity is defined as the amount of strong acid or base that can be added to a given volume of a buffer solution before the pH changes significantly (usually by one unit). It depends on the absolute concentrations of the buffer components.

Which conjugate acid-base pair is primarily responsible for maintaining the pH of human blood around 7.4?

$\text{HCl} / \text{Cl}^-$
$\text{H}_2\text{PO}_4^- / \text{HPO}_4^{2-}$
$\text{CH}_3\text{COOH} / \text{CH}_3\text{COO}^-$
$\text{H}_2\text{CO}_3 / \text{HCO}_3^-$

View Answer & Explanation

Correct Answer: Option D -

$\text{H}_2\text{CO}_3 / \text{HCO}_3^-$

Explanation:

The principal buffer system in human blood is the carbonic acid/bicarbonate ion system ( $\text{H}_2\text{CO}_3 / \text{HCO}_3^-$ ). This system neutralizes metabolic acids and bases to keep blood pH within the narrow range of 7.35–7.45.

Select the correct form of the Henderson-Hasselbalch equation for a weak acid ( $\text{HA}$ ) and its conjugate base ( $\text{A}^-$ ).

$\text{pH} = \text{p}K_a - \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)$
$\text{pH} = \text{p}K_a + \log \left( \frac{[\text{HA}]}{[\text{A}^-]} \right)$
$\text{pH} = \text{p}K_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)$
$\text{pH} = \text{p}K_a \times \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)$

View Answer & Explanation

Correct Answer: Option C -

$\text{pH} = \text{p}K_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)$

Explanation:

The Henderson-Hasselbalch equation is derived from the $K_a$ expression and is written as: $\text{pH} = \text{p}K_a + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right)$ . For a weak acid/conjugate base pair, this is $\text{pH} = \text{p}K_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)$ .

What is the pH of a buffer solution where the concentration of the weak acid ( $\text{HA}$ ) is equal to the concentration of its conjugate base ( $\text{A}^-$ )?

$\text{pH} = 7$
$\text{pH} = 1$
$\text{pH} = \text{p}K_a$
$\text{pH} = 14 - \text{p}K_a$

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Correct Answer: Option C -

$\text{pH} = \text{p}K_a$

Explanation:

According to the Henderson-Hasselbalch equation: $\text{pH} = \text{p}K_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)$ . If $[\text{A}^-] = [\text{HA}]$ , then the ratio is 1, and $\log(1) = 0$ . Therefore, $\text{pH} = \text{p}K_a$ .

Calculate the pH of a buffer solution containing $0.10\,\text{M}$ acetic acid and $0.10\,\text{M}$ sodium acetate. Given: $K_a$ for acetic acid is $1.8 \times 10^{-5}$ .

2.87
4.74
5.35
9.26

View Answer & Explanation

Correct Answer: Option B -

4.74

Explanation:

First, calculate $\text{p}K_a = -\log(1.8 \times 10^{-5}) \approx 4.74$ . Since the concentrations of acid and base are equal ( $0.10\,\text{M}$ each), $\text{pH} = \text{p}K_a + \log(1) = 4.74 + 0 = 4.74$ .

A buffer solution contains $0.50\,\text{M}$ $\text{HA}$ ( $\text{p}K_a = 5.0$ ) and $0.50\,\text{M}$ $\text{A}^-$ . If $0.01\,\text{mol}$ of solid NaOH is added to $1.0\,\text{L}$ of this buffer (assuming negligible volume change), what is the new pH?

4.98
5.00
5.02
6.00

View Answer & Explanation

Correct Answer: Option C -

5.02

Explanation:

Initial moles: $\text{HA} = 0.50$ , $\text{A}^- = 0.50$ . Added NaOH ( $0.01\,\text{mol}$ ) reacts with HA: $\text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O}$ .
New moles HA = $0.50 - 0.01 = 0.49$ .
New moles $\text{A}^-$ = $0.50 + 0.01 = 0.51$ .
$\text{pH} = 5.0 + \log(0.51/0.49) \approx 5.0 + 0.017 = 5.017 \approx 5.02$ .

You need to prepare a buffer with pH 5.00 using an acid with $\text{p}K_a = 4.70$ . What ratio of $[\text{A}^-]/[\text{HA}]$ is required?

0.30
0.50
2.00
1.00

View Answer & Explanation

Correct Answer: Option C -

2.00

Explanation:

Using $\text{pH} = \text{p}K_a + \log(\text{ratio})$ :
$5.00 = 4.70 + \log(\text{ratio})$
$0.30 = \log(\text{ratio})$
$\text{ratio} = 10^{0.30} \approx 1.995 \approx 2.00$ .

In a buffer solution, if the concentration of the conjugate base ( $[\text{A}^-]$ ) is ten times the concentration of the weak acid ( $[\text{HA}]$ ), the pH of the solution will be:

$\text{p}K_a + 1$
$\text{p}K_a - 1$
$\text{p}K_a + 10$
$\text{p}K_a$

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Correct Answer: Option A -

$\text{p}K_a + 1$

Explanation:

$\text{pH} = \text{p}K_a + \log \left( \frac{10[\text{HA}]}{[\text{HA}]} \right) = \text{p}K_a + \log(10)$ .
Since $\log(10) = 1$ , the pH is $\text{p}K_a + 1$ .

Buffer Solutions and Titrations Practice Questions & Answers

Module 12: Aqueous Equilibria

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