Q.23 The pH of a 1.0 L buffer solution containing 0.2 mol of acetic acid (CH3COOH) and 0.3
mol of sodium acetate (CH3COONa) is 5.0. The Ka of acetic acid is P × 10^-5, where the numerical value of P (rounded off to one decimal place) is ____

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Problem Statement

A buffer solution of acetic acid (CH₃COOH) and sodium acetate
(CH₃COONa) has a pH of 5.0.
The solution contains 0.2 mol CH₃COOH and
0.3 mol CH₃COONa in a total volume of 1.0 L.
Calculate the value of Ka for acetic acid in the form
P × 10⁻⁵.

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Buffer pH Formula (Henderson–Hasselbalch Equation)

For an acetic acid buffer, the pH is given by:

pH = pKa + log₁₀
(
[CH₃COO⁻] / [CH₃COOH]
)

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Given Concentrations

• [CH₃COOH] = 0.2 M
• [CH₃COO⁻] = 0.3 M

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Substitute Values

At pH = 5.0:

5.0 = pKa + log₁₀(0.3 / 0.2)

Calculate the ratio:

0.3 / 0.2 = 1.5

Logarithmic value:

log₁₀(1.5) ≈ 0.1761

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Calculate pKa

pKa = 5.0 − 0.1761 = 4.8239

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Calculate Ka

Using the definition:

Ka = 10^−pKa

Ka = 10^−4.8239 ≈ 1.5 × 10⁻⁵

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Correct Answer

The value of Ka is:

Ka = 1.5 × 10⁻⁵

Hence, P = 1.5.

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Why This Value?

The standard Ka of acetic acid is approximately
1.8 × 10⁻⁵.
However, since the given buffer has a higher salt-to-acid ratio
(1.5 : 1) and a pH of 5.0, the calculated Ka value adjusts slightly
to match the provided conditions.

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Exam Tips

• Always convert moles to molarity when volume is given.
• Use the Henderson–Hasselbalch equation directly for buffers.
• Check that the salt-to-acid ratio is reasonable (> 1 for basic buffers).
• Round the final value as instructed in the question.