Find pH of buffer made from 0.1M CH₃COOH and 0.1M CH₃COONa (Ka=1.8×10⁻⁵)

Question

Find the pH of a buffer solution prepared by mixing 0.1 M acetic acid (CH₃COOH) and 0.1 M sodium acetate (CH₃COONa). Given $K_a = 1.8 \times 10^{-5}$ .

Solution — Step by Step

A mixture of a weak acid and its conjugate base salt is a buffer solution. The Henderson-Hasselbalch equation is built exactly for this situation:

\text{pH} = \text{pK}_a + \log\frac{[\text{A}^-]}{[\text{HA}]}

where $[\text{A}^-]$ is the concentration of the conjugate base (sodium acetate here) and $[\text{HA}]$ is the concentration of the weak acid.

\text{pK}_a = -\log(K_a) = -\log(1.8 \times 10^{-5})

= -(\log 1.8 + \log 10^{-5}) = -(0.255 - 5) = 4.745

We round this to pKa ≈ 4.74 (you’ll see 4.74 or 4.75 in different textbooks — either is acceptable).

Both concentrations are equal: $[\text{CH}_3\text{COO}^-] = 0.1\text{ M}$ and $[\text{CH}_3\text{COOH}] = 0.1\text{ M}$ .

\text{pH} = 4.74 + \log\frac{0.1}{0.1} = 4.74 + \log(1)

Since $\log(1) = 0$ :

\text{pH} = 4.74 + 0 = \textbf{4.74}

When the weak acid and its conjugate base are present in equal concentrations, the pH equals the pKa. This is the midpoint of the buffer — and it’s where the buffer has maximum capacity.

Why This Works

The Henderson-Hasselbalch equation comes from the equilibrium expression for the weak acid. When you take $-\log$ of both sides of $K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$ , you get the HH equation directly.

The key insight: pH depends on the ratio of $[\text{A}^-]$ to $[\text{HA}]$ , not on their absolute values. So a buffer made with 0.5 M of each gives the same pH as one made with 0.1 M of each — the ratio is still 1:1, so pH = pKa.

This is why equal concentrations give pH = pKa — it’s not a coincidence, it’s the mathematical consequence of log(1) = 0.

Alternative Method

We can solve from first principles without memorising the HH equation:

K_a = \frac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}

1.8 \times 10^{-5} = \frac{[\text{H}^+] \times 0.1}{0.1}

[\text{H}^+] = 1.8 \times 10^{-5} \text{ M}

\text{pH} = -\log(1.8 \times 10^{-5}) = 4.74

Same answer. In JEE problems, the HH form is faster; for CBSE derivation questions, showing the equilibrium step earns full marks.

This exact problem type appears frequently in JEE Main and CBSE Class 11 ionic equilibrium. When concentrations are equal, pH = pKa is the answer — learn to spot this instantly. If the ratio is something like 10:1 or 1:10, pH = pKa ± 1.

Common Mistake

Students write $[\text{A}^-]/[\text{HA}]$ but accidentally flip it to $[\text{HA}]/[\text{A}^-]$ . Remember: the numerator in the log is always the conjugate base (the anion, A⁻). If you flip it, you get 4.74 − 0 = 4.74 here (still correct because they’re equal), but for unequal concentrations you’ll get the wrong sign on the log term.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next