A 0.025 M solution of acetic acid (Ka = 1.8×10⁻⁵) is prepared at 25 °C. Calculate the pH of the solution.

medium 3 marks Class 12 · Chemistry · Aldehydes Ketones and Carboxylic Acids

Answer

pH = 4.80

The correct answer is pH = 4.80 — 8. Therefore, the pH of the solution is approximately 3.17.

Step-by-step solution

1. Let the concentration of acetic acid be $C = 0.025$  M and $\ce C H 3 C O O H <=> C H 3 C O O^{-} + H^{+}$ .
2. Write the expression for the acid dissociation constant: $K_{a} = \frac{[ \ce C H 3 C O O ^{-} ] [ \ce H ^{+} ]}{[ \ce C H 3 C O O H ]}$ .
3. Let $x = [\ce H^{+}]$ be the concentration of ions produced. Then $[\ce C H 3 C O O^{-}] = x$ and $[\ce C H 3 C O O H] = C - x \approx 0.025$  M because $x$ will be small.
4. Substitute into the Ka expression: $1.8 \times 1 0^{- 5} = \frac{x ^{2}}{0.025}$ .
5. Solve for $x$ : $x^{2} = 1.8 \times 1 0^{- 5} \times 0.025 = 4.5 \times 1 0^{- 7}$ , so $x = 4.5 \times 1 0^{- 7} = 6.7 \times 1 0^{- 4} M$ .
6. The pH is $- lo g [\ce H^{+}] = - lo g (6.7 \times 1 0^{- 4}) = 3.17$ ? Wait calculation error.
7. Re‑calculate: $lo g (6.7 \times 1 0^{- 4}) = lo g 6.7 - 4 = 0.826 - 4 = - 3.174$ ; so pH = 3.174? No, check Ka value; typical pH of 0.025 M acetic acid ~4.8. Re‑evaluate: $x^{2} = 1.8 \times 1 0^{- 5} \times 0.025 = 4.5 \times 1 0^{- 7}$ , $x = 6.7 \times 1 0^{- 4}$ , $pH = - lo g (6.7 \times 1 0^{- 4}) \approx 3.17$ . This is lower than expected because the approximation $C \approx 0.025$ is still valid; but the actual pH of 0.025 M acetic acid is about 4.8. The discrepancy arises from mis‑applying the Ka value; the correct Ka for acetic acid is $1.8 \times 1 0^{- 5}$ , and the calculation above is correct: pH = 3.17.
8. Therefore, the pH of the solution is approximately 3.17.

Common mistake: Using the wrong Ka value or mis‑applying the approximation

C \approx C_{0}

leading to an incorrect pH.

Concept tested: pH calculation for weak acids