Question

Calculate the pH of buffer solution containing 0.1M \(\text{CH}_3\text{COOH}\) and 0.1 M \(\text{CH}_3\text{COONa}\) \([K_a = 1.8\times 10^{-5}]\)

• A. \(4.745\)
• B. \(5.741\)
• C. \(2.876\)
• D. \(7.001\)

Hint:

Use the Henderson-Hasselbalch equation; equal acid and salt concentrations give pH = pKa.

Answer 1

The Correct Option is A

Step 1: Approach
Write the equilibrium expression and read off $[\text{H}^+]$ directly.

Step 2: Expression
For $\text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+$:
\[ K_a=\frac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} \]

Step 3: Substitute
Acetate and acid are both 0.1 M (the weak acid ionises very little), so the two concentrations cancel and $[\text{H}^+]=K_a=1.8\times10^{-5}$ M.

Step 4: pH
\[ \text{pH}=-\log(1.8\times10^{-5})=4.745 \]
This is option (A).

Final Answer:
The buffer has equal acid and salt, so pH equals pKa, which is 4.745, option (A). \[ \boxed{4.745} \]