Discussion
4 g of NaOH were dissolved in 1 litre of a solution containing 1 mole of CH3COOH and 1 mole of CH3COONa. The [H+] in the resultant solution is
(Given: Ka (CH3COOH) = 1.1×10-5)
Answer: E
The hydrogen ion concentration is \(0.9 \times 10^{-5}\) M.
4 g of NaOH is 0.1 mol. It neutralises 0.1 mol of the acetic acid, forming 0.1 mol more acetate, so the buffer now holds 0.9 mol acid and 1.1 mol salt in 1 L. For an acidic buffer:
\[ \begin{aligned} [\mathrm{H^{+}}] &= K_{a} \times \frac{[\text{acid}]}{[\text{salt}]} \\ &= 1.1 \times 10^{-5} \times \frac{0.9}{1.1} = 0.9 \times 10^{-5}\ \text{M} \end{aligned} \]
The added base lowers \([\mathrm{H^{+}}]\) only slightly from \(1.1 \times 10^{-5}\) M, which is the buffer action. Every other option is above \(1.1 \times 10^{-5}\) M, which would mean adding a base made the solution more acidic.
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