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Question: The pH of 0.5 M aqueous solution of HF (K<sub>a</sub> = 2 × 10<sup>–4</sup>) is –...

The pH of 0.5 M aqueous solution of HF (Ka = 2 × 10–4) is –

A

2

B

4

C

6

D

10

Answer

2

Explanation

Solution

[H+] = CKa=0.5×2×10−4\sqrt{CK_{a}} = \sqrt{0.5 \times 2 \times 10^{- 4}} = 10–2 M

pH = – log10 [H+] = – log 10–2 = 2