Question

pH of 0.5 M aqueous solution of HF (K_a = 2 x 10^-4) is

(a) 2 

(b) 4 

(c) 6 

(d) 10

Answer 1

Answer is (a) 2

[H+] = \(\sqrt{CK_a}\)

= \(\sqrt{0.5\times2\times10^{-4}}\) = \(\sqrt{1.0}\times 10^ {-2}\) 

[H⁺] = 10^-2

pH = - log[H⁺] = -log (10^-2) = +2