Question

A weak acid, HA is found to be 10% ionized 0.01 M aqueous. Calculate the pH of a solution which is 0.1 M in HA and 0.05 M in NaA:

• A. 5.365
• B. 6.355
• C. 3.653
• D. 6.59

Answer 1

The correct option is C 3.653

Given,

α = 0.1 and C =0.01 M

$K_a=\frac{C{\alpha }^2}{1-\alpha }=\frac{0.01\times (0.1{)}^2}{1-0.1}=1.11\times {10}^{-4}$

$p{K}_a=-log(1.11\times {10}^{-4})$ = 3.9542

$pH=p{K}_a+log\frac{\left[salt\right]}{\left[acid\right]}$

$\Rightarrow pH=3.9542+log\left[\frac{0.05}{0.10}\right]$ = 3.653

$\therefore pH=3.653$

Hence, option C is correct .