$[H^+] = C, \alpha = 2 \times 10^{-3} \times 0.1$.
$ = 2 \times {10^{ - 3}} \times \frac{1}{{10}} = 2 \times {10^{ - 4}}$
$= 2 \times 10^{-4}$
$pH = - log [H^+] = -log(2 \times 10^{-4})$
$= -log 2 - log 10^{-4} = -0.3010 + 4 = 3.7$