Question

The $pH$ of a 0.005 M $H_{2}S{O}_4$ solution is:

• A. 3.3
• B. 5
• C. 2
• D. 4

Answer 1

Answer: (C)

2

$H_{2}S{O}_4$ is a strong diprotic acid. The equation for ionization is

$H_{2}S{O}_4\to 2H^{+}+S{O}_4^2-$

Each molecule of acid gives 2 molecules of $H^{+}$ .

The concentration of $H^{+}$ ion is 2 x 0.005 = 0.010M

$pH=-log\left[H^{+}\right]=-log\left[0.01\right]=2$.

$pH=2$

Answer is option $C$.