Question

The pH of $1.0M$ $NaHS{O}_4$ solution will be:

(Given that $K_1$ and $K_2$ for $H_{2}S{O}_4$ equal to $\infty$ and ${10}^{-2}$ respectively)

• A. nearly 1.0
• B. between 2 and 3
• C. between 1.2 and 1.8
• D. between 3 and 4

Answer 1

Answer: (A)

nearly 1.0

$HS{O}_4^{-}\leftrightharpoons S{O}_4^{2-}+H^{+}$

Initially $100$

Finally $1-xxx$

$K_2=\frac{x^2}{1-x}={10}^{-2}$

$\therefore x^2+{10}^{-2}x-{10}^{-2}=0$

Solving the quadratic equation, we get, $x=9.5\times {10}^{-2}$

$\therefore pH=-\log x=1.022$

Hence, $pH$ is nearly $1$.