Question

$H_{2}A$ is a diprotic acid for which $K_{a_1}={10}^{-7}and{K}_{a_2}={10}^{-11}$. The solution which will have a pH closest to 9 is:

• A. 0.1 M $H_{2}A$
• B. 0.1 M ${Na}_{2}A$
• C. 0.1 M NaHA
• D. 0.1 M NaHA + 0.1 M ${Na}_{2}A$

Answer 1

The correct option is A 0.1 M $H_{2}A$

The acid dissociates into,

$H_{2}A\to H^{+}+H{A}^{-}$

$0.1M$ $0.1M$

We know,

$pH=p{K}_a-log\left[H^{+}\right][HA{]}^{-}$

$⟹pH=p{K}_a-log(0.1{)}^2$

$⟹pH=-log\left({10}^{-7}\right)-log(0.1{)}^2$

$⟹pH=7+2=9$

So option A has pH close to 9