Question

What is the pH OF $0.1$ M $NaHC{O}_3$? $K_1$ = $5$ $\times$ ${10}^{-7}$,$K_2$ = $5$ $\times$ ${10}^{-11}$ for carbonic acids.

• A. 8.68
• B. 9.68
• C. 7.68
• D. 5.58

Answer 1

The correct option is B 9.68

$NaHC{O}_3+H_{2}O\to H_{2}C{O}_3+NaOH$

$K_b=\frac{K_w}{K_a}$

Where $K_a$ is equal to $K_1=4.5\times {10}^{-7}$

So,

$K_b=\frac{{10}^{-14}}{4.5\times {10}^{-7}}=2.22\times {10}^{-8}$

Dissociation is low,

$⟹2.22\times {10}^{-8}=\frac{[O{H}^{-}{]}^2}{0.1}$

Now,

$[OH{]}^{-}$ is equal to $4.7\times {10}^{-5}$

Hence,

$pOH=-log{\left[OH\right]}^{-}=4.32$

So,$pH=14-4.32=9.68$