Question

Sulphurous acid $\left(H_{2}S{O}_3\right)$ has $K_{a1}=1.7\times {10}^{-2}$ and $Ka2=6.4\times {10}^{-8}$. The pH of $0.588M{H}_{2}S{O}_3$ is ______ (Round off to the Nearest Integer).

Answer 1

$H_{2}S{O}_3$ is a dibasic acid.

Concentration, c = 0.588 M

$\Rightarrow$ pH of solution is due to first dissociation only dissociation only, since $K_{a1}>>K_{a2}$
$\Rightarrow$ First dissociation of $\left(H_{2}S{O}_3\right)$

$H_{2}S{O}_3\left(aq\right)\rightleftharpoons H^{+}\left(aq\right)+HS{O}_3^{-}\left(aq\right):K_{a1}=1.7\times {10}^{-2}$

t = 0 C
At 't' C - x x x

$\Rightarrow K_{a1}=\frac{1.7}{100}=\frac{\left[H^{+}\right]\left[HS{O}_3^{-}\right]}{\left[H_{2}S{O}_3\right]}$

$\Rightarrow \frac{1.7}{100}=\frac{x^2}{(0.58-x)}$

$\Rightarrow 1.7\times 0.588-1.7x=100x^2$

$\Rightarrow 100x^2+1.7x-1=0$

$\Rightarrow \left[H^{+}\right]=x=\frac{-1.7+\sqrt{{\left(1.7\right)}^2+4\times 100\times 1}}{2\times 100}=0.09186$

Therefore pH of sol is : $pH=-\log \left[H^{+}\right]$

$\Rightarrow pH=-log\left(0.09186\right)=1.036\approx 1$