Question

Sulphurous acid $\left(H_{2}S{O}_3\right)$ has $K{a}_1=1.7\times 10^{-2}$ and $K{a}_2=6.4\times 10^{-8}$. The $pH$ of $0.588M{H}_{2}S{O}_3$ is ________ .(Round off to the Nearest Integer)

Answer 1

Answer: 1

Sol. $H_{2}S{O}_3$ [Dibasic acid]

$c=0.588M$

$\Rightarrow pH$ of solution $P$ due to First dissociation only

since $K_a,\gg K{a}_2$

$\Rightarrow$ First dissociation of $H_{2}S{O}_3$

$H_{2}S{O}_3(aq)\rightleftharpoons H^{\oplus }(aq)+HS{O}_3^{-}(aq):k{a}_1=1.7\times 10^{-2}$



$\Rightarrow K{a}_1=\frac{1.7}{100}=\frac{\left[H^{\oplus }\right]\left[HS{O}_3^{-}\right]}{\left[H_{2}S{O}_3\right]}$

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

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

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

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

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

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