Question

Sulphurous acid $\left( H _{2} SO _{3}\right)$ has $Ka _{1}=1.7 \times 10^{-2}$ and $Ka _{2}=6.4 \times 10^{-8}$. The $pH$ of $0.588\,M H _{2} SO _{3}$ is ________ .(Round off to the Nearest Integer)

Answer 1

Sol. $H _{2} SO _{3}$ [Dibasic acid]

$c =0.588 M$

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

since $K _{ a }, \gg Ka _{2}$

$\Rightarrow $ First dissociation of $H _{2} SO _{3}$

$H _{2} SO _{3}( aq ) \rightleftharpoons H ^{\oplus}( aq )+ HSO _{3}^{-}( aq ): ka _{1}=1.7 \times 10^{-2}$



$\Rightarrow Ka _{1}=\frac{1.7}{100}=\frac{\left[ H ^{\oplus}\right]\left[ HSO _{3}^{-}\right]}{\left[ H _{2} SO _{3}\right]}$

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

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

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

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

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

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