Question

$100\text{mL}$ of $0.1\text{M NaOH}$ solution is titrated with $100\text{mL}$ of $0.05\text{M}{H}_{2}S{O}_4$ solution. The $pH$ of the resulting solution is : $(\text{for}{H}_{2}S{O}_4,K{a}_1=\infty ,K{a}_2={10}^{-2})$

• A. 7
• B. 7.2
• C. 7.4
• D. 6.8

Answer 1

The correct option is B 7.2

no of milimoles of $NaOH=100\times 0.1=10$
no of milimoles of $H_{2}S{O}_4=100\times 0.05=5$
now $2NaOH+H_{2}S{O}_4\to N{a}_{2}S{O}_4+2H_{2}O$
Since 1 mole $H_{2}S{O}_4$ rects with 2 mole of $NaOH$
the above reaction will go till complete neutralization and after that $[S{O}_4{]}^{2-}=\frac{5}{200}M$ and this will go anion hydrolysis
$S{O}_4^{2-}+H_{2}O\leftrightharpoons HS{O}_4^{-}+O{H}^{-}$
Where $HS{O}_4^{-}$ is conjugate acid with $Ka={10}^{-2}$
$pH=\frac{1}{2}[p{K}_w+p{K}_a+logC]$
$pH=\frac{1}{2}[14+2-1.6]=7.2$