Question

Two solutions, A and B each of $100$ $L$ was made by dissolving $4$ $g$ of $NaOH$ and $9.8$ $g$ of $H_{2}S{O}_4$ in water, respectively. The pH of the resultant solution obtained from mixing $40\text{L}$ of solution A and $10\text{L}$ of solution B is

Answer 1

Molarity of $NaOH$ (4 g in 100 L)=${10}^{-3}\text{M}$
Molarity of $H_{2}S{O}_4$ (9.8g in 100 L)=${10}^{-3}\text{M}$
Equivalent of $NaOH=M\times V\times n_f={10}^{-3}\times 40\times 1=0.04$
Equivalent of $H_{2}S{O}_4=M\times V\times n_f={10}^{-3}\times 10\times 2=0.02$
$M_{NaOH}\times (n_f{)}_{NaOH}\times V_{NaOH}-M_{H_{2}S{O}_4}\times V_{H_{2}S{O}_4}\times (n_f{)}_{H_{2}S{O}_4}=M.V_{\text{total}}$
${10}^{-3}\times 40\times 1-{10}^{-3}\times 10\times 2=M.50$
$M=4\times {10}^{-4}$
$pOH=-\log M$
$=4-2\log 2$
$=3.4$
$pH=14-3.4=10.6$