Question

Calculate the $pH$ of resultant solution at ${25}^{o}C$ when $10ml$ of $0.04M$ $H_2{SO}_4$ is diluted $1000$ times and $100ml$ of the diluted acid is mixed with $100ml$ of $0.001M$ $NaOH$ solution at ${25}^{o}C$.

• A. $8.2$
• B. $9.5$
• C. $10.7$
• D. $12.8$

Answer 1

The correct option is C $10.7$

Mole of $H^{+}$ ions in $0.04M{H}_{2}S{O}_4=0.04M\times 2\times 0.01l=0.0008mole$

Concentration of diluted solution $=\frac{0.0008}{1}=0.0008mole/l$

Moles of $H^{+}$ in 100ml $=0.0008\times 0.1=0.00008moles$

Moles of $O{H}^{-}$ in $0.001MNaOH=0.001M\times 0.1l=0.0001moles$

Concentration of $O{H}^{-}$ ions in mixed solution$=\frac{0.0001-0.00008mole}{0.2l}=0.0001mole/l$

$pOH=-log\left[{10}^{-4}\right]=4$

$pH=14-4=10$