Question

$0.12mL$ of a solution of $KOH$ ($59$% by mass $KOH$) specific gravity $1.5g/{cm}^3$ is diluted to $250mL$. Calculate the value of $(12-\text{pH of solution})$ (nearest integer).

Answer 1

First calculate the hydroxide ion concentration.
$\left[O{H}^{-}\right]=\frac{0.12mL\times \frac{1.5g}{1mL}\times \frac{59}{100}\times \frac{1000mL}{1L}}{\frac{56.1g}{1mol}\times 250mL}=7.57\times {10}^{-3}M$
Now calculate the pOH of solution.
$pOH=-log\left[O{H}^{-}\right]=-log7.57\times {10}^{-3}=2.20$.
Now calculate the pH of the solution $pH=14-2.20=11.80$.
Hence, the pH of the solution is 11.80.

Answer $=12-11.80=.20\approx 0$