Question

Calculate the pH of a solution obtained by mixing $5_{m}L$ of $0.1MN{H}_{4}OH$ with $250mL$ of $0.1MN{H}_{4}Cl$ solution. $K_b$ for $N{H}_{4}OH=1.8\times 10^{-5}$

• A. $5.0223$
• B. $6.3562$
• C. $7.5563$
• D. $8.6210$

Answer 1

Answer: (C)

$5mL$ of $0.1MN{H}_{4}OH=5\times 0.1$ millimoles $=0.5$ millimoles
$250mL$ of $0.1MN{H}_{4}Cl=250\times 0.1$ millimoles $=25$ millimoles
Total volume of solution after mixing $=255mL$
Now,
[Salt] $\left[N{H}_{4}Cl\right]=\frac{25}{255}M$
and [Base] $\left[N{H}_{4}OH\right]=\frac{0.5}{255}M$
Therefore, $pH$ and $pOH$ values can be calculated as
$pOH=p{K}_b+log\frac{\left[Salt\right]}{\left[Base\right]}$
$=-log\left(1.8\times 10^{-5}\right)+log\frac{25/255}{0.5/255}$
$=(5-0.2553)+1.6990=6.4437$
$pH=14-6.4437=7.5563$