Question

Calculate the pH of a solution obtained by mixing $5_ mL$ of $0.1 \,M \,NH_4OH$ with $250\, mL$ of $0.1\, M \,NH_4Cl$ solution. $K_b$ for $NH_4OH = 1.8 × 10^{-5}$

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

Answer 1

Answer: (C)

$5 \,mL$ of $0.1\, M\, NH_4OH = 5 × 0.1$ millimoles $= 0.5$ millimoles
$250\, mL$ of $0.1 M \,NH_4Cl = 250 × 0.1$ millimoles $= 25$ millimoles
Total volume of solution after mixing $= 255\, mL$
Now,
[Salt] $\left[NH_{4}Cl\right]=\frac{25}{255}M$
and [Base] $\left[NH_{4}OH\right]=\frac{0.5}{255}M$
Therefore, $pH$ and $pOH$ values can be calculated as
$pOH=pK_{b}+log \frac{\left[Salt\right]}{\left[Base\right]}$
$=-log\left(1.8\times10^{-5}\right)+log \frac{25/255}{0.5/255}$
$= (5 - 0.2553) + 1.6990 = 6.4437$
$pH = 14 - 6.4437 = 7.5563$