Question

What is the $pH$ of a $0.5M$ aqueous solution of $N{H}_{4}Cl$ with $0.2MN{H}_3$?
$(p{K}_a=9.25)$

• A. $8.85$
• B. $10.15$
• C. $5.15$
• D. $4.35$

Answer 1

Answer: (A)

$N{H}_4^{+}$ and $N{H}_3$ are conjugate acid-base pair.
For conjugate add-base pair $\to p{K}_a+p{K}_b=14$
$p{K}_b=14-9.25=4.75$
$pOH=p{K}_b+log\frac{[\text{Conjugate acid }N{H}_4^{+}]}{[\text{Weak base }N{H}_3]}$
$=4.75+log\frac{0.5}{0.2}$
$=4.75+0.40=5.15$
$pH=14-pOH=14-5.15=8.85$.