Question

What is the $pH$ of a $0.5\, M$ aqueous solution of $NH_4Cl$ with $0.2\, M \,NH_3$?
$ (pK_a = 9.25)$

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

Answer 1

Answer: (A)

$NH_{4}^{+} $ and $NH_3$ are conjugate acid-base pair.
For conjugate add-base pair $ \to pK_a + pK_b = 14$
$pK_b = 14 - 9.25 = 4.75$
$pOH = pK_b + log \frac{[{\text{Conjugate acid }} NH_{4}^{+}]}{[\text{Weak base } NH_3]}$
$ = 4.75 + log \frac{0.5}{0.2}$
$ = 4.75 + 0.40 = 5.15 $
$pH = 14 - pOH = 14 - 5.15 = 8.85$.