JEE PYQ: Ionic Equilibrium Question 24
Question 24 - 2019 (10 Apr Shift 2)
Type: MCQ
The pH of a 0.02 M $\text{NH}_4\text{Cl}$ solution will be [given $K_b(\text{NH}_4\text{OH}) = 10^{-5}$ and $\log 2 = 0.301$]
(a) 2.65
(b) 4.35
(c) 4.65
(d) 5.35
Show Answer
Answer: (c)
Solution
pH $= 7 - \frac{1}{2}\text{pK}_b - \frac{1}{2}\log C$
$= 7 - \frac{5}{2} - \frac{1}{2}(\log 2 \times 10^{-2}) = 7 - 2.5 - \frac{1}{2}(-1.699) = 5.35$
pH $= 5.35$
BETA
