Chemical and Ionic Equilibrium 2 Question 68

3. The pH of a $0.02 \mathrm{M} \mathrm{NH}{4} \mathrm{Cl}solutionwillbe[GivenK{b}\left(\mathrm{NH}_{4} \mathrm{OH}\right)=10^{-5}and\left.\log 2=0.301\right]$

(2019 Main, 10 April II)

(a) 4.65

(b) 2.65

(c) 5.35

(d) 4.35

Show Answer

Answer:

Correct Answer: 3. (c)

Solution:

  1. Key Idea $\mathrm{NH}{4} \mathrm{Cl}isasaltofweakbase\left(\mathrm{NH}{4} \mathrm{OH}\right)andstrongacid(\mathrm{HCl}).Onhydrolysis,\mathrm{NH}_{4} \mathrm{Cl}willproduceanacidicsolution(\mathrm{pH}<7)andtheexpressionof\mathrm{pH}$ of the solution is

pH=712(pKb+logC)

Given, Kb(NH4OH)=105

pKb=logKb=log(105)=5

C = concentration of salt solution =0.02M =2×102M

Now, pH=712(pKb+logC)

On substituting the given values in above equation, we get

=712[5+log(2×102)]

=712[5+log22]

=712[5+0.3012]=71.65=5.35