Question

A buffer solution have 0.1M NH₄Cl & 0.1M NH₄OH. If 0.05 mole HCl is added in the solution. The change in pH is x × 10^–1 (log3 = 0.48)

Answer 1

Answer is : (5)

Initially

\(pOH = pK_b + log \frac{[NH_4Cl]}{[NH_4OH]} = pK_b + log \frac{0.1}{0.1}\)

pOH = pKb

pH = 14 – pOH = 14 – pK_b …(i)

When 0.05 mole HCl is added

\(NH_4OH + H^+ \rightleftharpoons NH _{4}^{+} + H_2O\)

0.1           0.05      0.1

0.1–0.05  0           0.15

= 0.05

\(pOH = pK_b + log \frac{[NH _{4}^{+}]}{[NH_4OH]} = pK_b + log \frac{0.15}{0.5}\)

\(pH = 14 - pK_b - log \frac{0.15}{0.05}\)

\(\text{Change in pH} = 14 - pK_b - log \frac{0.15}{0.05} - 14 + pK_b\)

= –log3 = –0.48

Change in pH = 0.48 or 4.8 × 10^–1

^{\(= x= 4.8 \approx 5\)}