Question

What is the $\%$ dissociation of $0.01\, N\, NH _{4} OH$ in its buffer solution containing $0.01\, N\, NH _{4} Cl :$
$\left( K _{ b }\right.$ of $\left.NH _{4} OH =2 \times 10^{-5}\right)$

• A. $0.2 \%$
• B. $2 \times 10^{-3} \%$
• C. $2 \times 10^{-5}$
• D. $5 \times 10^{-3} \%$

Answer 1

Answer: (A)

$K _{ b }=\frac{(0.01 \alpha)(0.01 \alpha+0.01)}{0.01(1-\alpha)}$
$2 \times 10^{-5}=0.01 \alpha(1+\alpha)$
$\alpha=2 \times 10^{-3}$
$\% \alpha=2 \times 10^{-3} \times 100=0.2$