Question

Degree of dissociation of $N{H}_{4}OH$ in water is $1.8\times {10}^{-5}$, then hydrolysis constant of $NaOH$ in $N{H}_{4}Cl$ is:

• A. $1.8\times {10}^{-5}$
• B. $1.8\times {10}^{-10}$
• C. $5.55\times {10}^{-5}$
• D. $5.55\times {10}^{-10}$

Answer 1

The correct option is D $5.55\times {10}^{-10}$

Given

$k_h=?,$

$k_{\infty }={10}^{-14},$

$k_w=1.8\times {10}^{-5}$

$kh=\frac{k_w}{k_b}$

$=\frac{{10}^{-14}}{1.8\times {10}^{-5}}$
$=0.555\times {10}^{--9}$

$=5.55\times {10}^{-10}$