Chemistry - Ionic Equilibrium Question with Solution | TestHub

The question asks to calculate the percentage of hydrolysis in a 0.01 M aqueous solution of NaOCN, given that $K_b$ for $OC{N}^{-}$ is $10^{-10}$.

Here's how to solve this problem using the direct formula:

1. Write the hydrolysis reaction:

$OC{N}^{-}(aq)+H_{2}O(l)\rightleftharpoons HOCN(aq)+O{H}^{-}(aq)$

2. The hydrolysis constant, $K_h$, can be calculated using the following relationship:

$$K_h=\frac{K_w}{K_a}$$

Since we are given $K_b$ for $OC{N}^{-}$, we can find $K_a$ for its conjugate acid, HOCN, using:

$$K_w=K_a\cdot K_b$$

$$K_a=\frac{K_w}{K_b}=\frac{10^{-14}}{10^{-10}}=10^{-4}$$

Therefore,

$$K_h=\frac{K_w}{K_a}=K_b=10^{-10}$$

3. The degree of hydrolysis, $h$, can be approximated using the formula:

$$h=\sqrt{\frac{K_h}{C}}$$

where C is the concentration of the salt (NaOCN), which is 0.01 M.

$$h=\sqrt{\frac{10^{-10}}{0.01}}=\sqrt{10^{-8}}=10^{-4}$$

4. The percentage of hydrolysis is:

Percentage hydrolysis = $h\times 100$

Percentage hydrolysis = $10^{-4}\times 100=0.01\%$

Therefore, the percentage of hydrolysis is 0.01%.