Calculate degree of hydrolysis and pH of 0.25 NaCN solution at `25^@C`. `K_a=4.8xx10^(-10)`

Explanation

To solve the problem of calculating the degree of hydrolysis and pH of a 0.25 M NaCN solution at 25°C, we can follow these steps:

### Step 1: Understand the Components
NaCN is a salt formed from a strong base (NaOH) and a weak acid (HCN). In solution, NaCN will dissociate into Na⁺ and CN⁻ ions. The CN⁻ ion will undergo hydrolysis to produce OH⁻ ions.

### Step 2: Calculate the Degree of Hydrolysis (h)
The degree of hydrolysis (h) can be calculated using the formula:
\[ h = \frac{K_w}{K_a \cdot C} \]
where:
- \( K_w \) is the ion product of water, \( 1.0 \times 10^{-14} \) at 25°C,
- \( K_a \) is the acid dissociation constant of HCN, given as \( 4.8 \times 10^{-10} \),
- \( C \) is the concentration of the salt, which is 0.25 M.

Substituting the values:
\[ h = \frac{1.0 \times 10^{-14}}{4.8 \times 10^{-10} \cdot 0.25} \]

Calculating this gives:
\[ h = \frac{1.0 \times 10^{-14}}{1.2 \times 10^{-10}} = 8.33 \times 10^{-5} \]

### Step 3: Calculate the Concentration of OH⁻
The concentration of OH⁻ ions produced due to hydrolysis can be calculated as:
\[ [OH^-] = h \cdot C = 8.33 \times 10^{-5} \cdot 0.25 = 2.08 \times 10^{-5} \, \text{M} \]

### Step 4: Calculate pOH
Using the concentration of OH⁻, we can find pOH:
\[ pOH = -\log[OH^-] = -\log(2.08 \times 10^{-5}) \]

Calculating this gives:
\[ pOH \approx 4.68 \]

### Step 5: Calculate pH
Using the relationship between pH and pOH:
\[ pH = 14 - pOH \]
\[ pH = 14 - 4.68 = 9.32 \]

### Final Results
- Degree of Hydrolysis (h) = \( 8.33 \times 10^{-5} \)
- pH of 0.25 M NaCN solution = 9.32