The ionization constant of nitrous acid is `4.5xx10^(-4)`. Calculate the `pH` of `0.04 M` sodium nitrite solution and also its degree of hydrolysis.

Explanation

To solve the problem step by step, we will first calculate the pH of the 0.04 M sodium nitrite solution and then determine the degree of hydrolysis.

### Step 1: Understand the Hydrolysis of Sodium Nitrite
Sodium nitrite (NaNO2) is a salt derived from a weak acid (nitrous acid, HNO2) and a strong base (NaOH). When dissolved in water, it undergoes hydrolysis:
\[ \text{NO}_2^- + \text{H}_2\text{O} \rightleftharpoons \text{HNO}_2 + \text{OH}^- \]

### Step 2: Calculate the Hydrolysis Constant (K_h)
The hydrolysis constant \( K_h \) can be calculated using the formula:
\[ K_h = \frac{K_w}{K_a} \]
where:
- \( K_w = 1.0 \times 10^{-14} \) (the ion product of water at 25°C)
- \( K_a = 4.5 \times 10^{-4} \) (the ionization constant of nitrous acid)

Calculating \( K_h \):
\[ K_h = \frac{1.0 \times 10^{-14}}{4.5 \times 10^{-4}} \]
\[ K_h = 2.22 \times 10^{-11} \]

### Step 3: Set Up the Hydrolysis Equation
For the hydrolysis of the salt:
Let \( C \) be the initial concentration of sodium nitrite, which is 0.04 M. The degree of hydrolysis is represented by \( h \).

The expression for \( K_h \) is given by:
\[ K_h = \frac{[HNO_2][OH^-]}{[NO_2^-]} \]
Assuming \( h \) is small, we can approximate:
\[ [HNO_2] \approx h \]
\[ [OH^-] \approx h \]
\[ [NO_2^-] \approx C - h \approx C \]

Thus, the equation simplifies to:
\[ K_h = \frac{h^2}{C} \]

### Step 4: Solve for Degree of Hydrolysis (h)
Substituting the values:
\[ 2.22 \times 10^{-11} = \frac{h^2}{0.04} \]
\[ h^2 = 2.22 \times 10^{-11} \times 0.04 \]
\[ h^2 = 8.88 \times 10^{-13} \]
\[ h = \sqrt{8.88 \times 10^{-13}} \]
\[ h \approx 2.98 \times 10^{-7} \]

### Step 5: Calculate pH
To find the pH, we first need to calculate the concentration of hydroxide ions \( [OH^-] \):
Since \( [OH^-] = h \):
\[ [OH^-] \approx 2.98 \times 10^{-7} \]

Now, we can find the pOH:
\[ pOH = -\log[OH^-] \]
\[ pOH = -\log(2.98 \times 10^{-7}) \]
\[ pOH \approx 6.53 \]

Finally, we can find the pH:
\[ pH + pOH = 14 \]
\[ pH = 14 - pOH \]
\[ pH = 14 - 6.53 \]
\[ pH \approx 7.47 \]

### Summary of Results
- **pH of 0.04 M sodium nitrite solution**: approximately 7.47
- **Degree of hydrolysis (h)**: approximately \( 2.98 \times 10^{-7} \)