The ionization constant of nitrous acid is 4.5 × 10–4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis ?

To find the pH of a 0.04 M sodium nitrite solution and its degree of hydrolysis, we first need to understand the relationship between sodium nitrite and nitrous acid. Sodium nitrite (NaNO2) is the sodium salt of nitrous acid (HNO2), and when dissolved in water, it dissociates to release nitrite ions (NO2-). These nitrite ions can undergo hydrolysis, which affects the pH of the solution.

Step 1: Understanding Hydrolysis

When nitrite ions are in water, they can react with water molecules in a process called hydrolysis:

• NO2^- + H2O ⇌ HNO2 + OH^-

This reaction shows that nitrite ions can generate hydroxide ions (OH^-), which will make the solution basic. The extent of this reaction is determined by the ionization constant of nitrous acid (K_a), which is given as 4.5 × 10^-4.

Step 2: Calculating K_b for Nitrite Ion

To find the base dissociation constant (K_b) for the nitrite ion, we can use the relationship between K_a and K_b:

• K_w = K_a × K_b

Where K_w (the ion product of water) is 1.0 × 10^-14 at 25°C. Rearranging gives us:

• K_b = K_w / K_a

Substituting the values:

• K_b = (1.0 × 10^-14) / (4.5 × 10^-4) ≈ 2.22 × 10^-11

Step 3: Setting Up the Hydrolysis Equation

Now, we can set up the hydrolysis equilibrium expression for the nitrite ion:

• K_b = [HNO2][OH^-] / [NO2^-]

Let x be the concentration of OH^- produced at equilibrium. Initially, the concentration of NO2^- is 0.04 M, and at equilibrium, it will be (0.04 - x) M. The concentrations of HNO2 and OH^- will both be x M.

Step 4: Solving for x

Substituting into the K_b expression gives:

• 2.22 × 10^-11 = (x)(x) / (0.04 - x)

Assuming x is small compared to 0.04, we can simplify to:

• 2.22 × 10^-11 ≈ (x²) / 0.04

Solving for x:

• x² = 2.22 × 10^-11 × 0.04
• x² = 8.88 × 10^-13
• x = √(8.88 × 10^-13) ≈ 9.42 × 10^-7 M

Step 5: Finding pOH and pH

Now that we have the concentration of OH^-, we can find the pOH:

• pOH = -log(9.42 × 10^-7) ≈ 6.03

To find the pH, we use the relationship:

• pH + pOH = 14

Thus:

• pH = 14 - 6.03 ≈ 7.97

Step 6: Degree of Hydrolysis

The degree of hydrolysis (h) can be calculated using the formula:

• h = (x / C) × 100%

Where C is the initial concentration of the salt (0.04 M):

• h = (9.42 × 10^-7 / 0.04) × 100% ≈ 0.00236%

Final Results

In summary, the pH of the 0.04 M sodium nitrite solution is approximately 7.97, and the degree of hydrolysis is about 0.00236%. This indicates that while the nitrite ions do hydrolyze to some extent, the solution remains close to neutral due to the relatively low concentration of hydroxide ions produced.