Question

Calculate the percent ionization of 0.3 moles HNO₂ dissolved in 1 L of a 0.4 M NO₂⁻ solution. Assume no volume change due to the addition of HNO₂. The Kₐ of HNO₂ is 4.5x10⁻⁴. What is the pH of this solution? a) pH = 3.38

b) pH = 4.52
c) pH = 2.25
d) pH = 4.0

Answer 1

d) pH = 4.0

The percent ionization of 0.3 moles of HNO₂ dissolved in a 0.4 M NO₂⁻ solution is 133%. The pH of this solution is 4.0.

Step-by-step explanation:

To calculate the percent ionization of HNO₂, we will use the formula:

Percent Ionization = (H₃O⁺ concentration / Initial concentration of HNO₂) x 100%

The initial concentration of HNO₂ is 0.3 M.

Since the Kₐ value of HNO₂ is 4.5x10⁻⁴, we can assume that the concentration of H₃O⁺ is equal to the concentration of NO₂⁻.

Since the solution is 0.4 M NO₂⁻, the H₃O⁺ concentration is also 0.4 M.

Substituting these values into the formula, we get:

Percent Ionization = (0.4 M / 0.3 M) x 100% = 133%

To calculate the pH of this solution, we can use the formula:

pH = -log[H₃O⁺]

pH = -log(0.4) ≈ 0.40

Therefore, the correct option is d) pH = 4.0.