What is the pH of 0.0169 M sodium sulfite.
To determine the pH of a solution, we need to assess whether the compound is an acid or a base. Sodium sulfite (Na2SO3) is a salt composed of a strong base (sodium hydroxide) and a weak acid (sulfurous acid). As a result, it will hydrolyze in water to form a basic solution.
To calculate the pH, we need to consider the hydrolysis reaction of sodium sulfite:
Na2SO3 + H2O ⇌ 2NaOH + H2SO3
Since sodium hydroxide is a strong base that fully dissociates in water, its contribution to the pH is negligible. Therefore, we only need to focus on the hydrolysis of sulfurous acid (H2SO3).
The hydrolysis equation for sulfurous acid is:
H2SO3 + H2O ⇌ H3O+ + HSO3-
The equilibrium constant expression for this reaction is:
Kw/Ka = [H3O+][HSO3-]/[H2SO3]
Given the concentration of sodium sulfite (0.0169 M) and the stoichiometry of the reaction, we can calculate the initial concentration of sulfurous acid:
[H2SO3] = 0.0169 M / 2 = 0.00845 M
Since we assume the reaction reaches equilibrium, the concentration of H3O+ ions will be equal to the concentration of HSO3- ions:
[H3O+] = [HSO3-]
Now, let's denote the concentration of H3O+ as x. Therefore, the HSO3- concentration will also be x.
Using the equilibrium constant expression, we can write:
Kw/Ka = x^2 / 0.00845
The value of Kw/Ka for water at 25°C is 1.0 x 10^-14.
By rearranging the equation, we can solve for x:
x^2 = (1.0 x 10^-14) * 0.00845
x = √((1.0 x 10^-14) * 0.00845)
x ≈ 2.91 x 10^-8
Since the concentration of [H3O+] = [HSO3-] = x, we can conclude that the pH of the solution of 0.0169 M sodium sulfite is approximately -log(2.91 x 10^-8) = 7.54.