Question

What is the final pH of a solution with an initial concentration of 2.5mM Ascorbic acid (H2C6H6O6) which has the following Kas: 7.9x10-5 and 1.6x10-12

Answer 1

Answer:

pH = 3.39

Explanation:

The equilibrium in water of ascorbic acid (With its conjugate base) is:

H₂C₆H₆O₆(aq) + H₂O(l) ⇄ HC₆H₆O₆⁻(aq) + H₃O⁺(aq)

Where the acidic dissociation constant is written as:

Ka = 7.9x10⁻⁵ = [HC₆H₆O₆⁻] [H₃O⁺] / [H₂C₆H₆O₆]

H₂O is not taken in the Ka expression because is a pure liquid.

As initial concentration of H₂C₆H₆O₆ is 2.5x10⁻³M, the equilibrium concentration of each species in the equilibrium is:

[H₂C₆H₆O₆] = 2.5x10⁻³M - X

[HC₆H₆O₆⁻] = X

[H₃O⁺] = X

Replacing in the Ka expression:

7.9x10⁻⁵ = [X] [X] / [2.5x10⁻³M - X]

1.975x10⁻⁷ - 7.9x10⁻⁵X = X²

0 = X² + 7.9x10⁻⁵X - 1.975x10⁻⁷

Solving for X:

X = -0.00048566→  False solution, there is no negative concentrations

X = 0.00040666 → Right solution

As [H₃O⁺] = X, [H₃O⁺] = 0.00040666

pH is defined as -log [H₃O⁺];

pH = -log 0.00040666,

pH = 3.39