Question

A 75.0 mL sample of 0.020 M acetic acid (HC2H3O2) is titrated with 0.020 M NaOH. ? Determine the pH of the solution before the addition of any NaOH. (Ka of acetic acid is 1.8 x 10-5) HC2H3O2 (aq) + H2O (l) D C2H3O2-(aq) + H3O+ (aq) (Hint: before titration so acid only, use ICE table)

Answer 1

Answer:

pH = 3.23

Explanation:

Before the addition of any NaOH, the only you have is a 0.020M acetic acid solution. That is in equilibrium with water as follows:

HC₂H₃O₂(aq) + H₂O(l) ⇄ C₂H₃O₂⁻(aq) + H₃O⁺(aq)

The Ka of this reaction is:

Ka = 1.8x10⁻⁵ = [C₂H₃O₂⁻] [H₃O⁺] / [HC₂H₃O₂]

Where [] are concentrations in equilibrium of each species

As you have in solution just HC₂H₃O₂, the equilibrium concentrations will be:

[HC₂H₃O₂] = 0.020M - X

[C₂H₃O₂⁻] = X

[H₃O⁺] = X

Where X is reaction coordinate.

Repalcing in Ka expression:

1.8x10⁻⁵ = [C₂H₃O₂⁻] [H₃O⁺] / [HC₂H₃O₂]

1.8x10⁻⁵ = [X] [X] / [0.020M - X]

3.6x10⁻⁷ - 1.8x10⁻⁵X = X²

3.6x10⁻⁷ - 1.8x10⁻⁵X - X² = 0

Solving for X:

X = -0.0006M → False solution. There is no negative concentrations

X = 0.000591M → Right solution.

As:

[H₃O⁺] = X

[H₃O⁺] = 0.000591M

As pH = -log[H₃O⁺]

pH = 3.23