Question

What is the pH of an aqueous solution of 1 M CH3COOH (pKa=4)

Answer 1

Answer:

pH = 2.

Explanation:

A weak acid is in equilibrium with its ions in a solution, so it must have an equilibrium constant (Ka). And, pKa = -logKa

$Ka = 10^{-pKa}$

Ka = 10⁻⁴

So, for CH₃COOH the equilibrium must be:

CH₃COOH(aq) ⇄ H⁺(aq) + CH₃COO⁻(aq)

1 M                           0                0                Initial

-x                              +x              +x               Reacted

1-x                             x                x                 Equilibrium

And the equilibrium constant:

$Ka = \frac{[H+]x[CH3COO-]}{[CH3COOH]}$

$10^{-4} = \frac{x^2}{1-x}$

Supposing x << 1:

10⁻⁴ = x²

x = √10⁻⁴

x = 10⁻² M, so the supposing is correct.

So,

pH = -log[H⁺]

pH = -log10⁻²

pH = 2