Question

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

Answer 1

pH = 2.

Step-by-step 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 = ([H+]x[CH3COO-])/([CH3COOH])`

`10^(-4) = (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