What is the pH of a 0.05 M solution of acetic acid, given that Ka = 1.75 × 10-5?

To determine the pH of a 0.05 M solution of acetic acid, we can use the dissociation constant (Ka) of acetic acid, which is given as 1.75 × 10^-5.

Acetic acid (CH3COOH) is a weak acid that partially dissociates in water according to the equilibrium:

( CH_3COOH \rightleftharpoons H^+ + CH_3COO^- )

Setting up the expression for the acid dissociation constant (Ka):

Ka = (\frac{[H^+][CH_3COO^-]}{[CH_3COOH]})

Assuming that the initial concentration of acetic acid is 0.05 M and let ( x ) represent the concentration of hydrogen ions (([H^+])) that dissociate from the acetic acid. The concentrations at equilibrium will be as follows:

• ([CH_3COOH]) at equilibrium = (0.05 - x)

• ([H^+]) at equilibrium = (x)

• ([CH_3COO^-]) at equilibrium = (x)

Substituting into the Ka