A solution of ethanoic acid in water has a concentration of 3 g/dm^3. Given that the pKa of ethanoic acid is 4.76, calculate the pH of this solution.

Key facts to be noted by student:

Molar mass of ethanoic acid = 60 g/mol.

pK_a = -log(K_a)

pH =-log([H⁺])

Concentration of acid solution = 3/60 = 0.05 moldm^-3

k_a = [H⁺]² / [CH₃COOH]   therefore

[H⁺]² = 10^-4.76 / 0.05   --> [H⁺] = 0.0186 moldm^-3

ph = -log(0.0186)   ---> pH =1.73