Q3. A third beaker, C, contains 100.0 cm^3 of 0.0125 mol/dm^3 ethanoic acid ( Ka = 1.74 × 10^−5 mol/dm^3 at 25 ºC). Write an expression for Ka and use it to calculate the pH of the ethanoic acid solution in beaker C.

A3.   Ethanoic acid is a weak acid with the formula CH3COOH. It will dissociate into H+ and CH3COO- therefore the expression for Ka is [H+][CH3COO-]/[CH3COOH]   The [H+] = [CH3COO-] and so the expression becomes, Ka = [H+]^2 / [CH3COOH]   We know the Ka value and the [CH3COOH] from the question and so can rearrange the expression to find [H+] and hence the pH   [H+] = sqrt(Ka x [CH3COOH])    Ka = 1.74 x 10^-5 mol/dm^3, [CH3COOH] = 0.0125 mol/dm^3 so [H+] = 4.66 x 10-4   pH = -log[H+] = 3.33