Acrylic acid is a weak monoprotic acid with Ka = 5.5×10-5 M. NaOH(s) was gradually added to 1.00 L of 8.58×10-2 M acrylic acid.
Calculate the pH of the solution after the addition of 3.43×10-2 mol of NaOH(s).
pH = pKa + log([base]/[acid])
pH = -log(5.5*10^-5) + log((0.0343)/(0.0858-0.0343)) = 4.08
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