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.
1)Calculate the pH of the solution before the addition of the base.
Ka = 5.5*10-5 = x2 / (0.0858 –x) [rearrange and quadratic formula]
Gives x = 2.2*10-3 M = [H+]
So pH = -log(2.2*10-3) = 2.66
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