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 after the addition of 3.43×10-2 mol of NaOH(s).
[H+] = (8580-8.58v)/(1000+v) * 10^(-2), where v is the volume of NaOH added
v = 149 mL
[H+] = 0.06354
pH = 1.197
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