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 1.20×10-1 mol of NaOH(s).
Please note that the concentration of H+, is:
[H+] = (8850-12v)/(1000+v) *10^-2 millimol/ml
where v is the volume of the NaOH added
using the value of your Equilibrium constant, I get:
v= 30.36 mL of NaOH
so we get:
[H+]= 0.082 = 8.2 * 10^(-2)
so, we have:
pH = -log [H+] = 1.086
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