Acrylic acid is a weak monoprotic acid with K_{a} = 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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