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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 8.58×10-2 mol of NaOH(s). 

Feb 19th, 2015

[acid] = [base] , so all acrylic acid has been converted into acrylate. 

A- + H2O     =>    HA   +   OH -     ;    Kb = Kw / Ka = (10^-14) / (5.5*10^-5) = 1.8*10^-10

(1.8*10^-10) = (x^2) / (.0858 - x)

=> x = 3.9 * 10^-6 M = [OH-]

=> pOH = -log(3.9*10^-6) = 5.41

=> pH = 14 - 5.41 = 8.59 


Feb 19th, 2015

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Feb 19th, 2015
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