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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 1.20×10-1 mol of NaOH(s). 
Feb 19th, 2015

8.58e-2 mol A - 1.2e-1 = -0.0342 mol

This means that there are 0.0342 mol of excess NaOH

[OH-] = 0.0342 mol/ 1L =0.0342 M

pOH = -log(0.0342) = 1.466

pH = 14 - 1.466

pH = 12.534

Feb 19th, 2015

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Feb 19th, 2015
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Feb 19th, 2015
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