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questions are related to the titration of 25.00 mL of a 0.0800 M acetic acid solution with 0.0700 M KOH. temperature is 25 oC.

What is the initial pH of the analyte solution?

What is the pH when 6.00 mL of the KOH solution have been added?

Feb 16th, 2015

Ka = 1.8*10^-5 = x^2 / (0.0800 -x) [rearrange and quadratic formula]

Gives x = 1.2*10^-3 M = [H+]

So pH = -log(1.2*10^-3) = 2.92

AFTER ADDITION OF 6.00 mL KOH

mols OH- added = (0.0700 mol/L)*(0.006 L) = 4.2*10^-4

mols H+ present in solution = (1.2*10^-3 mol/L)*(0.025 L) = 3.0*10^-5

=> mols OH- present after addition = (4.2*10^-4) - (3.0*10^-5) = 3.9*10^-4

so [OH-] = (3.9*10^-4 mols) / (0.031 L) = 0.0126 M

therefore pOH = -log(0.0126) = 1.90

hence pH = 14 - 1.90 = 12.1

Feb 16th, 2015

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Feb 16th, 2015
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Feb 16th, 2015
Dec 4th, 2016
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