Description
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?
Explanation & Answer
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
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