Posted by Joycelyn on Saturday, December 1, 2012 at 7:16pm.
1. First calculate the volume KOH needed to reach the equivalcnce point. That will tell you if the volume(s) in the problem are BEFORE or AFTER the equivalence point.
HA + KOH ==> KA + H2O
2. For the beginning, zero mL.
.........HA ==> H^+ + A^-
I........0.1....0......0
C.........-x....x......x
E.......0.1-x....x......x
Ka = (H^+)(A^-)/(HA)
Substitute from the E line and solve for x = (H^+) then convert to pH.
b. All points before the eq. pt. I will use 5 mL KOH as an example.
100 mL HA x 0.1M = 10 millimoles
5 mL KOH x 1M = 5 mmoles KOH.
10-5 = 5 mmols HA remains.
5 mmols KA formed. (HA) = 5 mmols/105 mL = ?
(A^-) = 5 mmols/105 mL = ?
Substitute into the Ka expression and solve for (H^+), then convert to pH.
c. At the equivalence point the solution is KA and the A^- is a base.
(KA) = (100 x 0.1)/110 mL = about 0.09M
..........A^- + HOH ==> HA^- + OH^-
I.......0.09............0......0
C.........-x............x......x
E.......0.09-x..........x.......x
Kb for A^- = (Kw/Ka for HA) = (x)(x)/(0.09-x)
Substitute and solve for x = (OH^-), then convert to pH.
d. All points after the eq pt.
mmoles HA initially.
mmols KOH added.
KOH will be in excess; therefore,
mmoles KOH-mmoles HA = mmols KOH in excess. M KOH = mmol/mL and that gives you (OH^-). Then convert to pH.
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