chemistry
posted by Joycelyn on .
pls help.
A weak acid HA (pKa = 5.00) was titrated with 1.00 M KOH. The acid solution had a volume of 100.0 mL and a molarity of 0.100 M. Find the pH at the following volumes of base added and make a graph of pH versus Vb: Vb = 0, 1, 5, 9, 9.9, 10, 10.1, and 12 mL.

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.......0.1x....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.
105 = 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.......0.09x..........x.......x
Kb for A^ = (Kw/Ka for HA) = (x)(x)/(0.09x)
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 KOHmmoles HA = mmols KOH in excess. M KOH = mmol/mL and that gives you (OH^). Then convert to pH.