Question

20.00 mL of 0.160 M HC2H3O2 (Ka=1.8x10^-5) is titrated with 0.200 M NaOH.
Use the calculation to determine the:
a) pH of the solution before the titration begins? (2)
b) pH after 8.00 mL of NaOH has been added? (2)
c) pH at the equivalence point? (2)
d) pH after 20.00 mL of NaOH has been added? (2)

Answer 1

Before titration we have a 0.160 M solution of HC2H3O2 only.
................HC2H3O2 ==> H^+ + C2H3O2^-
I...............0.160 M..................0...........0
C................-x......................x............x
E.............0.160-x.....................x............x
Plug the E line into the Ka expression for HC2H3O2 and solve for x = (H^+). Convert to pH.

b. Start with 20.00 mL x 0.160 M HC2H3O2 = 3.2 millimols. Add 8.00 mL x 0.2 M NaOH = 1.6 millimols NaOH.
.....................HC2H3O2 + NaOH ==> NaC2H3O2 + H2O
I.........................3.2............0.....................0...................0
add...................................1.6...........................................
C.......................-1.6........-1.6..................+1.6..............+1.6
E........................1.6...........0......................1.6..................1.6
Ka = (H^+)(C2H3O2^-)/(HC2H3O2) = 1.8E-5
total volume is 20.00 + 8.00 = 28.00 mL and concn = millimoles/mL.
You can see from the ICE chart that (C2H3O2^-) = 1.6 mmols/28 mL and (HC2H3O2) = 1.6 mmols/28 mL. Plug those numbers in and solve for H^+ and convert to pH.
c. At the equivalence point you have only NaC2H3O2 because all of the acid has been neutralized. What volume of NaOH is required. You know there are 3.2 mmol HC2H3O2 to start so you must add 3.2 mmols NaOH. You are using 0.200 M NaOH so M = mmols/mL or 0.200 = 3.2/?mL. Solve for ?mL required. I get mL = 3.2/0.200 = 16 mL but check that. Now the total volume is 20.00 + 16.00 = 36.00 ml.
The pH at the equivalence point is determined by the hydrolysis of the sodium acetate.
...................C2H3O2^- + HOH ==> HC2H3O2 + OH^-
I...................3.2/36.00..........................0..................0
C...................-x.....................................x...................x
E................0.0889-x..............................x...................x
Follow this. We need Kb for C2H3O2^-. We don't have it be we can calculate it. It is Kb for acetate = Kw for water/Ka for HC2H3O2 (which you have). So Kb = Kw/Ka.
Kb = (C2H3O2^-)(OH^-)/(HC2H3O2^-). Plug in those values like this
Kb = 1E-14/1.8E-5 = (x)(x)/(0.0889-x) and solve for x. That gives you OH^-. convert to H+ then to pH.

d. You had 20.00 mL x 0.160 = 3.2 mmols to start. You add 20.00 mL of 0.200 M NaOH or 4.00 mmols NaOH. You will have how much NaOH left over? That is 4.00-3.20 = 0.80 millimols NaOH in the solution. The total volume is 20 + 20 = 40 mL so (NaOH) = (OH^-) = 0.8 mmols/40 mL = ?. Convert to H^+ then to pH.
Post your work if you get stuck on any of these.