Question

Consider a 20.0mL sample of 0.105M HC2H3O2 is titrated with 0.125M NaOH.
Ka=1.8x10^-5. Determine each of the following:

a) the initial pH

b) the pH at 5.0mL of added base

c) the pH at one-half of the equivalence point

d) the pH at the equivalence point

Answer 1

The secret to these titration problems is to know where you are on the titration curve.
a. at the beginning of the titration; therefore, you have a solution of acetic acid. Set up and ICE chart, substitute and solve for H^+ then convert to pH.

b. use the Henderson-Hasselbalch equation.
c. same as b but I can tell you the answer is pH = pKa.

d. The equivalence point pH is determined by the hydrolysis of the salt.
Set up the hydrolysis equation, write the K expression, set it equal to Kb for acetate (which is Kb = Kw/Ka) and solve for OH^-, convert to pOH, then to pH.
Post your work if you get stuck.

Answer 2

I understand everything except for part d.
Could you show me what you mean by writing the k expression.

Answer 3

Let's call acetate, Ac^- just to save some typing.
Ac^- + HOH ==> HAc (acetic acid) + OH^-

Kb for acetate = (Kw/Ka) where Ka is the acid constant for acetic acid.
Kb = (Kw/Ka) = (HAc)(OH^-)/(Ac^-)
Let HAc = x = OH, then
(Kw/Ka) = x^2/Ac^-.
Kw you know. Ka you know. Ac^- is the concn of acetate ion at the equivalence point which is M x L of the acid or base and that divided by the total volume. Solve for x, convert to pOH and pH.

Answer 4

Here is the K expression (for Kb).
Kb = (Kw/Ka) = (HAc)(OH^-)/(Ac^-)

Answer 5

is Ac^- equal to to 20 ml x the concentration? if so then I am getting the wrong answer according to my answer sheet. By the way what would my total volume be?

Answer 6

nvm i got it. Thanks for the help

Answer 7

for part d, how do you know the total volume at the equiv point?