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
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.
I understand everything except for part d.
Could you show me what you mean by writing the k expression.
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.
Here is the K expression (for Kb).
Kb = (Kw/Ka) = (HAc)(OH^-)/(Ac^-)