Consider the titration of 21.0 mL sample of 0.110 M HC2H3O2 with 0.130 M NaOH. what is the initial pH? what is the volume of added base required to reach equivalance point? what is the pH at 5.00 mL of added base? pH at one half o the equivalence point? pH at the equivalence point?

To work this menagerie of problems, you must recognize what is in the solution at each of the points on the titration curve.

a. Pure acetic acid.
b. mL x M = mL x M
c. Buffered solution. Use the Henderson-Hasselbalch equation.
d. see c.
e. The pH at the equivalence is determined by the concn of the salt formed and the hydrolysis of that salt.

If you have problems, take one at a time and let me know what you don't understand about it.

I can't figure out how to get the initial pH. Also, when I tried part B I used 21.0 mL*.100M=XmL*0.130 and it wasn't the correct answer.

I put in the wrong numbers for the second part.

HAc is acetic acid. At the beginning.

HAc ==> H^ + Ac^-

Ka = (H^+)(Ac^-)/(HAc)
Set up ICE chart, substitute into Ka expression and solve for H^+, then convert to pH.
(x)(x)/(0.110-x) = Ka
Solve for x and convert to pH.
For part B do you mean you wrote the numbers wrong in the original question or you punched in the wrong numbers in the calculator.
Clear that up before I work a problem that doesn't exist.

I see.

21.0 x 0.110 = x*0.130
Solve for x = mL NaOH needed to reach the equivalence point.

Thank you! Now I am having problems with part e.

At the equivalence point we have the salt, sodium acetate. The acetate ion is hydrolyzed.

Ac^- + HOH ==> HAc + OH^-

Kb = (Kw/Ka) = (HAc)(OH^-)/(Ac^-)
I would set up an ICE chart. You know Kw, Ka for acetic acid, (HAc) = x, (OH^-) = x, and (Ac^-) is the concn of the salt. That will be moles/L. moles = 21.0 x 0.110 = ?? and L = total volume which is 21.0 + mL of the 0.130 base which you calculated from part b, then convert mL to L. Solve for x in the Ka expression (which is OH^-), convert to pOH, then to pH. Acetate ion givea a pH of approximately 8.5 or so depending upon the concn.