Question

calculate the ph 25.00 mL solution of 0.125 M acetic acid (Ka= 1.75 x 10^-5) after the addition of each of the following volumes of 0.100 M KOH: 10.00 mL, Vep and 35.00 mL

Answer 1

KOH + HAc ==> H2O + KAc
Divide the problem up into
a. beginning (I've worked this but the problem doesn't ask for it.)
b. equivalence point
c. everything between a and b
d. everything after b.

What volume is b?
mols HAc = 0.025 x 0.125 = 0.003125
mols KOH = 0.003125(The coefficients are 1 mol KOH to 1 mol HAc.)
M KOH = mols KOH/L KOH solve for L; I obtained 31.25 mL. The pH is determined by the hydrolysis of the salt. Salt concn is M = mol/L = 0.003125/(25.00+ 31.25) = about 0.055M (that's an estimate)
.......Ac^- + HOH ==> HAc + OH^-
I....0.055.............0.....0
C.......-x.............x......x
E.....0.055-x .........x......x

Kb for Ac^- = (Kw/Ka for HAc) = (x)(x)/(0.055-x)
Solve for x = OH^- and convert to pH.

a........ HAc ==> H^+ + Ac^-
I.......0.125.....0......0
C..........-x.....x......x
E......0.125-x....x......x

Substitute into Ka expression and solve for x = (H^+) and convert pH.

c. Use the Henderson Hasselbalch equation for these.

d. mols HAc = ?
mols KOH = ? and this is greater than mols HAc.
Excess KOH = subtract mols KOH-mols HAc.
Excess = KOH = OH^- ? and convert to pH.