The pH will be that of the salt due to hydrolysis.
10.0 mL x 0.6M = ?mL x 0.200 M
?mL = 0.6 x 10/0.2 = 30 mL; therefore, the final volume will be 40 mL for the 6 millimoles for (NaZ) = 6/40 = 0.15M
.............Z + H2O ==> HZ + OH^-
I...........0.15..........0....0
C............-x............x.....x
E..........0.15-x........x.......x
Kb(for Z^-) = (Kw/Ka for Hz) = (HZ)(OH^-)/(Z^-)
Substitute and solve for x = OH^- then convert to pH.
What is the pH at the equivalence point in the
titration of 10.0 mL of 0.6 M HZ with 0.200
M NaOH? Ka = 7.0 × 10−6
for HZ.
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