You have three problems here. The secret to each is to recognize that the pH at the equivalence point is determined by the salt present at that point and what that salt does upon hydrolysis.
a. HBr + NaOH ==> NaBr + H2O
b. HL + NaOH ==> NaL + H2O
c. NaHCrO4 + NaOH ==> Na2CrO4 + H2O
a. Neither Na^+ nor Br^- is hydrolyzed since it is the salt of a strong acid and a strong base; therefore, you have a salt in pure water and the water has a pH of 7.
b. NaL is the salt of a weak acid and a strong base. First you want to find the concn of lactate ion. You don't list a value for volume for the acid so you can't do that. I'll just make up a number of say 0.2M and show you how to do it. The lactate ion is hydrolyzed as follows:
......L- + HOH ==> HL + OH^-
I....0.2...........0.....0
C.....-x...........x.....x
E...0.2-x..........x.....x
Kb for L^- = (Kw/Ka for HL) = (x)(x)/(0.2-x)
Solve for x = (OH-) and convert to pH.
c is the same.
Post your work if you get stuck.
Calculate the pH at the equivalence point in titrating 0.100 M solutions of each of the following with 0.038 M NaOH.
(a) hydrobromic acid (HBr)
(b) lactic acid (HC3H5O3)
(c) sodium hydrogen chromate (NaHCrO4)
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