chemistry question

Question

You are titrating a 700 mL sample of 0.15 M CCl₃COOH (pKa = 0.66) with 0.2 M NaOH. What will the pH be at the equivalence point?

Answer 1

To begin, let us look at the reaction during the titration:

CCl₃COOH + NaOH ==> CCl₃COONa + H2O

moles CCl₃COOH initially present = 700 ml x 1 L / 1000 ml x 0.15 mol / L = 0.105 mols CCl₃COOH

moles NaOH needed to reach equivalence = 0.105 mols CCl₃COOH x 1 mol NaOH / mol = 0.105 mol

volume NaOH needed = 0.105 mol NaOH x 1 L / 0.2 mols = 0.525 L

Final volume at equivalence = 0.700 L + 0.525 L = 1.225 L

At equivalence, there will be no CCl₃COOH or NaOH left. All has been converted to 0.105 mols CCl₃COO^-

Final [CCl₃COO-] = 0.105 mol / 1.225 L = 0.0857 M

Now we look at hydrolysis of this conjugate base:

CCl₃COO- + H2O ==> CCl₃COOH + OH^-

We need to find the Kb for CCl₃COO^- and we can do this from the given value of pKa

pKa + pKb = 14

pKb = 14 - 0.66

pKb = 13.34

Kb = 1x10^-13.34

Kb = 4.57x10^-14

4.57x10^-14 = [CCl₃COOH][OH^-] / [CCl₃COO^-]

4.57x10^-14 = (x)(x) / 0.0857 - x (assume x is very small relative to 0.0857 and ignore it)

4.57x10^-14 = (x)(x) / 0.0857

x = [OH^-] = 6.26x10^-8

pOH = -log [OH-] = -log 6.26x10^-8

pOH = 7.20

pH = 14 - pOH

pH = 6.80

(be sure to check all of the math)