Question

A 35.00-mL solution of 0.2500 M HF is titrated with a standardized 0.1120 M solution of NaOH at 25 C

1) the pH at 0.50 mL before the equivalence point
2) the pH at the equivalence point
3) the pH at 0.50 ml after the equivalence point.

Answer 1

1) the pH at 0.50 mL before the equivalence point

HF millimoles = 35 x 0.250 = 8.75

NaOH millimoles = 0.1120 x 0.5 = 0.056

HF pKa = 3.14

HF + NaOH --------------------> NAF   + H2O

8.75     0.056                             0             0 -------------------> initial

8.694      0                                  0.056       0.056 ----------------> after reaction

pH = pKa + log [NaF]/[HF]

pH = 3.14 + log (0.056/8.694)

pH = 0.95

2) the pH at the equivalence point

at equivalence point moles of acid = moles of base

millimoles of acid = 8.75

millimoles of base = 8.75

M x V = 8.75

0.112 x V = 8.75

V = volume of base = 78.12 ml

at this point only salt remains

NaF millimoles = 8.75

NaF concentration = 8.75 / total volume

                               = 8.75 / (35 + 78.12)

                                = 0.077 M

pH = 7 + 1/2 [pKa + logC]

pH = 7 + 1/2 [3.14 + log 0.077]

pH = 8.0

3) the pH at 0.50 ml after the equivalence point.

base volume = 78.12 + 0.5 = 78.62

millimoles of base = 0.112 x 78.62 = 8.8

HF + NaOH ----------------> NaF + H2O

8.75      8.8                           0          0

0            0.05

NaOH millimoles = 0.05

NaOH concentration = 0.05 / (35+78.62)

                     [OH-]             = 4.4 x 1)^-4 M

pOH = -log[OH-] = 3.36

pH + POH = 14

pH = 10.64