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A 35.00-mL solution of 0.2500 M HF is titrated with a standardized 0.1120 M solution of...

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.

Solutions

Expert Solution

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


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