Question

A 35.00−mL solution of 0.2500 M HF is titrated with a standardized 0.1606 M solution of...

A 35.00−mL solution of 0.2500 M HF is titrated with a standardized 0.1606 M solution of NaOH at 25

°

C.

(a) What is the pH of the HF solution before titrant is added?



(b) How many milliliters of titrant are required to reach the equivalence point?

mL

(c) What is the pH at 0.50 mL before the equivalence point?



(d) What is the pH at the equivalence point?



(e) What is the pH at 0.50 mL after the equivalence point?

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Homework Answers

Answer #1

HF , Ka = 6.3 x 10^-4

pKa = 3.20

a )

pH = 1/2 (pKa - log C)

    = 1/2 (3.20 - log 0.25)

pH = 1.91

b)

At equivalence point :

millimoles of acid = millimoles of base

35 x 0.250 = V x 0.1606

V = 54.48 mL

volume of base = 54.48 mL

c)

millimoles of acid = 35 x 0.25 = 8.75

millimoles of base = 53.98 x 0.1606 = 8.669

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

8.75 8.669    0 0

0.081 0    8.669   

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

pH = 3.20 + log (8.669 / 0.081)

pH = 5.23

d)

at equivalence point :

salt remains.

salt concentration = 8.75 / 35 + 54.48 = 0.0978 M

pH = 7 + 1/2 (pKa + log C)

     = 7 + 1/2 (3.20 + log 0.0978)

pH = 8.10

e )

pOH = 3.05

pH = 10.95

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