Question

A 35.00 mL solution of 0.2500 M HF is titrated with a standardized 0.1118 M solution of NaOH at 25 degrees C.

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

b.) How many mL of titrant are required to reach the equivalence point?

c.) What is the pH of 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?

Answer 1

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

HF -------------------> H+ + F-

0.25 0 0 -------------> initial

0.25-x x x --------------> equilibrium

Ka = x^2 / 0.25-x

6.3 x 10^-4 = x^2 / 0.25-x

x^2 + 6.3 x 10^-4 x - 1.58 x 10^-4 = 0

x = 0.0122

x = [H+] = 0.0122 M

pH = -log [H+] = -log (0.0122 )

pH = 1.91

b.) How many mL of titrant are required to reach the equivalence point?

millimoles of acid = millimoles of base

35 x 0.25 = V x 0.1118

V = 78.26 mL

volume of titrant = 78.26 mL

c.) What is the pH of 0.50 mL before the equivalence point?

millimoles of acid = 35 x 0.25 = 8.75

millimoles o base = 0.1118 x (78.26 -0.50) = 8.69

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

8.75 8.69 0 0

0.06 0 8.69 -

pH = pKa + log [NaF / HF]

pH = 3.20 + log (8.69 / 0.06)

pH = 5.36

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

here only salt NaF remains

its concentration = 8..75 / (35 + 78.26) = 0.0773 M

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

pH = 7 + 1/2 [3.2 + log 0.0773]

pH = 8.04

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

millimoles of base= 0.1118 x (78.26 + 0.5) = 8.805

base concentration = 8.805 - 8.75 / (35 + 78.76) = 4.83 x 10^-4 M

[OH-] = 4.83 x 10^-4 M

pOH = 3.3

pH + pOH = 14

pH = 10.70