Question

Consider the titration of a 25.0 −mL sample of 0.110 M HC2H3O2 with 0.130 M NaOH.

Ka(H2C2H3O2) = 1.8x10^-5

A) Initial PH

B)  the volume of added base required to reach the equivalence point

C)   the pH at 4.00 mL of added base

D)   the pH at one-half of the equivalence point

Answer 1

A ) initial pH

CH3COOH --------------------> CH3COO- + H+

0.110                                          0                 0

0.110 -x                                       x                x

Ka = x^2 / 0.110 -x

1.8 x 10^-5 = x^2 / 0.110 -x

x^2 + 1.8 x 10^-5 x - 1.98 x 10^-6 = 0

x = 1.40 x 10^-3

x = [H+] = 1.40 x 10^-3 M

pH = -log [H+] = -log (1.40 x 10^-3)

pH = 2.85

B)

at equivalence point

moles of acid = moles of base

250 x 0.110 = 0.130 x V

V = 211.5 mL

volume of base = 211.5 mL

here only salt CH3COONa remains

its concentration = C = 250 x 0.110 / (250 + 211.5)

                         C = 0.0596 M

it is the salt of weak acid and strong base . so pH

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

pH = 7 + 1/2 [4.74 + log 0.0596]

pH = 8.76

C)

millimole of acid = 27.5

millimoles of base = 0.130 x 4 = 0.52

CH3COOH + NaOH --------------------> CH3COONa + H2O

27.5                0.52                                0                     0

26.98                0                                  0.52                  0.52

pH = pKa + log [salt /acid]

pH = 4.74 + log (0.52 / 26.98)

pH = 3.02

D )

at half equivalenece point pH = pKa

so

pH = 4.74