Question

A 25.0 mL sample of 0.150 M chloroacetic acid is titrated with a 0.150 M NaOH solution. What is the pH at the equivalence point? The of chloroacetic acid is 1.4 ×10-3.

Answer 1

The answer is: pH = 7.86. See solution below:

Let HA represent chloroacetic acid.

HA + NaOH => NaA + H2O

Volume of NaOH required = volume x molarity of HA/molarity of NaOH

= 25.0 x 0.150/0.150 = 25.0 mL

Initial molarity of A- = initial molarity of NaA

= moles of HA/total volume of solution

= (25/1000 x 0.150)/0.050 = 0.075 M

A-+H2O<=> HA + OH-

I0.0750 0

C-a+a +a

E0.075-aa a

Kb of A- = [HA][OH-]/[A-]

= a²/(0.075 - a) = Kw/Kb of HA

= 10^-14/1.4 x 10^-3 = 7.143 x 10^-12

a² + 7.143 x 10^-12a - 5.357 x 10^-13 = 0

The positive root of the quadratic equation is:

a = 7.319 x 10^-7

[OH-] = a = 7.319 x 10^-7 M

pOH = -log[OH-] = 6.14

pH = 14 - pOH = 7.86