Question

In an acid-base titration experiment, 50 mL of a 0.05 M solution of acetic acid (Ka = 1.75 x 10-5 ) was titrated with a 0.05M solution of NaOH at 25 oC. The system will acquire this pH after addition of 40 mL of the titrant:

Answer: 5.36

How do you solve this?

Answer 1

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

Moles of acetic acid = Molarity x Volume(in L) = 0.05 M x 0.05L = 2.5 x 10^-3 mol

Moles of NaOH = Molarity x Volume(in L) = 0.05 M x 0.04L = 2 x 10^-3 mol

So, the moles of CH3COONa formed = 2 x 10^-3 mol

Molarity of CH3COONa = 2 x 10^-3 mol / 0.09 L = 0.022 M

Now, let us find pH

         CH3COONa      <----------->         CH3COO- + Na+

         CH3COO- + H2O <------------>      CH3COOH + OH-

I         0.022                                                       0                 0

C       -x                                                              +x                +x

E        0.022 - x                                                  +x                +x

Given Ka of acetic acid = 1.75 x 10^-5

But, we need Kb of acetate ion.

We know that Kw = Ka x Kb

Kb = Kw / Ka

Kb = (1 x 10^-14) / (1.75 x 10^-5) = 5.7 x 10^-10

Kb = x^2 / (0.022-x) = 5.7 x 10^-10

As x is very small, we ignore it. So, 0.022-x = 0.022

x^2 = 5.7 x 10^-10 x (0.022)

x = 3.54 x 10^-6 M

Thus, [OH-] = x = 3.54 x 10^-6 M

pOH = -log [OH-] = -log(3.54 x 10^-6) = 5.45

pH = 14 - pOH

pH = 14 - 5.45 = 8.55