Chemistry, Acid Base Titration

Question

What is the pH if you have 1.0M weak acid with pKa = 5.55?

If you add 20.0mL of this weak acid with 20.0mL of 1.0M

NaOH, a strong base, what would be the resulting pH?

Please explain the steps thanks.

Answer 1

Let the weak acid be HA

HA ==> H⁺ + A^-

Ka = [H⁺][A^-] / [HA]

Ka = 1x10^-5.55 (from pKa) = 2.82x10^-6

2.82x10^-6 = (x)(x) / (1.0 -x) and assuming x is small relative to 1.0, ignore it in denominator

2.82x10^-6 = x² /1.0

x = [H⁺] = 1.68x10^-3 M (very small relative to 1.0 so above assumption was valid)

pH = -log [H⁺] = -log 1.68x10^-3

pH = 2.77

When adding 20 ml of HA to 20 ml 1.0 M NaOH, the reaction that takes place is as follows:

HA + NaOH ==> NaA + H2O

initial moles HA = 20.0 ml x 1 L / 1000 ml x 1.0 mol/L = 0.02 moles HA

initial moles NaOH = 20.0 ml x 1 L / 1000 ml x 1.0 mol/L = 0.02 mols NaOH

Final moles HA = 0 an Final mols NaOH = 0. All has been converted to NaA + H2O

You can see this in the following ICE table:

HA + NaOH ==> NaA + H2O

0.02....0.02...............0................Initial

-0.02...-0.02...........+0.02...........Change

0..........0...................0.02..........Equilibrium

At equilibrium, we have 40 ml (0.04L) so [A^-] = 0.02 mol / 0.04L = 0.5 M

So, to find the pH of this solution, we must look at the hydrolysis of the NaA. Since this is a neutralization reaction between a weak acid and strong base, the pH at equivalence will be >7.

A^- + H2O ==> HA + OH^- ... hydrolysis of NaA

A^- is acting as a base so we need the Kb. This can be obtained from the Ka for HA

KaKb = 1x10^-14

Kb = 1x10^-14 / 2.82x10^-6 = 3.55x10^-9

Kb = 3.55x10^-9 = [HA][OH^-] / [A^-] [A-]

3.55x10^-9 = (x)(x) / 0.5 -x and assume x is small relative to 0.5 and ignore it in denominator

3.55x10^-9 = x² / 0.5

x = 4.2x10^-5 M = [OH^-]

pOH = -log [OH^-] = -log 4.2x10^-5

pOH = 4.38

pH = 14 - 4.38

pH = 9.62