Question

In a titration of a 100.0mL 1.00M HA weak acid solution with 1.00M NaOH, what is the pH of the solution after the addition of 122 mL of NaOH? Ka = 1.80 x 10-5 for HA. (3 significant figures) **Remember to calculate the equivalence volume and think about in which region along the titration curve the volume of base falls, region 1, 2, 3, or 4**

Answer 1

moles of NaOH = molarity x volume in Litres = 1.0 M x 0.122 L = 0.122 mol

moles of weak acid, HA = molarity x volume in Litres = 1.0 M x 0.1 L = 0.1 mol

Let,

concentration of OH- equilibrium = x

Then,

[HA] = 0.1 +x

[NaOH] = 0.122 -x

A- + H2O ---------------> HA + OH-

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

Kw/Ka = [OH-] [ 0.1 +x ] / [ 0.122-x]

1.0x 10^-14/1.80 x 10^-5 = [x] [ 0.1 +x ] /[ 0.122-x]

0.555 x 10-9 =  [x] [ 0.1 +x ] /[ 0.122-x]

Since x is very small,

0.555 x 10^-9 =  [x] [ 0.1 ] / [ 0.122]

x = 0.6771 x 10^-9

Hence,

[OH-] = 0.6771 x 10^-9

pOH = - log [OH- ] = -log [ 0.6771 x 10^-9] = 9.17

Therefore,

pH = 14-9.17 = 4.83

pH = 4.83