Question

A sample of carbonic acid (0.125 L, 0.199 M, pK_a1 = 6.35, pK_a2 = 10.33) was titrated with 1.75 M NaOH. Calculate the pH at the following points:

Before the titration.

At the 1st midpoint.

At the 1st stoichiometric (equivalence) point.

At the 2nd midpoint.

At the 2nd stoichiometric (equivalence) point.

Answer 1

1st mid point:

At first mid point: pH = pKa1 = 6.35 (because the number of mols of carbonic acid = number of mols of NaOH)

1st equivalence point:

The number of mols of salt formed = (0.125 L)(0.199 mol/L) = 0.0248 mol

Therefore -

              HCO₃^- (aq) + H₂O (l)   <----------> H₂CO₃ (aq) + OH^- (aq)

I(mol)    0.199             -                             0                   0

C             -x                                               +x                +x

Eq         (0.199-x)                                         x                 x

                K_b   = 2.238*10^-8 = x² / (0.199-x)

By solving, [OH^-] = 6.673*10^-5 M

pOH = -log [OH^- = -log (6.673*10^-5 ) = 4.17

Hence, pH = 14 - 4.17 = 9.83

2nd midpoint: pH = pKa2 = 10.33

2nd equivalence point:

The number of mols of salt formed = (0.125 L)(0.199 mol/L) = 0.0248 mol

Therefore -

              HCO₃^- (aq) + H₂O (l)   <----------> H₂CO₃ (aq) + OH^- (aq)

I(mol)    0.178             -                             0                   0

C             -x                                               +x                +x

Eq         (0.178-x)                                         x                 x

               

                      K_b   = 4.671*10^-11 = x² / (0.178-x)

By solving, [OH^-] = 2.88*10^-6 M

pOH = -log [OH^- = -log (2.88*10^-6 ) = 5.53

Hence, pH = 14 - 5.33 = 9.80

2nd midpoint: pH = pKa2 = 10.33