Given data:
- 10 g of a dibasic acid in 1000 g of water freezes at -0.15°C
- 10 mL of the acid is neutralized by 12 mL of 0.1 N NaOH solution
- Kf (H2O) = 1.86 mol-1kg

To find: pH of the solution of the acid

Solution:
1. Calculation of molality (m) of the solution:
Molar mass of dibasic acid = 2 x atomic mass of carbon + 4 x atomic mass of hydrogen + 4 x atomic mass of oxygen
= 2(12.01) + 4(1.01) + 4(16.00)
= 88.06 g/mol
Number of moles of dibasic acid in 10 g = 10/88.06 = 0.1136 mol
Molality (m) = number of moles of solute / mass of solvent in kg
= 0.1136 / 0.990 (1000 g - 10 g) = 0.1149 mol/kg

2. Calculation of freezing point depression (∆Tf):
∆Tf = Kf x m
= 1.86 x 0.1149
= 0.2141°C

3. Calculation of freezing point of the solution:
Freezing point of the solution = -0.15°C - 0.2141°C
= -0.3641°C

4. Calculation of moles of NaOH used:
1 N NaOH = 1 mole of NaOH in 1 liter of solution
0.1 N NaOH = 0.1 mole of NaOH in 1 liter of solution
12 mL of 0.1 N NaOH = 0.1 x 12/1000 = 0.0012 moles of NaOH

5. Calculation of moles of dibasic acid neutralized by NaOH:
As dibasic acid reacts with 2 moles of NaOH,
moles of dibasic acid neutralized = 0.0012 x 2 = 0.0024 mol

6. Calculation of remaining moles of dibasic acid:
Remaining moles of dibasic acid = 0.1136 - 0.0024 = 0.1112 mol

7. Calculation of concentration of dibasic acid:
Concentration (C) = number of moles / volume of solution in liters
= 0.1112 / 0.99 (1000 mL - 10 mL)
= 0.1123 mol/L

8. Calculation of pKa of dibasic acid:
pH = pKa + log([HCOO-]/[H2COO])
As dibasic acid is a diprotic acid, it will have two pKa values. We need to consider the first pKa value, which corresponds to the dissociation of the first proton.
The balanced equation for the dissociation of dibasic acid is:
H2A ⇌ H+ + HA-
pKa = -log(Ka)
Ka = [H+][HA-] / [H2A]
At the half-neutralization point, [HA-] = [H2A] and [H+]