Therefore, the final pH of the solution is approximately 13.745.
Step 1: Calculate moles of pure NaOH in the impure sample:
Molar mass of NaOH = 39.997 g/mol
Mass of pure NaOH = 3.5 g * 0.987 = 3.479 g
Moles of pure NaOH = 3.479 g / 39.997 g/mol = 0.0869 mol
Step 2: Calculate concentration of NaOH solution:
Volume of solution = 100 cm3 = 0.1 L
Concentration of NaOH = 0.0869 mol / 0.1 L = 0.869 mol/dm³
Step 3: Calculate moles of H+ ions added from diprotic acid
Volume of diprotic acid added = 25 cm3 = 0.025 L
Moles of H+ ions (assuming complete dissociation) = 0.025 L * 0.35 mol/dm³ * 2 (2 H+ ions per acid molecule) = 0.0175 mol
Step 4: Determine the limiting reagent and remaining reactants:
Compare moles of OH- ions (0.0869 mol) to moles of H+ ions (0.0175 mol)
OH- ions are in excess as 0.0869 mol > 0.0175 mol
The remaining OH- ions will react with the first dissociation step of the diprotic acid.
Step 5: Calculate moles of OH- ions remaining after reaction with the first dissociation step:
Moles of OH- ions consumed = 0.0175 mol H+ ions
Moles of OH- ions remaining = 0.0869 mol - 0.0175 mol = 0.0694 mol
Step 6: Calculate concentration of OH- ions in the final solution:
Final volume of solution = 100 cm3 + 25 cm3 = 125 cm3 = 0.125 L
Concentration of OH- ions = 0.0694 mol / 0.125 L = 0.555 mol/dm³
Step 7: Calculate pOH and pH:
pOH = -log10([OH-]) = -log10(0.555) = 0.255
pH = 14 - pOH = 14 - 0.255 = 13.745
Therefore, the final pH of the solution is approximately 13.745.