Question

A 5.00 mL saturated sample of borax solution was collected at 43.8 degrees Celsius. It required 16.39 mL of 0.2503 M HCl to titrate. Calculate the Ksp for this sample.

Answer 1

n(HCl) = C(HCl) × V(HCl) = 0.2503M × 0.01639L = 0.0041mol.

  Xmol                0.0041mol

Na₂B₄O₇×10H₂O + 2HCl = 2NaCl + 4H₃BO₃ + 5H₂O

 1 mol                    2mol

n_borax = Xmol = 0.0041/2 mol = 0.00205mol.

C_borax = n_borax/V_borax = 0.00205mol/0.005L = 0.41M.

 0.41M                   0.82M 0.41M

Na₂B₄O₇×10H₂O  ⇄ 2Na⁺ + B₄O₇^2- +10H₂O

    1M                        2M       1M

K_{sp borax} = [Na⁺]² × [B₄O₇^2-] = 0.82² × 0.41 = 0.2757 ≈ 0.28.