Answer to Question #92546 in Organic Chemistry for Lee

Question

A titration was performed on a 10.00-mL sample of water taken from an acidic lake. If it took 9.56 mL of 0.0412 mol/L NaOH(aq) to neutralize the sulphuric acid in the lake water sample, calculate the concentration of the sulphuric acid. What is the pH of this lake water?

Answer 1

n(NaOH) = C(NaOH) × V(NaOH) = 0.0412 mol/L × 0.00956 L = 3.94 × 10^-4 mol.

3.94 × 10^-4 mol           X mol

      2NaOH       +        H₂SO₄ = Na₂SO₄ + 2H₂O

        2 mol                  1 mol

n(H₂SO₄) = X = 1.97 × 10^-4 mol.

C(H₂SO₄) = n(H₂SO₄) / V(H₂SO₄) = 1.97 × 10^-4 mol / 0.01 L = 0.0197 mol/L.

0.0197 mol/L     Y mol/L

     H₂SO₄       =     2H⁺ + SO₄^2-

     1 mol/L        2 mol/L

C(H⁺) = Y = 0.0394 mol/L.

pH = -lg[C(H⁺)] = -lg(0.0394) = 1.405 ≈ 1.4.