Answer to Question #92546 in Organic Chemistry for Lee

Question #92546
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?
1
Expert's answer
2019-08-13T04:53:53-0400

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+ + SO42-

     1 mol/L        2 mol/L


C(H+) = Y = 0.0394 mol/L.

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

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