Answer to Question #155759 in Organic Chemistry for Israa

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? Full GRASS solution.

Answer 1

H2SO4(aq) + 2NaOH(aq) = Na2SO4(aq) + 2H2O(l)

1 mol H2SO4 = 2 mol NaOH

no of mol of NaOH consumed = M*V

           = 0.0412$\times$ 9.56

           = 0.394 mmol

no of mol of H2SO4 reacted = $\frac{0.394}{2}$ = 0.197 mmol

concentration of H2SO4 in water sample = n/v

           = $\frac{0.197}{10}$

           = 0.0197 M

[H3O+] in water sample = 2$\times$ 0.0197 = 0.0394 M

pH = -log[H3O+]

  = -log0.0394

  = 1.4