Titration Chemistry

Question

7 mL of a 0.025 M H2SO4 solution neutralizes 8 mL of an AL9OH)3 solution of an unknown concentration. What is the concentration of the base?

Answer 1

First, write the correctly balanced equation for the reaction taking place:

3H₂SO₄ + 2Al(OH)₃ ==> Al₂(SO₄)₃ + 6H₂O ... balanced equation

moles H₂SO₄ used = 7 ml x 1 L/1000 ml x 0.025 mol/L = 0.000175 moles = 1.75x10^-4 mol H₂SO₄

moles Al(OH)₃ present = 1.75x10^-4 mol H₂SO₄ x 2 mol Al(OH)₃ / 3 mol H₂SO₄ = 1.17x10^-4 moles Al(OH)₃

Concentration of Al(OH)₃ = mol/L = 1.17x10^-4 mol / 0.008 L = 0.0146 M = 1.46x10^-2 M