Question

A 25.00 mL solution of Ca(OH)2 was titrated to the stoichiometric point with 12.15 mL of 0.144 M HNO3 (aq). What was the initial concentration of Ca(OH)2 in the solution?

Answer 1

Ca(OH)_2(aq) + 2 HNO_3(aq)Ca(NO₃)_2(aq) + 2 H₂O_(l)

Moles of HNO₃ = volume of HNO₃ x molarity of HNO₃

                          = 12.15 mL x 0.144 M

                          = 1.75 mmol

                          = 1.75 x 10^-3 mol

From the balanced equation above, it is clear that 2 mol of HNO₃ is required to titrate 1 mol of Ca(OH)₂ to the stoichiometric point or equivalence point

Therefore, 1.75 x 10^-3 mol of HNO₃ is required to titrate = (1.75 x 10^-3)/2 = 0.875 x 10^-3 mol of Ca(OH)₂ to the stoichiometric point or equivalence point

Initial volume of Ca(OH)₂ = 25.00 mL = 0.025 L

Let us say that the initial concentration of Ca(OH)₂ = Z (M)

Now,

0.025 x Z = 0.875 x 10^-3

or, Z = (0.875 x 10^-3)/0.025

        = 35 x 10^-3 M

        = 0.035 M

Hence, the initial concentration of Ca(OH)₂ = 0.035 M