Question

2) A student performed the titrations done in this experiment. The student pipetted 25mL of solution that contained saturated calcium hydroxide solution and 0.05 M calcium ions into a flask and then titrated it with 0.103 M HCl. It required 2.77mL of HCl to reach the orange-red end point. Calculate the molar solubility of calcium hydroxide in this solution.

Answer 1

Answer:

5.706x10⁻³M

Explanation:

Based on product solubility of Ca(OH)₂:

Ca(OH)₂(s) ⇆ Ca²⁺ + 2OH⁻(aq)

Ksp = 5.5x10⁻⁶ = [Ca²⁺] [OH⁻]²

Where [Ca²⁺] is 0.05M and [OH⁻] is obtained from the reaction with HCl.

The molar solubility, S, will be:

S = [OH⁻]/2

The [OH-] is:

Moles HCl = Moles OH⁻

2.77x10⁻³L * (0.103mol / L) = 2.85x10⁻⁴ moles OH⁻ in 25mL = 0.025L:

2.85x10⁻⁴ moles OH⁻ / 0.025L = 0.0114M = [OH⁻]

And S is:

0.0114M/2 =

5.706x10⁻³M