Question

A 25.00 ml sample of Ca(OH)2 (aq) is titrated with 0.05668 M HCL. The titration requires 18.88 ml of HCL (aq) to teach the endpoint.

1) Write a net-ionic equation for the titration reaction.

2) Determine the molarity of the Ca(OH)2 (aq)

3) Express the molarity of the CA(OH)(aq) as a solubility in g Ca(OH)2/ 100 ml soln.

Answer 1

1). Reaction is

Ca(OH)₂ + 2HCl gives 2H₂O + CaCl₂

or

2OH^- + 2H⁺ gives 2H₂O

2) For neutralization of calcium hydroxide, 18.88 mL 0.05668 M HCl is used.

18.88 mL 0.05668 M HCl is contains (0.05668/1000)*18.88 moles = 0.00107 moles

From the balanced equation, it is clear that one calcium hydroxide reacts with 2 HCl

Thus, the number of moles of calcium hydroxide = 0.00107/2 = 0.00053

This in 25 mL

since Molarity is the number of moles in one-liter solution, we have to convert it to 1000 mL

ie (0.0053/25)*1000 = 0.0214 M

3). the solubility product of calcium hydroxide = 5.5×10^–6

Ca(OH)₂ gives Ca²⁺+ 2OH ^-   at equilibria

let x is the concentration of ca2+ and 2x for concntration of OH-

Ksp = [Ca²⁺][OH^-]²

Ksp = 4x³

5.5×10^–6 = 4x³

x = 0.0111199 M

This is the molar solubility.

Molar mass of calcium hydroxide is 74g/mole

solubility of calcium hydroxide in 1000 mL = 0.0111199 x74 = 0.8228 g

solubility of calcium hydroxide in 100 mL = 0.08228 g