Question

This question is for a laboratory experiment titled "Determination of the Solubility Product Constant of Calcium Hydroxide" where a saturated solution of Ca(OH)₂ in 0.02523M NaOH was titrated with HCl.

Part 2 Data - Saturated Solution of Ca(OH)₂ in 0.02523M NaOH:

Volume of Ca(OH)₂/NaOH aliquot: 25.00mL

Concentration of standard HCl: 0.1342M

Indicator Used: Bromothymol Blue

Average volume of HCl to reach end point: 10.26mL

K_sp from Part 1 of lab: 3.54x10^-5

a. Calculate the TOTAL [OH^-] in the saturated solution of Ca(OH)₂ in sodium hydroxide for the solution assigned (Ca(OH)₂ in 0.02523M NaOH).

b. Calculate the [OH^-] that comes from the dissolution of Ca(OH)₂. The total [OH^-] (calculated above) is the sum of the [OH^-] from the NaOH and the [OH^-] from Ca(OH)₂.

c. Calculate the solubility of Ca(OH)₂ in the NaOH solution (in mol/L).

d. Calculate the experimental K_sp of Ca(OH)₂ for the saturated solution of Ca(OH)₂ in NaOH.

Please explain steps/show equations! Thank you.

Answer 1

Solution.

a) The net reaction between a strong acid and a strong base is

At the equivalence point

b) The sodium hydroxide dissociates completely giving 0.02523 M of hydroxide-ions. Therefore, the [OH^-] that comes from the dissolution of Ca(OH)₂ is 0.05508-0.02523 = 0.02985 M.

c) The dissociation equation of calcium hydroxide is

Ca(OH)_{2 =} Ca²⁺+2OH^-

The concentration of Ca²⁺ ions is two times less than that of hydroxide-ions formed, namely,

[Ca²⁺] = 0.02985/2 = 0.014925 M;

d) The solubility product expression is

Ksp = [Ca²⁺][OH^-]² = (0.014925)(0.05508)² = 4.5*10^-5.