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

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

Part 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).

This is an example of a strong acid-strong base reaction

H⁺ + OH^-       H₂O

At the equivalence point,

Moles of H⁺ = Moles of OH^-

[OH^-] = [(0.1342) x (10.26)]/(25) = 0.05508 M

Part 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.

[OH^-] from the dissolution of Ca(OH)₂ = 0.05508 M - 0.0252 M = 0.0298 M.

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

Ca(OH)₂ Ca²⁺ + 2OH^-

Therefore,

[Ca²⁺] = 0.0298/2 = 0.0149 M

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

Ksp = [Ca²⁺][OH^-]² = (0.0149)(0.055)² = 4.5 x 10^-5