This question is for a lab experiment titled \"Determination of the Solubility P

Question

This question is for a lab experiment titled "Determination of the Solubility Product Constant of Calcium Hydroxide".

Given below are data from a previous year. The Ksp is independent of [NaOH] while the solubility of Ca(OH)2 decreases as the [NaOH] increases. Explain why this is the case.

[NaOH] = 0.0127 mol Ll NaOH] [NaOH] [NaOH] 0 mol L 0.0252 mol L-1 0.0509 mol L-1 SolubilityKpSolubilityKp SolubilityKp Solubility Kp / mol L1 2.35E-02 5.2E-051.94E-025.1E-05 .59E-02 5.1E-051.08E-025.7E-05 2.35E-02 5.2E-051.95E-025.2E-05 .59E-02 5.2E-051.09E-025.7E-05 2.36E-02 5.3E-051.89E-024.8E-051.50E-02 4.5E-051.08E-025.7E-05 2.38E-02 5.4E-051.90E-02 4.9E-05 2.34E-02 5.1E-051.90E-02 4.9E-05 2.42E-02 5.7E-05 2.41E-02 5.6E-05 2.35E-02 5.2E-05 2.31E-02 5.0E-05 2.43E-02 5.7E-05 2.31E-02 5.0E-05 / mol L1 / mol L1 / mol L 1.05E-02 5.4E-05 9.95E-03 5.0E-05

Explanation / Answer

Ksp represents the maximum amount of solubility: Ksp = [Ca+][OH-]^2 which is a constant number

if for some reason [OH-] increeases (like NaOH was added) then Ksp is still the same number but [Ca+] must decrease to keep Ksp a constant - thus more Ca(OH)2 goes into solid phase

The solubility of a sparingly soluble salt such as Ca(OH)2, is reduced in a solution that contains an ion in common with that salt (in this case, OH-). This is called the common ion effect and is an example of Le Chatelier's principle. The solubility of Ca(OH)2 can be describe by the following equation:

Ca(OH)2(s) <--> Ca2+(aq) + 2OH-(aq)

Thus adding OH- (via NaOH) will shift the reaction to the left where the solubility decreases.

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