Question

[NaOH] [NaOH) = 0 mol L-1 [NaOH] = 0.0127 mol L- [NaOH] = 0.0252 mol L-1 = 0.0509 mol L- Kse BER Solubility /mol L-1 Solubility / mol L- Solubility /mol L-1 Solubility /mol L- 2.35E-02| 5.2E-05 1.94E-02| 5.1E-05| 1.59E-02| 5.1E-05| 1.08E-02|| 5.7E-05 2.35E-02| 5.2E-05 1.95E-02 5.2E-05 1.59E-02| 5.2E-05 1.09E-02 5.7E-05 2.36E-02 5.3E-05 1.89E-02 4.8E-05 1.50E-02 4.5E-05 1.085-02 5.7E-05 2.38E-02 5.45-05 1.90E-02| 4.9E-05 1.05E-02 5.4E-05 2.34E-02 5.15-05 1.90E-02| 4.9E-05 9.95E-03 5.0E-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

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

Answer 1

The solubility of Ca (OH), steadily decreases with the increasing NaOH concentration. The solubility data obtained at a constant ionic strength (1: IM Nalch, OH) enabled the determined of the conditional solubility product op G(OH)2 (s) i Igl Caloude: - 4.10'16.02). The solubility of Calolle solution is decreased in which Nooh is dissolved, compared with that in bure water due to common ion effect. The Hydroxide ion is common ion. The dissolution of Nool to a solution of Ca (OH)2 will decrease the solubility of ColoH) . CalOH)2 is scoringly soluble salt i The dissociation of Calolla in solution is reforesented by following solubility equilibrium Cq (014) 2 - 2 + + 2OH- (aq) Noon is a strong electrolyte and completely dissociates in aqueous solution - NaOH(s) - Not (aq) + Olt (eqi)