Ca(OH)2 ==> Ca^2+ + 2OH^-
and Ksp = (Ca^2+)(OH^-)^2
Ksp is a constant. NaOH is a strong base and ionizes 100%; therefore, NaOH ==> Na^+ + OH^-. Le Chatelier's Principle tells you that when the OH^- increases, due to the NaOH, the Ksp equilibrium of Ca(OH)2 is forced to the left thereby decreasing the solubility of Ca(OH)2 in the solution.
So what would the explanation for this be? Its really confusing me.
The Ksp is independent of [NaOH] while the solubility of Ca(OH)2 decreases as the [NaOH] increases. Explain why this is the case.
( in the data table, the naoh concentration is 0 M, and the solubility is 2.30 x 10^-2, and ksp is around 5.5x10^-5; next table the concentration of naoh increases and the solubility and ksp go down)
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