Asked by Help101
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)
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)
Answers
Answered by
DrBob222
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.
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.
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