Calculate the value of Ksp at this temperature.

Question

At 25 ^oC the solubility of calcium hydroxide is 1.11 x 10^-2 mol/L. Calculate the value of K_sp at this temperature. Give your answer in scientific notation to 2 SIGNIFICANT FIGURES

Answer 1

Hi Oscar!

To find K_sp, we first need a chemical equation describing calcium hydroxide dissolving:

Ca(OH)₂→Ca²⁺+2OH^-

We also need the rate law for this equation:

K_c=[Ca²⁺][OH^-]²/[Ca(OH)₂]

The numerator in this rate law, [Ca²⁺][OH^-]², is K_sp. So we need to figure out what [Ca²⁺] and [OH^-] are. We can find them since we have the molarity (same thing as solubility) of the calcium hydroxide, 1.11*10^-2 mol/L. We always assume the chemical equation occurs in a 1 L solution, which makes the calculations easier. So, we are starting with 1.11*10^-2 moles of Ca(OH)₂. By looking at the coefficients in the equation (if not written then the coefficient is 1), we see that for every mole of Ca(OH)₂ put in, we get 1 mole of Ca²⁺ as a product. Therefore the moles of Ca²⁺ obtained is the number of moles of Ca(OH)₂ put in, 1.11*10^-2. Since this is still happening in a 1 L solution, [Ca²⁺]=1.11*10^-2 mol/L. As for OH^-, the equation says that for every mole of Ca(OH)₂ put in, we get 2 moles of OH^-. Therefore the moles of OH^- obtained is 2 times the number of moles of Ca(OH)₂ put in, or 2.22*10^-2, and like with [Ca²⁺], [OH^-]=2.22*10^-2 mol/L.

Therefore K_sp=(1.11*10^-2)(2.22*10^-2)²=5.5*10^-6 (do not plug in any measurement units).

I hope this helps!