What is the concentration of OH- and Ca ions and the Ksp?

Question

In the lab, you'll be titrating a saturated solution of Ca(OH)₂ using 0.05 M HCl to determine the K_sp for Ca(OH)₂ in water. This is possible since when you titrate to the equivalence point, the moles of OH^- = moles of H⁺; by using the volume and molarity of your known acid, you can determine the concentration of hydroxide from Ca(OH)₂.

If you used 31 mL of 0.05 M HCl to titrate 20 mL of Ca(OH)₂, what is the concentration of hydroxide ions in the original solution?

Start with writing the reaction equation between Ca(OH)₂ and HCl, and use stoichiometry. 

Give your answer to the nearest 0.001 M.

______ M

Once you know the concentration of the hydroxide ions in solution, you can calculate the concentration of calcium ions as well.

Assuming the concentration of OH^- ions in solution is 0.056 M, what is the concentration of Ca²⁺ ions?

Give your answer to the nearest 0.001 M.

________ M

Once you know the concentration of your hydroxide and calcium ions, you can easily calculate Ksp. 

Assuming the concentration of hydroxide ions in solution was 0.032 M, what is the K_sp value?

Give your answer to the 3 significant figures. 

____ x10___

Answer 1

Reaction:

Ca(OH)₂ + 2HCl ==> CaCl₂ + 2H₂O ... balanced equation (note mol ratio of HCl to OH- is 1:1 since you have 2 OH- from each Ca(OH)₂

Concentration of OH- in original solution:

moles HCl used = 31 mls x 1 L / 1000 mls x 0.05 mols / L = 0.00155 mols HCl

moles OH- present = 0.00155 mols HCl x 2 mol HCl / 2 mols OH- = 0.00155 mols OH-

concentration of OH- = 0.00155 mols OH- / 20 mls x 1000 mls / L = 0.0775 mols OH-/L = 0.078 M

If the [OH-] is 0.056 M, the [Ca²⁺ ] = 1/2 x 0.056 = 0.028 M since there are 2 OH- for each Ca(OH)₂

Assuming [Ca²⁺] = 0.028 M and assuming [OH-] = 0.032 M, then we can find Ksp as follows:

Ca(OH)₂(s) <==> Ca²⁺(aq) + 2OH-(aq)

Ksp = [Ca²⁺][OH-]²

Ksp = [0.028][0.032]²

Ksp = 2.87x10^-5 (3 sig. figs.)