Question

A sample of ca(oh)2 is allowed to stand in distilled water until a saturated solution forms. A 50.00 mL sample of this saturated solution is neutralized with 24.4 mL of 0.05 mol/L Hcl
2hcl+ ca(oh)2>cacl2+2h2o
Use this information to calculate the ksp for ca(oh)2

Answer 1

mols HCl used = M x L = ?
mols Ca(OH)2 = 1/2 that from the coefficients in the balanced equation (of the titrated sample which is 50 mL).
mols Ca(OH)2 in the original sample of 0.05L. Convert to mols/L, substitute into Ksp expression and solve for Ksp.

Answer 2

I don't get it could you explain it with the calculations

Answer 3

So I tell you HOW to do it and you ask me to do your work for you?

mols HCl used = M x L = <b>0.0244L*0.050M = 0.00122 mols</b>

mols Ca(OH)2 = 1/2 that from the coefficients in the balanced equation (of the titrated sample which is 50 mL).
mols Ca(OH)2 in the original sample of 0.05L.<b>0.00122*1/2 = 0.00061 mols Ca(OH)2 in the 50 mL.</b>

Convert to mols/L, substitute into Ksp expression and solve for Ksp.
<b>0.00061*(1000 mL/50 mL) = 0.0122M = [Ca(OH)2]. If [Ca(OH)2] = 0.0122 then (Ca^2+) = 0.0122 and (OH^-) = 2*0.0122 = 0.0244 and
Ksp = (Ca^2+)(OH^-)^2 =
(0.0122)(0.0244)^2 = ?
Check my work carefully.</b>