Question

Common ion effect;At some other temperature, a solid sample of Ca(OH)2 is shaken with 0.0100?
Common ion effect;At some other temperature, a solid sample of Ca(OH)2 is shaken with 0.0100 M CaCl2. Once equ?
At some other temperature, a solid sample of Ca(OH)2 is shaken with 0.0100 M CaCl2.
Once equilibrated, some solid Ca(OH)2 remains undissolved.
The solution is filtered and a 20.00 mL sample requires 38.80 mL of 0.0100 M HCl to neutralize it.
Calculate the value for Ksp of Ca(OH)2 from this data.
What is the solubility of Ca(OH)2 in this solution, in g/100mL
thanks in advance:)

Answer 1

I worked this today but I don't remember these numbers. I wonder if I misinterpreted the first post or if this one is actually different?
Ca(OH)2 + 2HCl ==> CaCl2 + 2H2O
moles HCl = M x L = 0.03880 x 0.01 =?
moles Ca(OH)2 = 1/2 that
mols Ca(OH)2 x molar mass = grams/20 mL.
grams x (100/20) = ? g/100 mL.

Ksp = (Ca^2+)(OH^-)^2

Ca^2+ = 0.01M
You have mols Ca(OH)2 from above. Calculate OH^- from that and divide by 0.02 to convert to M then substitute into Ksp expression.

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