The solubility of slaked lime, Ca(OH)2, in water is 0.185 g/100.0 mL. Calculate the number of moles of Ca(OH)2 in 1.10×101 mL of a saturated solution.?

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Answer 1 · 2017-11-05T12:08:05

In what volume....?

We interrogate the equilibrium....

$Ca(OH)_2(s) stackrel(H_2O)rarrCa^(2+) + 2HO^-$

For which, $K_"sp"=[Ca^(2+)][HO^-]^2=5.5xx10^-6$ according to [this site...](https://en.wikipedia.org/wiki/Calcium_hydroxide)

If we put $"solubility of calcium hydroxide"-=S$ , then substituting in the $K_"sp"$ expression....we get....

$K_"sp"=Sxx(2S)^2=4S^3$ ....and thus $S=""^(3)sqrt((K_"sp")/(4))$

$=""^3sqrt((5.5xx10^-6)/(4))=0.0111*mol*L^-1$

A gram solubility of ...........................

$0.0111*mol*L^-1xx74.09*g*mol^-1=0.824*g*L^-1$ ...which is different to the solubility you quoted. I wonder who has the correct value.