# 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.?

Nov 5, 2017

In what volume....?

#### Explanation:

We interrogate the equilibrium....

$C a {\left(O H\right)}_{2} \left(s\right) \stackrel{{H}_{2} O}{\rightarrow} C {a}^{2 +} + 2 H {O}^{-}$

For which, ${K}_{\text{sp}} = \left[C {a}^{2 +}\right] {\left[H {O}^{-}\right]}^{2} = 5.5 \times {10}^{-} 6$ according to this site...

If we put $\text{solubility of calcium hydroxide} \equiv S$, then substituting in the ${K}_{\text{sp}}$ expression....we get....

${K}_{\text{sp}} = S \times {\left(2 S\right)}^{2} = 4 {S}^{3}$....and thus S=""^(3)sqrt((K_"sp")/(4))

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

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

$0.0111 \cdot m o l \cdot {L}^{-} 1 \times 74.09 \cdot g \cdot m o {l}^{-} 1 = 0.824 \cdot g \cdot {L}^{-} 1$...which is different to the solubility you quoted. I wonder who has the correct value.