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...

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.