`Ca(OH)_2 harr Ca^(2+)+2OH^(-)`
If the solubility of `Ca(OH)_2` is xM then in the equilibrium solution we have;
`[Ca^(2+)] = x`
`[OH^-] = 2x`
`K_(sp) = [Ca^(2+)][OH^-]^2`
`8xx10^(-6) = x(2x)^2`
`8xx10^(-6) = 4x^3`
`x = 1.26xx10^(-2)`
Molar mass of `Ca(OH)_2` = 74g/mol
Solubility of `Ca(OH)_2 = 74xx1.26xx10^(-2) = 0.93g/L`
So the solubility of `Ca(OH)_2 ` is `0.93g/L`
Further Reading
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