`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)...

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`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**