Answer:
0.000241 M is the molar solubility of [tex]Mg(OH)_2[/tex] in pure water.
Explanation:
Solubility product of metal hydroxide =[tex] K_{sp}=5.61\times 10^{-11}[/tex]
[tex]M(OH)_2\rightleftharpoons M^{2+}+2OH^-[/tex]
S 2S
The expression of a solubility product is given by :
[tex]K_{sp}=[M^{2+}][OH^-]^2[/tex]
[tex]K_{sp}=S\times (2S)^2=4S^3[/tex]
[tex]5.61\times 10^{-11}=4S^3[/tex]
Solving for S:
[tex]S=0.000241 M[/tex]
0.000241 M is the molar solubility of [tex]Mg(OH)_2[/tex] in pure water.
