Question #151289

calculate the solubility of Mg(OH) in moles per liter if the solubility product of Mg(OH)2 is 5 x 10^12

Expert's answer

Mg(OH)^{2} ------------- Mg^{2+ }+2(OH)^{-}

Solubility product=[Mg^{2+}][OH^{-}]^{2}

Let the conc of Mg^{2+}= x

So (OH^{-}) =2x

So we have;

5×10^{12}=(x)(2x)^{2}

5×10^{12}=4x^{3}

x^{3}=1.25×10^{12}

x=1.08×10^{4}

Therefore the Solubility of Mg(OH)2=1.08×10^{4}

Learn more about our help with Assignments: Inorganic Chemistry

## Comments

Hasnain naqvi16.12.20, 15:57Sir ,thanks a lot .you make easy my work.

## Leave a comment