Question

Mg(OH)2 is a sparingly soluble salt with a solubility product, Ksp, of 5.61×10−11. It is used to control the pH and provide nutrients in the biological (microbial) treatment of municipal wastewater streams. What is the ratio of solubility of Mg(OH)2 dissolved in pure H2O to Mg(OH)2 dissolved in a 0.180 M NaOH solution? Express your answer numerically as the ratio of molar solubility in H2O to the molar solubility in NaOH.

Answer 1

Answer:

The ration of the molar solubility is 165068.49.

Explanation:

The solubility reaction of the magnesium hydroxide in the pure water is as follows.

$Mg(OH)_{2}\Leftrightarrow Mg^{2+}(aq)+2(OH)^{-}(aq)$

              $[Mg^{2+}][OH^{-}]$

Initial      0          0

Equili     +S       +2S

Final      S          2S

$K_{sp}=[Mg^{2+}][OH^{-}]$

$5.61\times 10^{-11}=(S)(2S)^{2}$

$S=(\frac{5.61\times 10^{-11}}{4})^{1/3}=2.41\times 10^{-4}M$

Solubility of $Mg(OH)_{2}$ in 0.180 M NaOH is a follows.

$Mg(OH)_{2}\Leftrightarrow Mg^{2+}(aq)+2(OH)^{-}(aq)$

              $[Mg^{2+}][OH^{-}]$

Initial      0          0

Equili     +S       +2S

Final      S          2S+0.180M

$K_{sp}=[Mg^{2+}][OH^{-}]$

$5.61\times 10^{-11}=(S)(2S+0.180)^{2}$

$S=1.46\times 10^{-9}M$

$Ratio\,of\,solubility=\frac{2.41\times 10^{-4}}{1.46\times 10^{-9}}=165068.49$

Therefore, The ration of the molar solubility is 165068.49.