Question

How do you calculate the solubility $ \left( \text{moles/liter} \right)$ of $ \text{Mg}{{\left( \text{OH} \right)}_{\text{2}}}$ in water? The$ {{K}_{sp}}=5.0\times {{10}^{-11}}$ ?

Answer 1

Hint Write the balanced ionization reaction of$ \text{Mg}{{\left( \text{OH} \right)}_{\text{2}}}$ .
We write the solubility product as the product of the solubility of the ions formed.
For a reaction,$ AB\rightleftharpoons {{A}^{+}}+{{B}^{-}}$
$ {{K}_{sp}}=\left[ {{A}^{+}} \right]\left[ {{B}^{-}} \right]$
$ {{K}_{sp}}=s\times s={{s}^{2}}$ , where s is the solubility of ions formed.

Complete step by step solution:
So in the question it is asked that how can we calculate the solubility of $ \text{Mg}{{\left( \text{OH} \right)}_{\text{2}}}$ in water in terms of moles per liter if the given solubility product value is $ 5.0\times {{10}^{-11}}$
If we have to know what is solubility and how to write the solubility product equation so that we could easily calculate the question and find the final answer.
So solubility of a substance is a parameter of how much or maximum amount of a substance can be dissolved in a specific known volume. Generally we calculate the solubility of the ions in the solution. The solution can be defined as the homogenous mixture of two or more substances; the solution consists of solute which is the dissolved substance in a medium and the solvent is the medium in which the solute substance is dissolved. The maximum amount of solute that can be dissolved in a volume of solvent which is known to us is called the solubility of that substance.
Here we have to find the solubility of magnesium hydroxide, for that lets write the balanced equilibrium dissociation equation of $ \text{Mg}{{\left( \text{OH} \right)}_{\text{2}}}$ .
The equation is as follows,
$ \text{Mg}{{\left( \text{OH} \right)}_{\text{2}}}_{\left( \text{s} \right)}\rightleftharpoons \text{Mg}_{\left( \text{aq} \right)}^{\text{2+}}\text{+2OH}_{\left( \text{aq} \right)}^{\text{-}}$
So from the equation we know that one mole of magnesium hydroxide produce one mole of magnesium ions but one mole of $ \text{Mg}{{\left( \text{OH} \right)}_{\text{2}}}$ has produced 2 moles of $ O{{H}^{-}}$ ions.
Now let’s write the solubility product for the reaction.
If the solubility of the ion is taken as ‘s’ the we can write the solubility of $ M{{g}^{2+}}$ is s and solubility of $ O{{H}^{-}}$ is 2s.
Hence we can write the solubility product equation as,
$ {{K}_{sp}}=s\times {{\left( 2s \right)}^{2}}=s\times 4{{s}^{2}}=4{{s}^{3}}$
$ {{K}_{sp}}=4{{s}^{3}}$ $ \text{Mg}{{\left( \text{OH} \right)}_{\text{2}}}_{\left( \text{s} \right)}\rightleftharpoons \text{Mg}_{\left( \text{aq} \right)}^{\text{2+}}\text{+2OH}_{\left( \text{aq} \right)}^{\text{-}}$
$ {{s}^{3}}=\dfrac{{{K}_{sp}}}{4}$
The value of$ {{K}_{sp}}=5\times {{10}^{-11}}$ , substitute this value in the above equation.
$ {{s}^{3}}=\dfrac{5\times {{10}^{-11}}}{4}=1.25\times {{10}^{-11}}$
$ s=\sqrt[3]{(1.25\times {{10}^{-11}})}=2.3207\times {{10}^{-4}}mole/L$

Note: Always be caution about writing the solubility product since if the coefficients are present it is written as the power of the concentration of ions and the coefficient value should be written as the power while writing the solubility of the ions.
If we have to find the solubility in terms of gram per liter, then multiply the molar mass with the solubility obtained. A determined parameter is the mass solubility.
The molar mass of $ \text{Mg}{{\left( \text{OH} \right)}_{\text{2}}}$ is $ 58.3197g/mol$
$ \text{Mass solubility=58}\text{.3197}\times \left( \text{2}\text{.3207}\times {{10}^{-4}} \right)=1.3534\times {{10}^{-2}}g/L$