Question

How do you calculate ${{K}_{sp}}$ from molar solubility?

Answer 1

Hint: Solubility product is the product of the ionic concentrations or activities of an electrolyte and the molar solubility is the number of moles of salts that can be dissolved in one litre of solution to form a saturated solution and then derive the required equation which gives the answer.

Complete step by step answer:
In the classes of physical chemistry, we have studied about the concept of molar solubility and also about the solubility product and some of the general definitions relating to it.
Let us now see how the solubility product can be calculated from the molar solubility.
- Firstly let us understand what solubility product is and what the molar solubility states.
- Solubility product is the maximum product of the ionic concentrations or the activities of an electrolyte that at one temperature will be able to continue in equilibrium with undissolved phase or in simple word say can say that${{K}_{sp}}$is equilibrium constant for a solid substance dissolving in aqueous solution.
- Molar solubility is defined as the number of moles of salts that can be dissolved in one litre of solution to form a saturated solution.
Let us now take the general dissociation equilibrium as:
\[{{X}_{n}}{{Y}_{m}}n.{{X}^{m+}}+m.{{Y}^{n-}}\]
- We shall assume that we have molar solubility as $s$mol${{L}^{-1}}$ for this particular salt in water at room temperature and this depicts that we can dissolve ‘s’ moles of\[{{X}_{n}}{{Y}_{m}}\] per litre of solution at this particular temperature.
Here, every mole of \[{{X}_{n}}{{Y}_{m}}\] that dissolves will produce n moles of\[{{X}^{m+}}\]that are cations and m moles of \[{{Y}^{n-}}\] that are anions.
This is nothing but the saturated solution forms will have \[\left[ {{X}^{m+}} \right]=n.s\] and \[\left[ {{Y}^{n-}} \right]=m.s\]
Thus, the solubility product for this equilibrium dissociation will be:
${{K}_{sp}}={{\left( n\times s \right)}^{n}}\times {{\left( m\times s \right)}^{m}}$
\[\Rightarrow {{K}_{sp}}={{n}^{n}}\times {{s}^{n}}\times {{m}^{m}}\times {{s}^{m}}\]
This is nothing but,
${{K}_{sp}}={{n}^{n}}\times {{m}^{m}}\times {{s}^{(n+m)}}$
Thus, this gives the required answer.

Note: Note that the molar solubility is related to the solubility product constant ${{K}_{sp}}$ as the higher the value of the solubility product, more soluble the compound will be and this says that they are directly related to each other.