Question

At \[25{}^\circ C\], the molar solubility of silver phosphate is $1.8\times {{10}^{-5}}mol{{L}^{-1}}$. How do you calculate $Ksp$ for this salt?

Answer 1

Hint: The solubility product of a salt is the value obtained by taking the product of the concentration of the ions which are involved in the reaction.
Molar solubility of a substance is the solubility of the ions or the number of ions which are soluble in a given unit mole of the solution in order to become a saturated solution.

Complete step by step answer: In order to solve this question, we must first write the chemical reaction which is involved in this reaction.
\[A{{g}_{3}}P{{O}_{4}}(s)\rightleftharpoons 3A{{g}^{+}}(aq)+P{{O}^{-3}}_{4}(aq)\]
Now, we will first define the solubility product of the ions in a solution. So, it is the maximum product of the concentration of the ions which are involved in the process, or the activity of the given electrolyte which is made up of ions in the aqueous solution which at a certain specific temperature would be able to establish and continue the equilibrium at the undissolved phase, or in other words, it can be defined as the maximum solubility product formed by taking the product of concentration of the ions in the aqueous solution. $Ksp$ is the symbol which represents the equilibrium constant of the solid substance which is dissolving in the aqueous solution.
On the other hand molar solubility of a substance is the number of moles of the salt in a given reaction which can be dissolved in a unit litre of the solution in order to form a saturated solution.
The value of molar solubility is given in the question as, $1.8\times {{10}^{-5}}mol{{L}^{-1}}$ which is equal to the product of concentrations of the constituent ions in the product part of the reaction. We get,
\[Ksp=1.8\times {{10}^{-5}}M={{[A{{g}^{+}}]}^{3}}(aq)\times [PO_{4}^{3-}](aq)\]
Now, we will replace the ions of the reactant with a letter ‘X’ as it is the value of $Ksp$ which we are supposed to find. We get,
\[1.8\times {{10}^{-5}}M={{(3X)}^{3}}\times (X)\]
Now, we will take the cube of three from the right hand side of the equation to the left hand side in order to get the value of ‘X’. we get,
\[\dfrac{1.8\times {{10}^{-5}}M}{27}~=~6.67\times {{10}^{-7}}M\] on the left hand side of the equation and on the right hand side we get ${{X}^{4}}$ after multiplying the cube of ‘X’ with the single ‘X’. So, the equation becomes,
\[6.67\times {{10}^{-7}}M={{X}^{4}}\]
So, now, in order to get the value of ‘X’ we will take the fourth root of the value on the left hand side, and we get,
\[{{(6.67\times {{10}^{-7}}M)}^{\dfrac{1}{4}}}=2.9\times {{10}^{-2}}M\]
So, the value of $Ksp$ is \[2.9\times {{10}^{-2}}M\] which is the required answer.

Note: The solubility product constant of a solution is related to the molar solubility of a substance as, with increase in solubility constant the molar solubility also increases. So, we could say that both are directly proportional to each other.