Question

If the solubility product of $Z{{r}_{3}}{{(P{{O}_{4}})}_{4}}$ is denoted by ${{K}_{sp}}$ and its molar solubility is denoted by S, then which of the following relation between S and ${{K}_{sp}}$ is correct?
A. $S={{(\dfrac{{{K}_{sp}}}{929})}^{\dfrac{1}{9}}}$
B. $S={{(\dfrac{{{K}_{sp}}}{216})}^{\dfrac{1}{7}}}$
C. $S={{(\dfrac{{{K}_{sp}}}{144})}^{\dfrac{1}{6}}}$
D. $S={{(\dfrac{{{K}_{sp}}}{6912})}^{\dfrac{1}{7}}}$

Answer 1

The compound given is $Z{{r}_{3}}{{(P{{O}_{4}})}_{4}}$, which is an ionic compound. Write the ions formed when the compound dissociates in an aqueous solution. Then write in terms of molar solubility and find the expression for ${{K}_{sp}}$.

Complete step by step solution:
In order to solve the question let us know about molar solubility and solubility products. The solubility product constant can be called as the constant of the equilibrium, which is used for the dissociation of a solid substance into an aqueous solution or water. It is denoted by the symbol ${{K}_{sp}}$. Molar solubility, which is somehow related to the solubility product, is actually the number of moles of the solute that can be dissolved in 1 litre of solution before saturation of the solution takes place. Now, let us write the ${{K}_{sp}}$ for the compound we have in our question:
${{K}_{sp}}=[Z{{r}^{4+}}][P{{O}_{4}}^{3-}]$
We can also write the dissociation of the compound as:
$\begin{aligned} & Z{{r}_{3}}{{(P{{O}_{4}})}_{4}} \leftrightarrow 3Z{{r}^{4+}}+4P{{O}_{4}}^{3-} \\\ & \,\,\,\,\,\,\,s\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3s\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4s \\\ \end{aligned}$, taking solubility to be s, it is multiplied by the number of moles
In the final expression for ${{K}_{sp}}$, we will shift the number of moles to the power:
$\begin{aligned} & {{K}_{sp}}={{(3s)}^{3}}\times {{(4s)}^{4}} \\\ & {{K}_{sp}}=27{{s}^{3}}\times 256{{s}^{4}} \\\ & {{K}_{sp}}=6912{{s}^{7}} \\\ \end{aligned}$
So, we can rearrange the terms and write the final expression as:
$s={{(\dfrac{{{K}_{sp}}}{6912})}^{\dfrac{1}{7}}}$
So, we get the correct answer option as D, which is the required answer.

So, the molar solubility obtained from the given data is (D)- ${{(\dfrac{{{K}_{sp}}}{6912})}^{\dfrac{1}{7}}}$.

Note: Generally, solubility of salts can give us an idea about which type of salts they are(based on solubility)

Type 1	Soluble	Solubility more than 0.1M
Type 2	Slightly soluble	0.01M
Type 3	Sparingly soluble	Solubility less than 0.1M.