Question

# Solubility of calcium phosphate (molecular weight= M) in water at $25{}^\circ C$ is W g per 100mL. Its solubility product at $25{}^\circ C$ will be approximately:(A) ${{10}^{9}}{{\left( \dfrac{W}{M} \right)}^{5}}$ (B) ${{10}^{7}}{{\left( \dfrac{W}{M} \right)}^{5}}$ (C) ${{10}^{5}}{{\left( \dfrac{W}{M} \right)}^{5}}$ (D) ${{10}^{3}}{{\left( \dfrac{W}{M} \right)}^{5}}$

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Hint: Solubility of a salt is the amount of salt soluble in the given solution at given temperature. Solubility product is the product of the concentration of the ions produced when a salt dissolves in the solution.

We should first write the reaction of solubility of calcium phosphate. The reaction for solubility of calcium phosphate is:

$C{{a}_{3}}{{(P{{O}_{4}})}_{2}}\rightleftharpoons 3C{{a}^{2+}}+2PO_{4}^{3-}$

The above reaction is dissociation of calcium phosphate. From the above reaction we can say that if solubility of calcium phosphate is S, so after their dissociation solubility of calcium ion will be 3S and that of phosphate ions will be 2S.
So, we can write that
\begin{align} & C{{a}_{3}}{{(P{{O}_{4}})}_{2}}\rightleftharpoons 3C{{a}^{2+}}+2PO_{4}^{3-} \\ & \,\,\,\,\,\,\,\,S\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3S\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2S \\ \end{align}
In the question it is given that there is 100 mL solution. This solution has W grams of calcium phosphate. So, we can say that 100 grams have W grams calcium phosphate. So, we will now find it in 1000 mL.
100 mL has W grams of calcium phosphate.

1000 mL will have $\dfrac{W\times 1000}{100}$ = $W\times 10$ gm

This is the weight of solution.
Now, we will find the value of solubility by dividing it with M:

So, solubility $S=\dfrac{W\times 10}{M}$

Now, we will find the solubility product (${{K}_{sp}}$) .
So, we can write the following equation as three calcium ions and two phosphate ions get dissociated when the salt gets dissolved.
\begin{align} & {{K}_{sp}}={{[C{{a}^{2+}}]}^{3}}{{[P{{O}_{4}}^{3-}]}^{2}} \\ & {{K}_{sp}}={{(3S)}^{3}}{{(2S)}^{2}} \\ & {{K}_{sp}}=108{{S}^{5}} \\ \end{align}
Thus, we obtained that ${{K}_{sp}}=108{{S}^{5}}$
Now, we will put the value of S in this.

${{K}_{sp}}=108{{\left( \dfrac{W\times 10}{M} \right)}^{5}}$
So, we can write it as ${{K}_{sp}}=108\times {{10}^{5}}{{\left( \dfrac{W}{M} \right)}^{5}}$
Thus, we can say that ${{K}_{sp}}=108\times {{10}^{5}}{{\left( \dfrac{W}{M} \right)}^{5}}\sim {{10}^{7}}{{\left( \dfrac{W}{M} \right)}^{5}}$

So, from this we can say that the correct answer of this question is option B.

Note: While calculating the solubility product of any salt, take in consideration the number of cations and anions it will give upon dissociation. Here, we are already given the solubility of the salt as W, so the temperature will not be used in the calculation of the solubility product.