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Question

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(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) ${{10}^{3}}{{\left( \dfrac{W}{M} \right)}^{5}}$

Answer

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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 \[{{K}_{sp}}=108\times {{10}^{5}}{{\left( \dfrac{W}{M} \right)}^{5}}\sim {{10}^{7}}{{\left( \dfrac{W}{M} \right)}^{5}}\]