So many solutions!

This question is an illustration of the Common-ion effect .

Solubility equilibrium of calcium phosphate:

$Ca_{3}(PO_{4})_{2(s)} \rightleftharpoons 3Ca^{2+} + 2PO_{4}^{3-}$

The expression for the solubility product is:

$K_{sp} = [Ca^{2+}]^3 [PO_{4}^{3-}]^2 = \beta$

In the first situation, the net amount (in mol/L) of calcium phosphate that will be dissociated when the system is in chemical equilibrium is $S_1$ . It follows from the stoichiometry that $[Ca^{2+}] = 3S_1$ and $[PO_{4}^{3-}] = 2S_1$ . Plugging into the the $K_{sp}$ expression:

$(3S_1)^3 (2S_1)^2 = \beta$

$S_1 = 2^{- \frac{2}{5}} 3^{- \frac{3}{5}} \beta ^{\frac{1}{5}}$

Now we need to find $S_2$ . Since the calcium nitrate solution has the same concentration of dissociated particles (in this case, ions) as a 4.5 mol/L NaCl solution:

$4.5 \hspace{2 mm} mol/L \hspace{2 mm} NaCl \Rightarrow 9 \hspace{2 mm} mol/L \hspace{2 mm} ions \Rightarrow 3 \hspace{2 mm} mol/L \hspace{2 mm} Ca(NO_3)_2 \Rightarrow 3 \hspace{2 mm} mol/L\hspace{2 mm} Ca^{2+}$

In the second situation, there are two sources of calcium ions: the ones from the $Ca(NO_3)_{2(aq)}$ ( $3 \hspace{2 mm} mol/L$ ) and the ones from the dissociation of $Ca_{3}(PO_{4})_{2(s)}$ ( $3S_2$ ). Therefore, $[Ca^{2+}] = 3S_2 + 3$ and $[PO_{4}^{3-}] = 2S_2$ . Again, plugging into the the $K_{sp}$ expression:

$(3S_2 + 3)^3 (2S_2)^2 = \beta$

Because calcium phosphate has a very low solubility, the amount of calcium ions provided is very small, in fact, much smaller than 1 mol/L. This allows us to make the following approximation:

$S_2 \ll 1 \hspace{2 mm} mol/L \Rightarrow 3S_2 + 3 \approx 3$

The first equation we found for $S_2$ would be pretty nasty to solve. Using the approximation, it's much easier to find $S_2$ as a function of $\beta$ :

$(3)^3 (2S_2)^2 = \beta$

$S_2 = 2^{-1} 3^{- \frac{3}{2}} \beta ^{\frac{1}{2}}$

$\frac{S_{1}}{S_{2}} = 2^{- \frac{2}{5} + 1} 3^{- \frac{3}{5} + \frac{3}{2}} \beta ^{\frac{1}{5} - \frac{1}{2}} = 2^{\frac{3}{5}} 3^{ \frac{9}{10}} \beta ^{- \frac{3}{10}}$

$a = \frac{3}{5}$ , $b = \frac{9}{10}$ and $c = \frac{3}{10}$

$(\frac{b+c}{a})^{4} = (\frac{\frac{9}{10} + \frac{3}{10}}{\frac{3}{5}})^{4} = 2^4 = \boxed{16}$

In order to have an idea of how drastically the presence of a soluble calcium salt reduces the solubility of calcium phosphate, let's plug the numerical value of $K_{sp,Ca_{3}(PO_{4})_{2(s)}} = \beta = 2.07 \times 10^{-33}$ at 25°C into the expression for $\frac{S_{1}}{S_{2}}$ :

$\frac{S_{1}}{S_{2}} = 2^{\frac{3}{5}} 3^{ \frac{9}{10}} (2.07 \times 10^{-33}) ^{- \frac{3}{10}} \approx 2.6 \times 10^{10}$

For solubility $\ce{Ca3(PO4)2}$ in water:

• $\ce{Ca3(PO4)2 <=> 3Ca^{2+} + 2PO4^{3-}}$
• the solubility product $\beta = [3\ce{Ca}^{2+}]^3 [2\ce{PO}_4^{3-}]^2 = 3^3S_1^3 \dot{} 2^2S_1^2 = 2^23^3S_1^5$
• $\Rightarrow S_1 = \left(\dfrac{\beta}{2^23^3}\right)^\frac{1}{5}$

Since the $\ce{Ca(NO3)2}$ ( $3$ ions) solution is isotonic with $\text{4.5 M}$ $\ce{NaCl}$ ( $2$ ions) solution, this means that its molarity $M_{\ce{Ca(NO3)2}} = \dfrac{2}{3} \times 4.5 = \text{3 M}$

For solubility $\ce{Ca3(PO4)2}$ in the $\ce{Ca(NO3)2}$ solution:

$\begin{aligned} \Rightarrow \beta & = \left( 3+[3\ce{Ca}^{2+}] \right)^3 [2\ce{PO}_4^{3-}]^2 \\ & = (\color{#3D99F6}{3+3S_2})^3 \dot{} 2^2 S_2^2 \quad \quad \small \color{#3D99F6}{\text{Since }3 \gg 3S_2} \\ & \color{#3D99F6}{\approx 3^3} \dot{} 2^2 S_2^2 = 2^23^3S_2^2 \\ \Rightarrow S_2 & = \left(\frac{\beta}{2^23^3}\right)^\frac{1}{2} \end{aligned}$

Therefore, $\dfrac{S_1}{S_2} = \left(\dfrac{\beta}{2^23^3}\right)^\frac{1}{5}\left(\dfrac{2^23^3}{\beta}\right)^\frac{1}{2} = 2^{\frac{6}{10}}3^{\frac{9}{10}}2^{-\frac{3}{10}}$

$\Rightarrow \left(\dfrac{b+c}{a}\right)^4 = \left(\dfrac{9+3}{6}\right)^4 = \boxed{16}$