Question

Solubility of calcium phosphate (molecular mass, M) in water is W g per $100$mL at ${\text{2}}{{\text{5}}^{\text{o}}}{\text{C}}$. Its solubility product at ${\text{2}}{{\text{5}}^{\text{o}}}{\text{C}}$ will be approximately:
A. ${\text{1}}{{\text{0}}^{\text{9}}}{\left( {\dfrac{{\text{W}}}{{\text{M}}}} \right)^{\text{5}}}$
B. ${\text{1}}{{\text{0}}^{\text{7}}}{\left( {\dfrac{{\text{W}}}{{\text{M}}}} \right)^{\text{5}}}$
C. ${\text{1}}{{\text{0}}^5}{\left( {\dfrac{{\text{W}}}{{\text{M}}}} \right)^{\text{5}}}$
D. ${\text{1}}{{\text{0}}^3}{\left( {\dfrac{{\text{W}}}{{\text{M}}}} \right)^{\text{5}}}$

Answer 1

${{\text{K}}_{{\text{sp}}}}$ represents the solubility product constant at equilibrium. When a solid substance dissolves into water it dissociates into ion. The product of solubilities of the ions with the number of each ion in power is known as the solubility product. First we will determine the molarity and then the equilibrium constant.

Complete solution:
The solubility product constant represents the product of concentrations of constituting ions of an ionic compound at equilibrium each raised number of ions as power.
An ionic compound dissociates into the water as follows:
${{\text{A}}_{\text{x}}}{{\text{B}}_{\text{y}}}\, \rightleftarrows \,\,{\text{x}}{{\text{A}}^{{\text{ + Y}}}}\,{\text{ + }}\,{\text{y}}{{\text{B}}^{ - x}}$
Where,
${\text{AB}}$ is an ionic compound.
The general representation for the solubility product of an ionic compound is as follows:
${{\text{K}}_{{\text{sp}}}}\,{\text{ = }}\,{\left( {{\text{x}}{{\text{A}}^{{\text{ + y}}}}\,} \right)^{\text{x}}}\, \times \,\,{\left( {{\text{y}}{{\text{A}}^{ - x}}\,} \right)^{\text{y}}}$
${{\text{K}}_{{\text{sp}}}}\,{\text{ = }}\,{\text{x}}{{\text{S}}^{\text{x}}}\,{ \times }\,{\text{y}}{{\text{S}}^{\text{y}}}$
Where,
${{\text{K}}_{{\text{sp}}}}$is the solubility product constant.
${\text{S}}$is the solubility of each ion.
$x$ and $y$ represents the number of ions.
The compound calcium phosphate dissociates into the water as follows:
${\text{C}}{{\text{a}}_3}{{\text{(P}}{{\text{O}}_4})_2} \rightleftarrows \,\,3\,{\text{C}}{{\text{a}}^{{\text{2 + }}}}{\text{ + }}\,\,2{\text{PO}}_4^{3 - }$
The solubility product for calcium phosphate is represented as follows:
${{\text{K}}_{{\text{sp}}}}\, = \,\,{\left[ {{\text{3}}\,{\text{C}}{{\text{a}}^{{\text{2 + }}}}} \right]^3}{\left[ {{\text{2PO}}_4^{3 - }} \right]^2}$
${{\text{K}}_{{\text{sp}}}}\, = \,\,{\left[ {{\text{3}}\,{\text{S}}} \right]^3}{\left[ {{\text{2S}}} \right]^2}$……$(1)$
Solubility tells the concentration of compounds in solution. Here, we take the solubility in units of molarity.
The formula of molarity is as follows:
${\text{molarity}}\,{\text{ = }}\,\dfrac{{{\text{mole}}\,{\text{of solute}}}}{{{\text{volume}}\,{\text{of}}\,{\text{solution}}}}$
The formula of mole is as follows:
${\text{mole}}\,{\text{ = }}\,\dfrac{{{\text{mass}}\,{\text{(W)}}}}{{{\text{molar}}\,{\text{mass (M)}}}}$
On substituting the mole in molarity formula we get,
${\text{molarity}}\,{\text{ = }}\,\dfrac{{{\text{W/M}}\,\,{\text{of solute}}}}{{{\text{volume}}\,{\text{of}}\,{\text{solution}}\,\,{\text{(L)}}}}$
Here, we take the volume in litre.
We will multiply the molarity formula with $1000$ to convert the volume in mL as follows:
${\text{molarity}}\,{\text{ = }}\,\dfrac{{{\text{W/M}}\,\,{\text{of solute}}}}{{{\text{volume}}\,{\text{of}}\,{\text{solution(mL)}}}} \times 1000$
The given volume of the solution is $100$mL . So,
${\text{molarity}}\,{\text{ = }}\,\dfrac{{{\text{W/M}}\,\,{\text{of solute}}}}{{{\text{100}}}} \times 1000$
${\text{molarity}}\,{\text{ = }}\,\dfrac{{{\text{W/M}}\,\,{\text{of solute}}}}{{\text{1}}} \times 10$
We can write the molarity for ${\text{C}}{{\text{a}}^{{\text{2 + }}}}$ ions as,
${\text{C}}{{\text{a}}^{2 + }}\,{\text{ = }}\,\dfrac{{{\text{W}} \times \,\,{\text{10}}}}{{{\text{M}}\,}}$……$(2)$
We can write the molarity for ${\text{PO}}_4^{3 - }$ ions as,
${\text{PO}}_4^{3 - }\,{\text{ = }}\,\dfrac{{{\text{W}} \times \,\,{\text{10}}}}{{{\text{M}}\,}}$……$(3)$
Now, we will substitute the molarities of ${\text{C}}{{\text{a}}^{{\text{2 + }}}}$ and ${\text{PO}}_4^{3 - }$ ions from equation $(2)$ and$(3)$ in equation $(1)$,
${{\text{K}}_{{\text{sp}}}}\, = \,\,{\left[ {{\text{3}}\, \times \dfrac{{{\text{W}} \times \,\,{\text{10}}}}{{{\text{M}}\,}}} \right]^3}{\left[ {{\text{2}} \times \dfrac{{{\text{W}} \times \,\,{\text{10}}}}{{{\text{M}}\,}}} \right]^2}\,$
${{\text{K}}_{{\text{sp}}}}\, = \,\,{\left[ {\dfrac{{3 \times \,\,{\text{10}}}}{{1\,}}} \right]^3}{\left[ {\dfrac{{2 \times \,\,{\text{10}}}}{{1\,}}} \right]^2}\,{\left( {\dfrac{{\text{W}}}{{{\text{M}}\,}}} \right)^5}$
${{\text{K}}_{{\text{sp}}}}\, = \,\,108{\text{00000}}\,\,{\left( {\dfrac{{\text{W}}}{{{\text{M}}\,}}} \right)^5}$
We can assume that $108{\text{00000}}\, \approx \,100{\text{00000}}$
${{\text{K}}_{{\text{sp}}}}\, = \,\,{10^7}\,\,{\left( {\dfrac{{\text{W}}}{{{\text{M}}\,}}} \right)^5}$
So, the solubility product of calcium phosphate at ${\text{2}}{{\text{5}}^{\text{o}}}{\text{C}}$ will be ${\text{1}}{{\text{0}}^{\text{7}}}{\left( {\dfrac{{\text{W}}}{{\text{M}}}} \right)^{\text{5}}}$ approximately.

Therefore, option (B), ${\text{1}}{{\text{0}}^{\text{7}}}{\left( {\dfrac{{\text{W}}}{{\text{M}}}} \right)^{\text{5}}}$ is correct.

Note: By the addition of atomic mass the molar mass is determined. Solubility tells the concentration of solid compounds dissolved in water. Concentration in the determination of solubility products is taken in terms of molarity. The formula of solubility product constant depends upon the number of ions produced by the ionic compounds in the water. Molar solubility represents the number of ions dissolved per liter of solution. Here, solubility represents the number of ions dissolved in a given amount of solvent.