Question

The solubility of calcium phosphate in water is x mol h at 25 Degree C. Its solubility product is equal to:
A. \[108{{x}^{2}}\]
B. \[36{{x}^{3}}\]
C. \[36{{x}^{5}}\]
D. \[108{{x}^{5}}\]

Answer 1

Hint: All phosphates are insoluble aside from those of sodium, potassium and ammonium. Some hydrogen phosphates, for example, \[C{{a}_{3}}{{(P{{O}_{4}})}_{2}}\], are solvent. I) All sulfides are insoluble aside from those of ammonium, sodium, calcium, potassium, magnesium, barium and strontium. These are somewhat hydrolyzed in water .

Step by step answer: The solubility product, by definition, is the equilibrium constant for dissolution of a solid ionic compound to yield ions in solution (usually water). It is abbreviated Ksp.
The equation for dissolution of calcium phosphate\[C{{a}_{3}}{{(P{{O}_{4}})}_{2}}\] in water is:
\[C{{a}_{3}}{{(P{{O}_{4}})}_{2}}\]= \[3Ca(2+)+2P{{O}_{4}}(3-)\]
and the equilibrium constant will be equal to:
Ksp = \[\left[ Ca(2+) \right]\hat{\ }3*\left[ P{{O}_{4}}(3-) \right]\hat{\ }2\]
Now, if you want to solve problems like how much calcium phosphate will dissolve in water you need to replace \[\left[ Ca{{(}^{2+}}) \right]\]and \[\left[ P{{O}_{4}}{{(}^{3+}}) \right]\]with something more beneficial. For every mole of\[C{{a}_{3}}{{(P{{O}_{4}})}_{2}}\] when it dissolves, will yield 3 moles of (=3x) and 2 moles of\[P{{O}_{4}}(3-)\]\[(=2x)\] so the Ksp expression becomes:
\[\left[ Ca{{(}^{2+}}) \right]\hat{\ }3*\left[ P{{O}_{4}}{{(}^{3+}}) \right]\hat{\ }2=>(3x)\hat{\ }3*(2x)\hat{\ }2=108x\hat{\ }5\]
Calcium phosphate is \[C{{a}_{3}}{{(P{{O}_{4}})}_{2}}\] ​.
 \[C{{a}_{3}}{{(P{{O}_{4}})}_{2}}3C{{a}^{2+}}+2P{{O}_{4}}^{3-}\]
                                           \[3x\] \[2x\]
\[{{K}_{sp}}={{\left[ 3x \right]}^{3}}{{\left[ 2x \right]}^{2}}=108{{x}^{5}}\]

So, the correct option is D. \[108{{x}^{5}}\]

Additional Information: Elementary calcium reacts with water. Calcium compounds are more or less water soluble. Calcium carbonate has a solubility of, \[14mg/L\] which is multiplied by a factor five in presence of carbon dioxide. Calcium phosphate solubility is\[20mg/L\], and that of calcium fluoride is\[16mg/L\].

Note: At \[{{25}^{\circ }}C\] and \[pH7.00\], Ksp for calcium phosphate is \[2.07*{{10}^{-33}}\] , indicating that the concentrations of \[C{{a}^{2+}}\] and \[P{{o}_{4}}^{3-}\] ions in solution that are in equilibrium with solid calcium phosphate are very low.