Question
Question: The solubility of calcium phosphate in water is x mol h at 25 Degree C. Its solubility product is eq...
The solubility of calcium phosphate in water is x mol h at 25 Degree C. Its solubility product is equal to:
A. 108x2
B. 36x3
C. 36x5
D. 108x5
Solution
All phosphates are insoluble aside from those of sodium, potassium and ammonium. Some hydrogen phosphates, for example, Ca3(PO4)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 phosphateCa3(PO4)2 in water is:
Ca3(PO4)2= 3Ca(2+)+2PO4(3−)
and the equilibrium constant will be equal to:
Ksp = [Ca(2+)] ^3∗[PO4(3−)] ^2
Now, if you want to solve problems like how much calcium phosphate will dissolve in water you need to replace [Ca(2+)]and [PO4(3+)]with something more beneficial. For every mole ofCa3(PO4)2 when it dissolves, will yield 3 moles of (=3x) and 2 moles ofP{{O}_{4}}(3-)$$$$(=2x) so the Ksp expression becomes:
[Ca(2+)] ^3∗[PO4(3+)] ^2=>(3x) ^3∗(2x) ^2=108x ^5
Calcium phosphate is Ca3(PO4)2 .
Ca3(PO4)23Ca2++2PO43−
3x 2x
Ksp=[3x]3[2x]2=108x5
So, the correct option is D. 108x5
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 is20mg/L, and that of calcium fluoride is16mg/L.
Note: At 25∘C and pH7.00, Ksp for calcium phosphate is 2.07∗10−33 , indicating that the concentrations of Ca2+ and Po43− ions in solution that are in equilibrium with solid calcium phosphate are very low.