Question

What is the solubility product $\left(K_{ sp }\right)$ of calcium phosphate in pure water? $[S=$ molar solubility $]$

• A. $108\, S ^{5}$
• B. $72 \,S^{3}$
• C. $6 \,S^{5}$
• D. $121\, S ^{2}$

Answer 1

Answer: (A)

$ \because$ Solubility product $\left(K_{ sp }\right)=x^{x} \cdot y^{y} \cdot S^{x+y}$

where, $x$ and $y$ are number of $Ca ^{2+}$ and $PO _{4}^{3-}$ ions in $1\, L$ of pure water respectively.

i.e. In $Ca _{2}\left( PO _{4}\right)_{3}$

Here, $x=2$ and $y=3$

Therefore, $K_{ sp }=2^{2} \cdot 3^{3} \,S ^{2+3}$

$ K_{ sp }=108\, S^{5}$