Question

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

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

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 $C{a}^{2+}$ and $P{O}_4^{3-}$ ions in $1L$ of pure water respectively.

i.e. In $C{a}_2{\left(P{O}_4\right)}_3$

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

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

$K_{sp}=108S^5$