Question

Solubility of calcium phosphate (molecular mass, $M$) in water is $W$g per $100mL$ at ${25}^{o}C$. Its solubility product at ${25}^{o}C$ will be approximately:

• A. ${10}^9{\left(\frac{W}{M}\right)}^5$
• B. ${10}^7{\left(\frac{W}{M}\right)}^5$
• C. ${10}^5{\left(\frac{W}{M}\right)}^5$
• D. ${10}^3{\left(\frac{W}{M}\right)}^5$

Answer 1

Answer: (B)

${10}^7{\left(\frac{W}{M}\right)}^5$

$C{a}_3(P{O}_4{)}_2\rightleftharpoons 3C{a}^{2+}+2P{O}_4^{3-}$

$3S$ $2S$

$K_{sp}=\left[C{a}^{2+}{]}^3\right[P{O}_4^{3-}{]}^2=\left(3S{)}^3\right(2S{)}^2=108S^5$

$S=\frac{w\times 1000}{M\times 100}=\frac{w\times 10}{M}$

$K_{sp}=108(\frac{w\times 10}{M}{)}^5={10}^7(\frac{w}{M}{)}^5$