Question

Solubility of calcium phosphate (molecular mass, M) in water is W g per 100 mL at 25°C. Its solubility product at 25°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

The correct option is B

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

$S=\frac{\frac{W}{M}}{100}\times 1000=10\frac{W}{M}mol/litre$.
$C{a}_3(P{O}_4{)}_2\rightleftharpoons 3C{a}^{2+}+2(P{O}_4{)}^{3-}$
$3S+2S$
$K_{sp}=\left(2S{)}^3\right(2s{)}^2$
$=108s^5$
Thus, $=108{\left(\frac{10W}{M}\right)}^5$
$={10}^7{\left(\frac{W}{M}\right)}^5$ (approximately)