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

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Solution

Q.
Solubility of calcium phosphate (molecular mass, M) in water is Wg per 100 mL at 25 °C. Its solubility product at 25 °C will be approximately. [2024]

1 $10^{7} {(\frac{W}{M})}^{3}$

2 $10^{3} {(\frac{W}{M})}^{5}$

3 $10^{7} {(\frac{W}{M})}^{5}$

4 $10^{5} {(\frac{W}{M})}^{5}$

Ans.
(3)

Mass of calcium phosphate dissolved in 100 mL water = W g

Mass of calcium phosphate dissolved in 1L (1000mL) water $(W_{{Ca}_{3} {({PO}_{4})}_{2}}) = 10 w g$

Molar mass of calcium phosphate $(M_{{Ca}_{3} {({PO}_{4})}_{2}}) = M g m o l^{- 1}$

Molar solubility (s) is the moles of calcium phosphate dissolved in 1 L water = $\frac{W_{{Ca}_{3} {({PO}_{4})}_{2}}}{M_{{Ca}_{3} {({PO}_{4})}_{2}}} = \frac{10 w}{M}$

   ${Ca}_{3} {({PO}_{4})}_{2} ⇌ \underset{3 s}{3 {Ca}^{2 +}} + \underset{2 s}{{2 PO}_{4}^{3 -}}$

  Solubility product,

   $K_{s p} = {[{Ca}^{2 +}]}^{3} {[{PO}_{4}^{3 -}]}^{2} = {[3 s]}^{3} {[2 s]}^{2}$ $= 108 s^{5} = 108 \times {(\frac{10 w}{M})}^{5} ≅ 10^{7} \times {(\frac{w}{M})}^{5}$

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