Ionic Equilibrium Question 2
Question 2 - 2024 (01 Feb Shift 2)
Solubility of calcium phosphate (molecular mass, M) in water is $W _g$ per $100 mL$ at $25^{\circ} C$. Its solubility product at $25^{\circ} C$ will be approximately.
(1) $10^{7}\left(\frac{W}{M}\right)^{3}$
(2) $10^{7}\left(\frac{W}{M}\right)^{5}$
(3) $10^{3}\left(\frac{W}{M}\right)^{5}$
(4) $10^{5}\left(\frac{W}{M}\right)^{5}$
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Answer (2)
Solution
$S=\frac{W \times 10}{M}$
$S=\frac{W \times 1000}{M \times 100}=\frac{W \times 10}{M}$
$K _{sp}=(3 s)^{3}(2 s)^{2}$
$=108 s^{5}$
$=108 \times 10^{5} \times\left(\frac{W}{M}\right)^{5}$
$=1.08 \times 10^{7}\left(\frac{W}{M}\right)^{5}$
