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