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The solubility of \(\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2\) in water is y moles / litre. Its solubility product is
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Verified Answer
The correct answer is:
\(108 \mathrm{y}^5\)
\(\begin{aligned}
& \text {Hints: } \mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2(\mathrm{~s}) \rightleftharpoons 3 \underset{3 \mathrm{~s}}{\mathrm{Ca}^{2+}}(\mathrm{aq})+\underset{2 \mathrm{~s}}{2 \mathrm{PO}_4}{ }^{3-}(\mathrm{aq}) \\
& \mathrm{~K}_{\mathrm{Sp}}=\left[\mathrm{Ca}^{2+}\right]^3 \cdot\left[\mathrm{PO}_4^{3-}\right]^2 \\
& =(3 \mathrm{~s})^3 \cdot(2 \mathrm{~s})^2 \\
& =27 \mathrm{~s}^3 \times 4 \mathrm{~s}^2 \\
& =108 \mathrm{~s}^5 \\
&
\end{aligned}\)
& \text {Hints: } \mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2(\mathrm{~s}) \rightleftharpoons 3 \underset{3 \mathrm{~s}}{\mathrm{Ca}^{2+}}(\mathrm{aq})+\underset{2 \mathrm{~s}}{2 \mathrm{PO}_4}{ }^{3-}(\mathrm{aq}) \\
& \mathrm{~K}_{\mathrm{Sp}}=\left[\mathrm{Ca}^{2+}\right]^3 \cdot\left[\mathrm{PO}_4^{3-}\right]^2 \\
& =(3 \mathrm{~s})^3 \cdot(2 \mathrm{~s})^2 \\
& =27 \mathrm{~s}^3 \times 4 \mathrm{~s}^2 \\
& =108 \mathrm{~s}^5 \\
&
\end{aligned}\)
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