TS EAMCET · Chemistry · Ionic Equilibrium
What is the solubility product \(\left(K_{\mathrm{sp}}\right)\) of calcium phosphate in pure water? \([S=\) molar solubility]
- A \(108 \mathrm{~S}^5\)
- B \(72 \mathrm{~S}^3\)
- C \(6 s^5\)
- D \(121 \mathrm{~S}^2\)
Answer & Solution
Correct Answer
(A) \(108 \mathrm{~S}^5\)
Step-by-step Solution
Detailed explanation
\(\because\) Solubility product \(\left(K_{\text {sp }}\right)=x^x \cdot y^y \cdot s^{x+y}\) where, \(x\) and \(y\) are number of \(\mathrm{Ca}^{2+}\) and \(\mathrm{PO}_4^{3-}\) ions in \(1 \mathrm{~L}\) of pure water respectively. i.e.…
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