JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
The molar solubility(s) of zirconium phosphate with molecular formula \(\left(\mathrm{Zr}^{4+}\right)_3\left(\mathrm{PO}_4^{3-}\right)_4\) is given by relation :
- A \(\left(\frac{\mathrm{K}_{\mathrm{sp}}}{9612}\right)^{\frac{1}{3}}\)
- B \(\left(\frac{\mathrm{K}_{\mathrm{sp}}}{6912}\right)^{\frac{1}{7}}\)
- C \(\left(\frac{\mathrm{K}_{\mathrm{sp}}}{8435}\right)^{\frac{1}{7}}\)
- D \(\left(\frac{\mathrm{K}_{\mathrm{sp}}}{5348}\right)^{\frac{1}{6}}\)
Answer & Solution
Correct Answer
(B) \(\left(\frac{\mathrm{K}_{\mathrm{sp}}}{6912}\right)^{\frac{1}{7}}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{cc}\mathrm{Zr}_3\left(\mathrm{PO}_4\right)_4(\mathrm{~s}) \rightleftharpoons & 3 \mathrm{Zr}^{+4}(\mathrm{aq})+4 \mathrm{PO}_4^{-3}(\mathrm{aq}) \\ - & 3 \mathrm{~s} \quad 4 \mathrm{~s}\end{array}\)…
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