ExamBro
ExamBro
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 :

  1. A \(\left(\frac{\mathrm{K}_{\mathrm{sp}}}{9612}\right)^{\frac{1}{3}}\)
  2. B \(\left(\frac{\mathrm{K}_{\mathrm{sp}}}{6912}\right)^{\frac{1}{7}}\)
  3. C \(\left(\frac{\mathrm{K}_{\mathrm{sp}}}{8435}\right)^{\frac{1}{7}}\)
  4. D \(\left(\frac{\mathrm{K}_{\mathrm{sp}}}{5348}\right)^{\frac{1}{6}}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\left(\frac{\mathrm{K}_{\mathrm{sp}}}{6912}\right)^{\frac{1}{7}}\)

Step-by-step Solution

Detailed explanation

\(\begin{array}{cc}\mathrm{Zr}_3\left(\mathrm{PO}_4\right)_4(\mathrm{~s}) \rightleftharpoons & 3 \mathrm{Zr}^{+4}(\mathrm{aq})+4 \mathrm{PO}_4^{-3}(\mathrm{aq}) \\ - & 3 \mathrm{~s} \quad 4 \mathrm{~s}\end{array}\)…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app