The molar solubility(s) of zirconium phosphate with molecular formula \( (\mathrm{Zr}^{4+})_{3}(\mathrm{PO}_{4}^{3-})_{4} \) is given by relation:
\(\displaystyle \left( \frac{K_{sp}}{8435} \right)^{1/7} \)
\(\displaystyle \left( \frac{K_{sp}}{6912} \right)^{1/7} \)
\(\displaystyle \left( \frac{K_{sp}}{5348} \right)^{1/6} \)
\(\displaystyle \left( \frac{K_{sp}}{9612} \right)^{1/3} \)