The molar solubility(s) of zirconium phosphate with molecular formula (Zr4+)3 \(\left(\mathrm{PO}_{4}^{3-}\right)_{4}\) is given by relation :
1
\(\left(\frac{\mathrm{K}_{\text {sp }}}{6912}\right)^{\frac{1}{7}}\)
2
\(\left(\frac{\mathrm{K}_{\mathrm{sp}}}{5348}\right)^{\frac{1}{6}}\)
3
\(\left(\frac{\mathrm{K}_{\mathrm{sp}}}{8435}\right)^{\frac{1}{7}}\)
4
\(\left(\frac{\mathrm{K}_{\text {sp }}}{9612}\right)^{\frac{1}{3}}\)