The solubility ' $\mathrm{S}$ ' of $\mathrm{Zr}_3\left(\mathrm{PO}_4\right)_4$ in terms of its solubility product, $\mathrm{K}_{\mathrm{sp}}$ is

• A

  $\left(\frac{\mathrm{K}_{\mathrm{sp}}}{144}\right)^{\frac{1}{4}}$
• B

  $\left(\frac{\mathrm{K}_{\mathrm{sp}}}{6912}\right)^{\frac{1}{5}}$
• C

  $\left(\frac{\mathrm{K}_{\mathrm{sp}}}{1728}\right)^{\frac{1}{6}}$
• Correct answer
  D

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

Hence, the correct answer is option: 'D'.