The solubility ' $\mathrm{S}$ ' of $\mathrm{Zr}_3\left(\mathrm{PO}_4\right)_4$ in terms of its solubility…
- $\left(\frac{\mathrm{K}_{\mathrm{sp}}}{144}\right)^{\frac{1}{4}}$
- $\left(\frac{\mathrm{K}_{\mathrm{sp}}}{6912}\right)^{\frac{1}{5}}$
- $\left(\frac{\mathrm{K}_{\mathrm{sp}}}{1728}\right)^{\frac{1}{6}}$
- $\left(\frac{\mathrm{K}_{\mathrm{sp}}}{6912}\right)^{\frac{1}{7}}$
Solution
Asked in: AP EAMCET 2017 (25 Apr Shift 2)