Which of the following will be the most effective in the...

According to Hardy-Schulze rule, coagulation power of ions is directly proportional to charge on ion. $\because \mathrm{Fe}(\mathrm{OH})_{3}$ is positively charged colloid. $\therefore$ It will be coagulated by anion. (a) $\mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2} \rightleftharpoons 3 \mathrm{Mg}^{2+}+2 \mathrm{PO}_{4}^{3-}$ (b) $\mathrm{BaCl}_{2} \rightleftharpoons \mathrm{Ba}^{2+}+2 \mathrm{Cl}^{-}$ (c) $\mathrm{NaCl} \rightleftharpoons \mathrm{Na}^{+}+\mathrm{Cl}^{-}$ (d) $\mathrm{KCN} \rightleftharpoons \mathrm{K}^{+}+\mathrm{CN}^{-}$ Because $\mathrm{PO}_{4}^{3-}$ has highest charge among the given anions, therefore, $\mathrm{Mg}_{3}$\left(\mathrm{PO}_{4}\right)_{2}$is the most effective in the coagulation of$\mathrm{Fe}(\mathrm{OH})_{3} solution.