Which of the following will be the most effective in the coagulation of `Fe(OH)_(3)` sol ?

Options

A. `Mg_(3) (PO_4)_2`
B. `BaCl_2`
C. `NaCl`
D. KCN

Answer

A

Explanation

According to Hardy-Schulze rule, coagulation power of ions is directly proportional to charge onion.
`:. Fe(OH)_3` is positively charged colloid
`:.` It will be coagulated by anion.
(a) `Mg_(3)(PO_4)hArr3Mg^(2+)++2PO_4^(3-)`
(b) `BaCl_(2) hArr Ba^(2+) +2Cl^(-)`
(c) `NaCl hArr Na^(+) +Cl^(-)`
(d) `KCN hArr K^(+) CN^(-)`
`:. PO_(4)^(3-) ` has highest charge among the given anions.
`:. Mg(PO_4)_(2)` is the most effective in coagulation of `Fe(OH)_3` sol.