Problem 1: Arrange $K_{4}$[Fe(CN){6}], $MgSO_{4}$, KCl, and $AlCl_{3}$ in order of coagulating power for Fe(OH){3} sol.

Step 1: Identify the charge on Fe(OH){3} sol. Fe(OH){3} sol is positively charged ($Fe^{3+}$ ions adsorbed from excess $FeCl_{3}$).

Step 2: Coagulating ions must be NEGATIVELY charged (anions).

• $K_{4}$[Fe(CN){6}] → provides [Fe(CN){6}]^{4-} (charge = 4−)
• $MgSO_{4}$ → provides $SO_{4}^{2-}$ (charge = 2−) and $Mg^{2+}$ (charge = 2+); coagulating anion = $SO_{4}^{2-}$
• KCl → provides $Cl^{-}$ (charge = 1−)
• $AlCl_{3}$ → provides $Cl^{-}$ (charge = 1−); same as KCl for coagulation of positive sol

Step 3: Rank anions by valency (charge magnitude): [Fe(CN)_{6}]^{4-} (4) > $SO_{4}^{2-}$ (2) > $Cl^{-}$ (1) = $Cl^{-}$ (1)

Step 4: Order of coagulating power: $K_{4}$[Fe(CN)_{6}] > $MgSO_{4}$ > KCl = $AlCl_{3}$ (since both provide $Cl^{-}$ for positive sol)

Problem 2: Which of $BaCl_{2}$, KCl, or $MgCl_{2}$ is most effective for coagulating $As_{2}S_{3}$ sol?

Step 1: $As_{2}S_{3}$ sol is NEGATIVELY charged ($S^{2-}$ ions adsorbed).

Step 2: Coagulating ions must be CATIONS.

• $BaCl_{2}$ → $Ba^{2+}$ (charge = 2+)
• KCl → $K^{+}$ (charge = 1+)
• $MgCl_{2}$ → $Mg^{2+}$ (charge = 2+)

Step 3: Rank: $Ba^{2+}$ = $Mg^{2+}$ (both 2+) > $K^{+}$ (1+). $BaCl_{2}$ = $MgCl_{2}$ > KCl. Most effective: $BaCl_{2}$ or $MgCl_{2}$ (equal, both divalent cations).