Question

Identify the correct pair of ions which are most effective towards the coagulation of sols \(Fe_2O_3\cdot xH_2O\) and CdS respectively.

• A. \(PO_4^{3-},\,Al^{3+}\)
• B. \(Al^{3+},\,PO_4^{3-}\)
• C. \([Fe(CN)_6]^{4-},\,Al^{3+}\)
• D. \(Al^{3+},\,[Fe(CN)_6]^{4-}\)

Hint:

Hardy-Schulze Rule: Greater the charge on the oppositely charged ion, greater is its coagulating power. \[ Al^{3+}>Ba^{2+}>Na^+ \] \[ [Fe(CN)_6]^{4-}>PO_4^{3-}>SO_4^{2-} \]

Answer 1

The Correct Option is C

Step 1: Recall the Hardy-Schulze rule.
To coagulate (clump) a charged sol, we add an ion of the opposite charge. The higher the charge on that opposite ion, the stronger its coagulating power.
Step 2: Find the charge of the ferric oxide sol.
The hydrated ferric oxide sol $Fe_2O_3\cdot xH_2O$ is positively charged. So it needs negative ions (anions) to coagulate it.
Step 3: Pick the best anion.
Among the anions, $[Fe(CN)_6]^{4-}$ carries a $4-$ charge, which is higher than $PO_4^{3-}$ at $3-$. So $[Fe(CN)_6]^{4-}$ is the most effective for the ferric oxide sol.
Step 4: Find the charge of the CdS sol.
The CdS sol is negatively charged. So it needs positive ions (cations) to coagulate it.
Step 5: Pick the best cation.
Among the cations, $Al^{3+}$ carries the highest positive charge $3+$. So $Al^{3+}$ is the most effective for the CdS sol.
Step 6: State the pair.
So the best ions are $[Fe(CN)_6]^{4-}$ for the ferric oxide sol and $Al^{3+}$ for the CdS sol. \[ \boxed{[Fe(CN)_6]^{4-},\ Al^{3+}} \]