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

Concept: According to the Hardy-Schulze rule, the coagulating power of an ion depends upon the charge carried by the ion opposite in charge to the colloidal particles. Higher the charge, greater the coagulating power.

Step 1: Determine charge on \(Fe_2O_3\cdot xH_2O\) sol. Hydrated ferric oxide sol is: \[ \boxed{\text{Positively charged}} \] Therefore, coagulation requires anions. Among the given anions: \[ PO_4^{3-} \] and \[ [Fe(CN)_6]^{4-} \] the higher charged ion is \[ [Fe(CN)_6]^{4-} \] Hence it is most effective.

Step 2: Determine charge on CdS sol. CdS sol is: \[ \boxed{\text{Negatively charged}} \] Therefore coagulation requires cations. Among given cations: \[ Al^{3+} \] has the highest positive charge. Thus it possesses maximum coagulating power.

Step 3: Apply Hardy-Schulze rule. For \(Fe_2O_3\cdot xH_2O\): \[ [Fe(CN)_6]^{4-} \] For CdS: \[ Al^{3+} \]

Step 4: Final conclusion. \[ \boxed{ [Fe(CN)_6]^{4-},\; Al^{3+} } \] Hence, \[ \boxed{\text{Option (C)}} \]