Question

When \(\mathrm{FeCl_3}\) solution is added to hot water, it forms a sol \(P\). However, when \(\mathrm{FeCl_3}\) solution is added to \(\mathrm{NaOH}\) solution it forms a sol \(Q\). What are \(P\) and \(Q\) respectively?

• A. \(\mathrm{Fe_2O_3\cdot xH_2O/Cl^-;\;Fe_2O_3\cdot xH_2O/OH^-}\)
• B. \(\mathrm{Fe_2O_3\cdot xH_2O/H^+;\;Fe_2O_3\cdot xH_2O/Na^+}\)
• C. \(\mathrm{Fe_2O_3\cdot xH_2O/Fe^{3+};\;Fe_2O_3\cdot xH_2O/OH^-}\)
• D. \(\mathrm{Fe_2O_3\cdot xH_2O/OH^-;\;Fe_2O_3\cdot xH_2O/Fe^{3+}}\) \bigskip

Hint:

The charge on a colloidal sol depends on the ion preferentially adsorbed on its surface.

• Excess metal ion \(\rightarrow\) Positive sol.
• Excess anion (e.g. \(\mathrm{OH^-}\)) \(\rightarrow\) Negative sol.

Answer 1

The Correct Option is C

Step 1: Formation of sol \(P\). On boiling \(\mathrm{FeCl_3}\) solution with water, \[ \mathrm{FeCl_3+3H_2O \rightarrow Fe(OH)_3+3HCl} \] The hydrated ferric oxide particles preferentially adsorb \(\mathrm{Fe^{3+}}\) ions and become positively charged. Hence, \[ P=\mathrm{Fe_2O_3\cdot xH_2O/Fe^{3+}} \]

Step 2: Formation of sol \(Q\). When \(\mathrm{FeCl_3}\) is added to excess NaOH, \[ \mathrm{FeCl_3+3NaOH \rightarrow Fe(OH)_3+3NaCl} \] The precipitated ferric hydroxide adsorbs excess \(\mathrm{OH^-}\) ions and forms a negatively charged sol. Hence, \[ Q=\mathrm{Fe_2O_3\cdot xH_2O/OH^-} \]

Step 3: Final conclusion. Therefore, \[ \boxed{P=\mathrm{Fe_2O_3\cdot xH_2O/Fe^{3+}},\; Q=\mathrm{Fe_2O_3\cdot xH_2O/OH^-}} \] Hence, the correct option is \(\boxed{(C)}\).