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: P: FeCl₃ + boiling water → Fe(OH)₃ sol adsorbs Fe³⁺ → positive. Q: FeCl₃ + excess NaOH → Fe(OH)₃ adsorbs OH⁻ → negative.

Step 2: Write the final answer. \(\boxed{P=\mathrm{Fe_2O_3\cdot xH_2O/Fe^{3+}},\; Q=\mathrm{Fe_2O_3\cdot xH_2O/OH^-}}\)