Question

$FeC{l}_3$ is added separately to excess of boiling water and excess of $NaOH$ solution; the solution obtained contain the dispersed particle to:

• A. $Fe(OH{)}_3/H^{+};Fe(OH{)}_3/O{H}^{-}$
• B. $Fe(OH{)}_3/F{e}^{3+};Fe(OH{)}_3/O{H}^{-}$
• C. $FeC{l}_3/F{e}^{3+};FeC{l}_3/O{H}^{-}$
• D. $Fe(OH{)}_3/F{e}^{3+};Fe(OH{)}_3/N{a}^{+}$

Answer 1

The correct option is B $Fe(OH{)}_3/F{e}^{3+};Fe(OH{)}_3/O{H}^{-}$

$\Rightarrow$ Addition of $FeC{l}_3$ to hot water produce a positive charged solution due to absorption of $F{e}^{3+}$ .

$\Rightarrow$ A negative charged solution is when $FeC{l}_3$ is added to $NaOH$ solution.

$FeC{l}_3+2NaOH⟶Fe(OH{)}_3+3NaCl$

The absorption of $O{H}^{-}$ ions forms an electric double layer.

Option B $Fe(OH{)}_3/F{e}^{3+},Fe(OH{)}_3/O{H}^{-}$ is the answer.