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How do we differentiate between $Fe^{3+}$ and $Cr^{3+}$ in groupIII?

(a) By adding excess of $NH_4OH$
(b) By increasing $NH_4^+$ ion concentration
(c) By decreasing $OH^-$ ion concentration
(d) By increasing $NH_4^+$ and decreasing $OH^-$ ion concentration
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  • $OH^-$ ions precipitate both $Fe^{3+}$ and $Cr^{3+}$ as their hydroxides. However, $Cr(OH)_3$ is soluble in excess of $NH_4OH.$
Answer: By increasing $NH_4^+$ and decreasing $OH^-$ ion concentration
$Fe^{3+} + 3OH^- \longrightarrow Fe(OH)_3 \downarrow$
$Cr^{3+} + 3OH^- \longrightarrow Cr(OH)_3 \downarrow$
$Cr(OH)_3 (s) + 6NH_3 (aq) \longrightarrow [Cr(NH_3)_6](OH)_3 (aq)$ (complex compound hydroxides)
answered Jan 28, 2014 by mosymeow_1
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