Question

In addition to aqueous \[NaOH\] to a salt solution. a white gelatinous precipitate is formed which dissolves in excess alkali. The salt solution contains
(A) Chromium ions
(B) Aluminium ions
(C) Barium ions
(D) Iron ions

Answer 1

Hint: Addition of sodium hydroxide to the solution leads to formation of hydroxides of respective ions. Alkalis are those bases which are soluble in water or dissolve in water. For example, \[NaOH\].

Complete Step by Step Solution:
Option A: chromium on addition of aqueous \[NaOH\] gives a green-coloured precipitate which is amphoteric in nature.
$CrC{l_3} + 3NaOH \to Cr{(OH)_3} + 3NaCl$
Chromium chloride is a green solution which reacts with colourless sodium hydroxide to form green colored precipitate chromium hydroxide and colourless sodium chloride.
$Cr{(OH)_3} + 3NaOH \to N{a_3}[Cr{(OH)_6}]$
in excess aqueous\[NaOH\], Chromium hydroxide dissolves and gives a green colour solution by forming sodium hexahydroxychromium (III) complex.

Option B: aluminium ions react with sodium hydroxide to form aluminium hydroxide which settles down as white precipitate. Aluminium hydroxide when treated with excess \[NaOH\] gives \[NaAl{O_2}\]as the following reaction.
\[A{l^{3 + }} + 3O{H^ - } \to Al{(OH)_{3}}\]
\[Al{(OH)_3} + NaOH \to NaAl{O_2} + 2{H_2}O\]
This white precipitate is also known as a white gelatinous precipitate.

Option C: any salt of barium say barium chloride reacts with sodium hydroxide to form barium hydroxide. Barium hydroxide is usually found forming stable complex with water and has the formula \[{\mathbf{Ba}}{\left( {{\mathbf{OH}}} \right)_{\mathbf{2}}}\cdot{\mathbf{8}}{{\mathbf{H}}_{\mathbf{2}}}{\mathbf{O}}\]
$BaC{l_2} + NaOH \to Ba{(OH)_2} + NaCl$

Option D: iron ions also react with sodium hydroxide to form iron hydroxide. The oxidation state of iron in iron compounds decides whether iron Di hydroxide is formed or iron tri hydroxide.
But the oxidation state of iron is retained in the hydroxide and no redox reaction takes place.
$ F{e^{ + 2}} + 2O{H^ - } \to Fe{(OH)_2} \\$
$ F{e^{ + 3}} + 3O{H^ - } \to Fe{(OH)_3} \\ $

Thus, the correct answer is B.

Note: A white gelatinous precipitate is obtained in case of zinc also when zinc reacts with \[NaOH\]. Gelatinous means gel-like substance having jelly-like texture which is poorly soluble in water and floats due to low density. It also appears like a blue lake when it absorbs blue litmus hence the test is known as lake test.