First slide

Alkali metals

Question

Select the species in which the precipitate formed on adding NaOH further dissolves in excess of NaOH.

Moderate

A

${ZnCl}_{2}$
B

${AlCl}_{3}$
C

${FeCl}_{3}$
D

${CrCl}_{3}$

Solution

$(a) {ZnCl}_{2} + 2 NaOH \to {⟶ Zn ({OH}_{2}) ↓}_{white ppt} + 2 NaCl$
$Zn (OH)_{2} + 2 NaOH ⟶ {Na}_{2} {ZnO}_{2}_{(soluble sodium zincate)} + 2 H_{2} O$
$(b) Al {Cl}_{3} + 3 NaOH \to {⟶ Zn ({OH}_{2}) ↓}_{white ppt} + 3 NaCl$
$Al (OH)_{3} + NaOH ⟶ {NaAIO}_{2}_{(sodium metaaluminate)} + 2 H_{2} O$
Note : ${NaAIO}_{2}$ is also written as $Na [Al (OH)_{4}]$
$(c) {FeCl}_{3} + 3 NaOH ⟶ \underset{Brown ppt}{Fe (OH)_{3} ↓} + 3 NaCl$
$Fe (OH)_{3}$ is insoluble in excess of $NaOH .$
$(d) {CrCl}_{3} + 3 NaOH ⟶ \underset{Green ppt}{Cr (OH)_{3} ↓} + 3 NaCl$
$Cr (OH)_{3}$ is insoluble in excess $NaOH .$
Note : $Cr (OH)_{3}$ can be made soluble in $NaOH$ if $H_{2} O_{2}$
$Cr (OH)_{3} + 2 NaOH + H_{2} O_{2} ⟶ \underset{\begin{matrix} Soluble \\ (yellow) \end{matrix}}{{Na}_{2} {CrO}_{4}} + H_{2} O$
This is redox reaction.

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App