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ICSE Class 10 Chemistry Question 6 of 10

Analytical Chemistry — Question 3

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Question

Question 3

What do you observe when caustic soda solution is added to the following solution, first a little and then in excess :

(a) FeCl3

(b) ZnSO4

(c) Pb(NO3)2

(d) CuSO4

Write balanced equations for these reactions.

Answer

(a) When caustic soda solution is added to FeCl3 dropwise, a reddish brown ppt is obtained, which is insoluble in excess of NaOH:

FeCl3yellow+3NaOHcolourless⟶Fe(OH)3↓reddish brown ppt+3NaClcolourless\underset{\text{yellow}}{{\text{FeCl}_3}} + \underset{\text{colourless}}{3\text{NaOH}} \longrightarrow \underset{\text{reddish brown ppt}}{\text{Fe(OH)}_3↓} + \underset{\text{colourless}}{3\text{NaCl}}

(b) When caustic soda solution is added to Zinc sulphate dropwise, a white gelatinous ppt is obtained, which dissolves in excess of NaOH:

ZnSO4colourless+2NaOHcolourless⟶Zn(OH)2↓white gelatinous ppt+Na2SO4 colourless\underset{\text{colourless}}{{\text{ZnSO}_4}} + \underset{\text{colourless}}{2\text{NaOH}} \longrightarrow \underset{\text{white gelatinous ppt}}{\text{Zn(OH)}_2↓} + \underset{\text{ colourless}}{\text{Na}_2\text{SO}_4}

Zn(OH)2+2NaOH excess⟶Na2ZnO2↓colourless+2H2O{\text{Zn(OH)}_2} + \underset{\text{ excess}}{2\text{NaOH}} \longrightarrow \underset{\text{colourless}}{\text{Na}_2\text{ZnO}_2↓} + 2\text{H}_2\text{O}

(c) When caustic soda solution is added to Pb(NO3)2 dropwise, a chalky white ppt is obtained, which dissolves in excess of NaOH:

Pb(NO3)2colourless+2NaOHcolourless⟶Pb(OH)2↓white ppt+2NaNO3 colourless\underset{\text{colourless}}{{\text{Pb(NO}_3)_2}} + \underset{\text{colourless}}{2\text{NaOH}} \longrightarrow \underset{\text{white ppt}}{\text{Pb(OH)}_2↓} + \underset{\text{ colourless}}{2\text{NaNO}_3}

Pb(OH)2+2NaOHexcess⟶Na2PbO2↓sodium plumbite - colourless+2H2O{\text{Pb(OH)}_2} + \underset{\text{excess}}{2\text{NaOH}} \longrightarrow \underset{\text{sodium plumbite - colourless}}{\text{Na}_2\text{PbO}_2↓} + 2\text{H}_2\text{O}

(d) When caustic soda solution is added to CuSO4 dropwise, a pale blue ppt is obtained, which is insoluble in excess of NaOH:

CuSO4blue+2NaOHcolourless⟶Cu(OH)2↓pale blue ppt+Na2SO4colourless\underset{\text{blue}}{{\text{CuSO}_4}} + \underset{\text{colourless}}{2\text{NaOH}} \longrightarrow \underset{\text{pale blue ppt}}{\text{Cu(OH)}_2↓} + \underset{\text{colourless}}{\text{Na}_2\text{SO}_4}