Write balanced equation:

(a) Reaction of sodium hydroxide solution with Iron (III) chloride solution

(b) Copper sulphate solution with ammonium hydroxide solution

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

$\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 ammonia solution is added dropwise to cupper sulphate, a pale blue ppt of copper hydroxide is obtained.

$\underset{\text{Blue}}{{\text{CuSO}$4}} + 2\text{NH}4\text{OH} \longrightarrow \underset{\text{Pale Blue ppt}}{\text{Cu(OH)}2} ↓ + \underset{\text{Colourless in solution}}{(\text{NH}4)2\text{SO}4}

On adding excess of ammonia solution, the ppt dissolves and a deep blue solution is obtained.

$\text{Cu(OH)}$2 + (\text{NH}4)2\text{SO}4 + 2\text{NH}4\text{OH} \longrightarrow \underset{\text{Tetraammine copper (II) sulphate}}{[(\text{Cu(NH}3)4]\text{SO}4} + 4\text{H}_2\text{O}