Question

State the effect of adding a small amount of a] sodium hydroxide b] ammonium hydroxide followed by an excess in each case to samples of each of the salt solutions: Zinc Nitrate solution and Calcium Nitrate solution ?

Answer 1

Hint: Take sodium hydroxide and ammonium hydroxide cases separately and find out the reaction and products formed according to the reactivity series and all the possible types of reaction, Also consider the usage of excess material in the reaction.

Complete step by step answer:
Sodium hydroxide (NaOH)

Zinc Nitrate solution:- When sodium hydroxide reacts with zinc nitrate , then double displacement reaction takes place and sodium nitrate along with zinc hydroxide ( a white precipitate) is formed. The reaction is as follows
$2NaOH + Zn{(N{O_3})_2} \to 2NaN{O_3} + Zn{(OH)_2}( \downarrow )$
When excess of NaOH is used the white ppt. becomes soluble .
$Zn{(OH)_2} + 2NaOH \to 2N{a^ + } + {[Zn{(OH)_2}]^{2 - }}$

Calcium Nitrate solution:- When sodium hydroxide reacts with calcium nitrate , then a double displacement reaction takes place , sodium nitrate and calcium hydroxide is formed , out of which calcium hydroxide precipitates down as white precipitate. The reaction is $2NaOH + Ca{(N{O_3})_2} \to Ca{(OH)_2}( \downarrow ) + 2NaN{O_3}$.
When NaOH is used in excess , white precipitation becomes insoluble.
Ammonium hydroxide ($N{H_4}OH$)

Zinc Nitrate solution:- When Ammonium hydroxide reacts with zinc nitrate , then double displacement reaction takes place and Ammonium nitrate along with zinc hydroxide ( a white precipitate) is formed.

The reaction is as follows
$2N{H_4}OH + Zn{(N{O_3})_2} \to 2N{H_4}N{O_3} + Zn{(OH)_2}( \downarrow )$
When excess of $N{H_4}OH$ is used, the white ppt. becomes soluble .
$Zn{(OH)_2} + 2N{H_4}OH \to 2N{H_4}^ + + {[Zn{(OH)_2}]^{2 - }}$

Calcium Nitrate solution:- When Ammonium hydroxide reacts with calcium nitrate , then no visible reaction takes place , as Ammonium hydroxide is a weak base which means on ionisation it will form low concentration of [OH-] ions and because the concentration of OH- ions are low it cannot react with any calcium salts to precipitate the hydroxide of calcium.$N{H_4}OH + Ca{(N{O_3})_2} \to {\text{No Reaction}}$.
Even when $N{H_4}OH$ is used in excess , no changes are observed.

Note: Students must be thorough with all the types of reaction , for e.g. Ionic reaction , Displacement reaction , Double Displacement reaction, etc. One must also be well versed with the reactivity series of metals , which helps in understanding and making the reaction much easier.