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Question: In III group precipitation, \[N{H_4}Cl\] is added before adding \[N{H_4}OH\] to. A. Decrease conce...

In III group precipitation, NH4ClN{H_4}Cl is added before adding NH4OHN{H_4}OH to.
A. Decrease concentration Of OHO{H^ - }
B. Prevent interference of PO43PO_4^{3 - }
C. Increase concentration of ClC{l^ - }
D. Increase of concentration of OHO{H^ - } ion.

Explanation

Solution

To solve the above question we must be aware of the conditions of precipitation. The ammonium chloride (NH4Cl)\left( {N{H_4}Cl} \right) is added before adding ammonium hydroxide (NH4OH)\left( {N{H_4}OH} \right) due to the solubility product and the common ion effect. We will consider the following factors while considering the statement.

Complete step by step answer:
First, we will understand the reactions involved in the III group analysis. Here, ammonium chloride (NH4Cl)\left( {N{H_4}Cl} \right) is a strong electrolyte and the ammonium hydroxide (NH4OH)\left( {N{H_4}OH} \right) is a weak base. Now we know that ammonium chloride (NH4Cl)\left( {N{H_4}Cl} \right) is a strong electrolyte so it will completely ionize or dissociate in a solution. Now ammonium hydroxide (NH4OH)\left( {N{H_4}OH} \right) is a weak base so it will not dissociate or completely ionize in the solution.
Now we will consider the reactions for strong electrolytes and weak base. So with the help of the reactions and the nature of solutes, we will be able to understand the reason that salt qualitative analysis NH4ClN{H_4}Cl is added before adding NH4OHN{H_4}OH. So the reactions are as follows. NH4ClNH4++ClN{H_4}Cl \to NH_4^ + + C{l^ - } and NH4OHNH4++OHN{H_4}OH \to NH_4^ + + O{H^ - }
Here, ammonium chloride will decompose and ionizes completely as it is a strong electrolyte. Whereas ammonium hydroxide will not completely dissociate as it is a weak base. Here is the common ion effect due to the presence of higher group cation (NH4+)\left( {NH_4^ + } \right) in both ammonium chloride and ammonium hydroxide. Due to the common ion effect, the concentration (OH)\left( {O{H^ - }} \right) decreases because otherwise, it will lead to the precipitation of higher group cation.

So, the correct answer is Option A.

Note: The Common ion effect is responsible for a decrease in the solubility product of a precipitate. The solubility product of higher group cations is greater than their ionic product. Therefore, they are not precipitated.