Question

In the dissociation of $N{H_4}OH$, if excess of $N{H_4}Cl$ is added before adding $N{H_4}OH$, the concentration of:
(A) $NH_4^ + $ ions increases and $O{H^ - }$ ions decreases
(B) Both $NH_4^ + $ ions and $O{H^ - }$ ions increases
(C) $NH_4^ + $ ions decreases and $O{H^ - }$ ions increases
(D) Both $NH_4^ + $ ions and $O{H^ - }$ ions decreases

Answer 1

Hint: $N{H_4}OH$ is a weak base and does not dissociate completely, while the $N{H_4}Cl$ is a strong electrolyte and undergoes complete dissociation. Since both of them have $NH_4^ + $ ion in common, the common ion effect takes place thus suppressing the dissociation of $N{H_4}OH$.

Complete step by step answer:
-First of all we will see how dissociation of $N{H_4}OH$ occurs. It dissociates into $NH_4^ + $ and $O{H^ - }$ ions as shown below:
                                $N{H_4}OH \rightleftharpoons NH_4^ + + O{H^ - }$

According to the law of mass action the above equation at equilibrium can be expressed by the equation:
  $K = \dfrac{{\left[ {NH_4^ + } \right]\left[ {O{H^ - }} \right]}}{{\left[ {N{H_4}OH} \right]}}$
We should keep in mind that $N{H_4}OH$ is a weak base and so it does not undergo complete ionisation.

-The $N{H_4}Cl$ acts as a strong electrolyte and thus undergoes complete ionisation. It’s ionisation reaction would be:
                               $N{H_4}Cl \rightleftharpoons NH_4^ + + C{l^ - }$
-In the above two ionisation reactions we can see that the ion $NH_4^ + $ is common in both. Hence this leads to the common ion effect. The common ion effect is the decrease in the solubility of some ionic precipitate due to the addition of a soluble compound into the solution which has an ion in common with the precipitate. This occurs in accordance with Le Chatelier’s principle for the equilibrium reaction of the ionic association or dissociation.
Hence according to the common ion effect the addition of excess of $N{H_4}Cl$ before adding $N{H_4}OH$ leads to increased concentration of $NH_4^ + $ ions and hence the dissociation of $N{H_4}OH$ gets suppressed in accordance with the Le Chatelier’s principle. Hence this leads to decreased concentration of $O{H^ - }$ ions.
So, the correct answer is “Option A”.

Note: We should know that if a small amount of HCl is added to this solution,${H^ + }$ ions of HCl get neutralized by the $O{H^ - }$ ions already present and thus more of $N{H_4}OH$ molecules will get ionized to compensate the loss of $O{H^ - }$ ions. Thus pH practically remains constant.