Question

Assertion: On addition of $\text{N}{{\text{H}}_{4}}\text{Cl}$ to $\text{N}{{\text{H}}_{4}}\text{OH}$, pH decreases but remains greater than 7.
Reason: Addition of $\text{NH}_{4}^{+}$ ion decreases ionization of $\text{NH}_{4}^{+}$, thus ${{\left[ \text{OH} \right]}^{-}}$ decreases and also pH decreases.
A. Both Assertion and Reason are correct but Reason is the correct explanation for Assertion
B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
C. Assertion is correct but Reason is incorrect
D. Assertion is incorrect but Reason is correct
E. Both Assertion and Reason are incorrect.

Answer 1

The concept behind this question is ‘common ion effect’. As, both $\text{N}{{\text{H}}_{4}}\text{Cl}$ and $\text{N}{{\text{H}}_{4}}\text{OH}$ dissociates to give the common ion as $\text{NH}_{4}^{+}$ relating with Le-chatelier principle. $\text{N}{{\text{H}}_{4}}\text{Cl}$ is a strong electrolyte and $\text{N}{{\text{H}}_{4}}\text{OH}$ is a weak base.

Complete step by step answer:
Let us understand common ion effect from the basic using Le-chatelier principle and law of mass action;
As we know that ammonium hydroxide is a weak base. It dissociates as $\text{N}{{\text{H}}_{4}}\text{OH}\rightleftharpoons \text{NH}_{4}^{+}+\text{O}{{\text{H}}^{-}}$ and ammonium chloride dissociates completely because it is a strong electrolyte $\text{N}{{\text{H}}_{4}}\text{Cl}\to \text{NH}_{4}^{+}+\text{C}{{\text{l}}^{-}}$.
The excess of $\text{N}{{\text{H}}_{4}}\text{Cl}$ is added in the solution of $\text{N}{{\text{H}}_{4}}\text{OH}$ ; the common ion effect occurs. Common-ion effect is the decrease in solubility of an ionic compound by the addition of a soluble compound with an ‘ion’ in common with the ionic compound to the solution.
So, at equilibrium, the ions present are $\text{NH}_{4}^{+},\text{O}{{\text{H}}^{-}}\text{and C}{{\text{l}}^{-}}$. The amount of $\text{NH}_{4}^{+}$ ions in the solution has increased because it is produced from the two sources. The equilibrium constant will be written by law of mass action as $\text{K=}\dfrac{\left[ \text{NH}_{4}^{+} \right]\left[ \text{O}{{\text{H}}^{-}} \right]}{\left[ \text{N}{{\text{H}}_{4}}\text{OH} \right]}$. Le-chatelier principle states that if a system in chemical equilibrium is directed to a disturbance it tends to change in a way that opposes the change. Here, the change is the increased amount of concentration of $\text{NH}_{4}^{+}$ ions. If right-hand side concentration is increased, the left-hand side should also increase in order to balance the sides, but K is dependent only on temperature. So, it won’t change without changing the temperature. In order to neutralise the change, the concentration of $\left[ \text{O}{{\text{H}}^{-}} \right]$ ions is decreased. It means that equilibrium is moved backward and ionization of $\text{N}{{\text{H}}_{4}}\text{OH}$ is decreased. So, a new equilibrium is reached.
$\text{pOH= }-\text{log}\left[ \text{O}{{\text{H}}^{-}} \right]$, if $\left[ \text{O}{{\text{H}}^{-}} \right]$ ions are decreased, then, pOH will increase. But pH will decrease as $\text{pH=14}-\text{pOH}$. The change in pH is not very large, there is a small change observed in its value.

The correct answer of this question is option ‘a’ that both Assertion and Reason are correct but Reason is the correct explanation for Assertion.

Note:
Here, the Le-chatelier principle can be applied because dissociation of ammonium hydroxide follows equilibrium as it is a weak base and does not dissociate completely. It is written as $\text{N}{{\text{H}}_{4}}\text{OH}\rightleftharpoons \text{NH}_{4}^{+}+\text{O}{{\text{H}}^{-}}$. That’s why it follows the common ion concept also. If $\left[ \text{O}{{\text{H}}^{-}} \right]$ ions is decreased, it is not that pOH will decrease because there is negative sign present before the formula $\text{pOH= }-\text{log}\left[ \text{O}{{\text{H}}^{-}} \right]$.