Common Ion Effect - Ionic Equilibrium

Question

Question

In which of the aqueous solutions of the following, dissociation of $\ce{NH4OH}$ will be minimum?

A) $\ce{NaOH}$

B) $\ce{H2O}$

C) $\ce{NH4Cl}$

D) $\ce{NaCl}$

My Thoughts

My book says that the answer is option C '$\ce{NH4Cl}$' giving the reason as common ion effect.

But I think that option A '$\ce{NaOH}$' also has a common ion as $\ce{OH-}$.

What should be the right answer and how do we compare which will cause more suppression by common ion effect?

I get why $\ce{H2O}$ will not suppress dissociation of $\ce{NH4OH}$ that much since its equilibrium constant is very low (of the order of $10^{-14}$). But then why not $\ce{NaOH}$ (which is a very strong base and thus will almost completely dissociate into constituent ions)? Which one out of $\ce{NaOH}$ and $\ce{NH4Cl}$ is a more stronger electrolyte?

Is it because $\ce{NH4Cl}$ will form a buffer with $\ce{NH4OH}$?

I am feeling really confused about this. Any help will be highly appreciated!

Final Question

I now understand that both $\ce{NaOH}$ and $\ce{NH4Cl}$ will cause decrease in dissociation of $\ce{NH4OH}$. Hence my final question is this: Which will cause more decrease in dissociation and why?

Answer 1

Let's consider an aqueous solution, the concentration of which is $\pu{1 M}$ in $\ce{NH3}$ and $\pu{1 M}$ in $\ce{NaOH}$. Thus, following equilibrium would be taken place:

$$\ce{NH3 (aq) + H2O <=> NH4+ (aq) + OH- (aq)}\tag1$$

$$\ce{NaOH (aq) -> Na+ (aq) + OH- (aq)}\tag2$$

The $\mathrm{p}K_\mathrm{b}$ of equilibrium $(1)$ is 4.75, thus $K_\mathrm{b} = \pu{1.78E{-5}}$. In pure ammonia solution, from equation $(1)$:

$$K_\mathrm{b} = \pu{1.78E{-5}} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]}\tag3$$

If ionized amount at equilibrium is $\alpha$, then

$$K_\mathrm{b} = \pu{1.78E{-5}} = \frac{\alpha \times \alpha}{1-\alpha} = \alpha^2 \ \Rightarrow \ \therefore \ \alpha = \sqrt{\pu{1.78E{-5}}} = \pu{4.22E{-3}}$$

Assumptions: $\alpha \lt\lt 1$, and thus $1-\alpha \approx 1$, and $\alpha \gt \gt \pu{1.00E{-7}}$ and autoionization of water can be ignored. At the end, since $[\ce{NH4+}] = [\ce{OH-}] = \alpha = \pu{4.22E{-3}}$, both of these assoumtions are correct.

Now consider, if you have $\ce{NaOH}$ in your solution. Since it is a strong base, it completely dissociate according to equation $(2)$. Thus, there is a common ion in this solution: $[\ce{OH-}] = \pu{1 M}$. Hence, from the equation $(3)$:

$$K_\mathrm{b} = \pu{1.78E{-5}} = \frac{\beta \times (1+\beta)}{1-\beta} = \beta \ \Rightarrow \ \therefore \ \beta = \pu{1.78E{-5}} $$

Hence ($\alpha \gt \beta$), the ionization amount of $\ce{NH3}$ in presence of the common ion $\ce{OH-}$ is less than that of the solution when no common ions are present.

In similar way, you can prove the ionization amount of $\ce{NH3}$ is larger in the presence of the common ion $\ce{NH4+}$ Using following equilibria:

$$\ce{NH3 (aq) + H2O <=> NH4+ (aq) + OH- (aq)}\tag1$$

$$\ce{NH4Cl (aq) -> NH4+ (aq) + Cl- (aq)}\tag4$$

$$\ce{NH4+ (aq) + H2O <=> H3O+ (aq) + NH3 (aq)}\tag5$$