Common Ion Effect - Ionic Equilibrium

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?

• Existance of NH4OH in water is a myth. There is no dissociation like $\ce{NH4OH(aq) <=> NH4+(aq) + OH-(aq)}$ but the equilibrium $\ce{NH3(aq) + H2O <=> NH4+(aq) + OH-(aq)}$, or the true dissociation of the conjugated acid: $\ce{NH4+(aq) + H2O <=> NH3(aq) + H3O+(aq)}$ – Poutnik Aug 14 at 8:50
• @Poutnik True, but I don't think that's relevant here. – Ivan Neretin Aug 14 at 8:54
• How can someone dissociate NH4OH in non-aqeous NaCl, NH4Cl or NaOH? Are you trying to dissociate it in their liquid phases? – Habib Aug 14 at 8:56
• @Ivan Neretin Well, partly true, as it is relavant to discourage its use at any occurance. – Poutnik Aug 14 at 8:56
• @Habib First you would need to have NH4OH. – Poutnik Aug 14 at 8:57

1 Answer

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:

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

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

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

• Thanks for spending time on answering my question. I get that addition of NaOH will cause the dissociation of NH4OH to decrease due to common ion 'OH-'. But why did you say that addition of NH4+ ions from NH4Cl will cause ionisation amount of NH3 to increase? – InfiniteCool23 Aug 15 at 10:09
• Wouldn't addition of NH4+ ions cause decrease of OH- ions and hence reduce the ionisation amount? – InfiniteCool23 Aug 15 at 10:17
• Final Question I now understand that both NaOH and NH4Cl will cause decrease in dissociation of NH4OH. Hence my final question is this: Which will cause more decrease in dissociation and why??? – InfiniteCool23 Aug 15 at 12:32