# Why can't this mixture be used as a buffer solution?

Explain, using a chemical equation to aid your answer, why a mixture of $50~\mathrm{cm^3}$ of $0.10~\mathrm{mol\:dm^{-3}}$ ammonia solution and $50~\mathrm{cm^3}$ of $0.10~\mathrm{mol\:dm^{-3}}$ hydrochloric acid cannot be used as a buffer solution.

So writing the equation down: $$\ce{NH4OH + HCl -> NH4Cl + H2O}$$

there are $0.005~\mathrm{mol}$ of each reactant. From my notes, I think there has to be double the amount of the weak base, but I don't really understand why. Even after googling explanations, I can't find one that answers my question. Why do you need double the amount (if this is correct) of the weak base? An answer to the question above would also be appreciated.

You have an example of an alkaline buffer solutions here. They are commonly made from a weak base ($\ce{NH3}$ aqueous solution) and one of its salts ($\ce{NH4Cl}$) and work in a little simplified picture as follows.

Ammonia is a weak base, so the position of the equilibrium for the following reaction is well to the left $$\ce{NH3_{(aq)} + H2O_{(aq)} <<=> NH4+_{(aq)} + OH^{-}_{(aq)}} \, ,$$ and once you add some $\ce{NH4Cl}$ it pushes the equilibrium of the reaction above even more to the left (see Le Chatelier's Principle). As the result you have both lots of ammonia (due to the equilibrium of the above reaction being well to the left) and of ammonium ions (coming from $\ce{NH4Cl}$). And you will still have some hydroxide ions due to the reaction above, so that solution will be alkaline. The actual $pH$ depends on in which molar proportions the ingredients were mixed. Usually, $\ce{NH3_{(aq)}}$ and $\ce{NH4Cl}$ mixed in equal molar proportions, the solution would have a pH of 9.25.

When some acid is added to such buffer it will mostly be removed by the following process $$\ce{NH3_{(aq)} + H+_{(aq)} <=>> NH4+_{(aq)}} \, ,$$ and when some base is added it will mostly be consumed by the following reaction with ammonium ions $$\ce{NH4+_{(aq)} + OH^{-}_{(aq)} <=>> NH3_{(aq)} + H2O_{(aq)}} \, ,$$ where in both cases the equilibrium is well to the right.

So, as you see, both the weak base and its salt should present in the solution. Thus, there should be some excess of weak base when you mix it with a strong acid, otherwise there will be no weak base at the end of neutralization reaction. The idea that one should have double amount of a weak base comes from the fact that, as I mentioned, usually a weak base and its salt are mixed in equal molar proportions.

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