# Understanding a passage related to salt buffer

Passage from my textbook:

Are reactions possible among any of the solution components; is so, what is their stoichiometry?

Suppose that you are asked to calculate $[\ce{OH-}]$ in a solution that is prepared to be $\pu{0.10 M}~\ce{NaOH}$ and $\pu{0.20 M}~\ce{NH4Cl}$. Before you answer that $[\ce{OH-}]=\pu{0.10M}$, consider whether a solution can be simultaneously $\pu{0.10M}$ in $\ce{OH-}$ and $\pu{0.20M}$ in $\ce{NH4+}$ . It cannot; any solution containing both $\ce{NH4+}$ and $\ce{OH-}$ must also contain $\ce{NH3}$. The $\ce{OH-}$ and $\ce{NH4+}$ react in a $1:1$ mole ratio until $\ce{OH-}$ is almost totally consumed:$$\ce{NH4+ +OH- ->NH3 + H2O}$$ You are now dealing with the buffer solution $\pu{0.10M}~\ce{NH3}-\pu{0.10M}~\ce{NH4+}$

I am having trouble understanding the third last sentence in the paragraph that starts with "2." The statement reads: "It cannot; any solution containing both $\ce{NH4+}$ and $\ce{OH-}$ must also contain $\ce{NH3}$. The $\ce{OH-}$ and $\ce{NH4+}$ react in $1:1$ mole ratio until $\ce{OH-}$ is almost totally consumed"

• To see if you really understand it, you should probably try some similar problems, such as what the pH is if 0.1M NaOH and 0.1M NH4+ are added in equal concentrations, and do the problems in the book. It will help you realize that the assumption is only momentary, and that there are more parts to the problem – Andy Apr 12 '15 at 20:38

There is no such thing as $\ce{NH4OH}$ . They would react to form water and ammonia. Thus, to say that both exist in the solution at such high concentrations would be mistaken. They would react. Consider the reaction between ammonium and hydroxide.

$\ce{NH4+ + OH- -> NH3 + H2O}$

However, do note that at low concentrations, they can exist according to dissociation. However that is not what your passage is saying

*On your test, it may be good to assume the reaction goes to completion and then solve for the dissociation of ammonia. That is most likely what is expected of you.

*I thought of a more understandable analogy:

Think about $\ce{OH-}$ and $\ce{H+}$ in solution at 25 celsius. If one increases, the other decreases! According to ${Kw}$. Adding a lot of acid and base together would simply neutralize, they cannot exist at any higher than ${Kw}$

• So if we were asked to calculate the [OH-] in a solution that is prepared to be 0.10M NaOH and 0.2M NH4Cl, is saying that [OH-] = 0.1M wrong? – Nerdingout Apr 12 '15 at 20:10
• Yes, you should assume that the base completely reacts with NH4Cl to create a solution of .1M NH4 and the reacted NH3. SInce both are identical, you could use the relation pH=pKa. But the pOH would not be 1. This is closely related to the concept of a buffer, which you could look into – Andy Apr 12 '15 at 20:12
• I see! But if the OH- and the NH4+ are reacting in a 1:1 mole ratio, how is it possible that they can continue reacting until the OH- is totally consumed, if they are in equal concentrations? – Nerdingout Apr 12 '15 at 20:14
• Well, in that case, we "pretend" they completely react because by observing the K of ammonia, we see ammonia is more favored so more of it would exist. Then, we would need to use the K value to calculate it. So we first assume it goes to completion and use that to find what happens "afterwards", which is more close to the experimental value – Andy Apr 12 '15 at 20:33
• Saying that "the $OH^-$ is totally consumed" is wrong and misleading. Most of the $OH^-$ is consumed and very little remains (I'm guessing less than $10^{-4} M$). This is because water can disassociate to produce more $OH^-$. – LDC3 Apr 12 '15 at 23:51