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In my textbook there is the following problem:

How many moles of $\ce{NH4Cl}$ must be added to $2.0~\mathrm{L}$ of $0.10~\mathrm{M}$ $\ce{NH3}$ to form a buffer with $\mathrm{pH}=9$? Assume the addition does not change the volume of the solution significantly.

The solution given is as follows:

The equilibrium between $\ce{NH3}$ and $\ce{NH4+}$ is given by $$\ce{NH3 + H2O <=> NH4+ + OH-}.$$ We know the $K_\mathrm{b}$ value and $\ce{[OH-]}$ from the $\ce{pH}$ and $\ce{[NH3]}$ is given, thus we can solve for $\ce{[NH4+]}$. The solution given says that the final answer is given by this concentration multiplied by the total volume $2~\mathrm{L}$.

I don't follow the logic in the solution. The $\ce{[NH4+]}$ we found was the total concentration of $\ce{NH4+}$ in the solution after adding the salt. But before adding $\ce{NH4Cl}$, there was already some $\ce{NH4+}$ in the solution (we don't know how much), and thus to find the number of moles $\ce{NH4Cl}$ we need to add, we must subtract the number of moles $\ce{NH4+}$ already in the solution from the answer we got in the textbook solution.

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    $\begingroup$ the amount of $\ce{NH4+}$ beforehand is negligible because ammonia is a fairly weak base $\endgroup$ – bon Mar 23 '15 at 18:18
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$\ce{NH3}$ is such a weak base ($\text{K}_{\text{b}}=1.8\times 10^{-5}$) that the initial concentration of $\ce{NH4^+}$ can be considered negligible. You can consider the initial concentration of $\ce{NH4^+}$, but the answer you will get is the same to a considerable number of significant figures.

By considering the equilibrium of $\ce{NH3}$ dissociating to form $\ce{NH4^+}$ and $\ce{OH^-}$ it is easy to find the initial concentration of $\ce{NH4^+}$.

$\text{K}_{\text{b}} = \frac{[\ce{NH4^+}][\ce{OH^-}]}{[\ce{NH3}]}$

Since $[\ce{NH4^+}] = [\ce{OH^-}]$

$\text{K}_{\text{b}}\times [\ce{NH3}] = [\ce{NH4^+}]^{2}$

Plugging in values gives:

$[\ce{NH4^+}] = 1.34\times 10^{-3} (3\text{s.f.})$

Which equates to $2.68\times 10^{-3}\text{mol}$ in the $2\text{L}$.

If I'm right this should be orders of magnitude smaller than the textbook answer, and therefore negligible.

Hope this helps.

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  • $\begingroup$ This makes sense. Can you prove to me mathematically that the initial concentration of $NH_4^+$ is negligible? How would we calculate it? Is it not simply $0.1 M$ by stoichiometry? $\endgroup$ – Joshua Benabou Mar 23 '15 at 21:22
  • $\begingroup$ I added a calculation of this to my answer, but it's not just 0.1M; the concentration is governed by an equilibrium. $\endgroup$ – Ivan Mar 23 '15 at 21:41
  • $\begingroup$ Oops I was assuming complete dissociation, which is not the case since (as you said) $NH_3$ is a weak base. Thanks again. $\endgroup$ – Joshua Benabou Mar 23 '15 at 21:50

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