# How many moles NH4Cl must be added to NH3 to create buffer with pH=9?

How many moles of $$\ce{NH4Cl}$$ must be added to $$\pu{2.0 L}$$ of $$\pu{0.10 M}$$ $$\ce{NH3}$$ 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 $$\mathrm{pH}$$ 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 $$\pu{2 L}$$.

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).

Thus to find the number of moles $$\ce{NH4Cl}$$ we need to add, shouldn't we subtract the number of moles $$\ce{NH4+}$$ already in the solution from the answer we got in the textbook solution?

• the amount of $\ce{NH4+}$ beforehand is negligible because ammonia is a fairly weak base – bon Mar 23 '15 at 18:18

$$\ce{NH3}$$ is such a weak base ($$K_{\mathrm{b}}=1.8\times 10^{-5}$$) 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 $$\ce{NH4+}$$.

$$K_{\mathrm{b}} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]}$$

Since \begin{align}[\ce{NH4+}] &= [\ce{OH-}] \tag{1}\\ K_{\mathrm{b}}\cdot [\ce{NH3}] &= [\ce{NH4+}]^{2} \tag{2} \end{align} Plugging in values gives:

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

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

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

• 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? – Joshua Benabou Mar 23 '15 at 21:22
• I added a calculation of this to my answer, but it's not just 0.1M; the concentration is governed by an equilibrium. – Ivan Mar 23 '15 at 21:41
• Oops I was assuming complete dissociation, which is not the case since (as you said) $NH_3$ is a weak base. Thanks again. – Joshua Benabou Mar 23 '15 at 21:50