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

Question

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?

Answer 1

$\ce{NH3}$ 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.