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?