Calculate the $\mathrm{pH}$ of a buffered solution containing $\pu{0.5 M}$ ammonia and $\pu{0.5 M}$ ammonium chloride when $\pu{0.15 M}$ $\ce{HCl}$ is added into it. The $\mathrm{p}K_\mathrm{b}$ of ammonia is $4.75$.
This is what I think shoud be going on in the buffer solution:
\begin{align} \ce{NH3(aq) +H2O(l) &<=> NH4+(aq) +OH-(aq)} \\ \ce{NH4Cl(aq) &-> NH4+(aq) +Cl-(aq)} \end{align}
This is where I get stuck thinking that the question told me that I have a buffered solution containing $\pu{0.5 M}$ $\ce{NH3}$ and $\pu{0.5 M}$ $\ce{NH4Cl}$ solution, but is that even possible? How can you make a $\pu{0.5 M}$ ammonia solution? Would not that ammonia exist as ammonium $\ce{NH4+}$ and whether the equations I wrote are correct. I am just so confused!
Sorry for the delay but I'm having exams. Okay, I'll try to do this now by not using the Henderson equation.
Firstly, ammonia will refer to ammonium hydroxide, i.e. $\ce{NH3 -> NH4OH}$:
\begin{array}{cccc} &\ce{&NH4OH &<=> &NH4+ &+ &OH-} \\ &\mathrm{I}: &\pu{0.5 M} & &\pu{0 M} & & \pu{0 M} \\ &\mathrm{E}: &(0.5 - x)\,\pu{M} & &x\,\pu{M} & &x\,\pu{M} \end{array}
Now for the salt:
\begin{array}{cccc} &\ce{&NH4Cl &<=> &NH4+ &+ &Cl-} \\ &\mathrm{I}: &\pu{0.5 M} & &x\,\pu{M} & &\pu{0 M} \\ &\mathrm{E}: &\pu{0 M} & &(0.5 + x)\,\pu{M} & &\pu{0.5 M} \end{array}
$$K_\mathrm{b} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH4OH}]} = \frac{(x + 0.5)x}{(0.5 - x)}$$
and since $K_\mathrm{b} = 1.778 \times 10^{-5}$, I can find the value of $x$ to be $1.7778 \times 10^{-5}$.
$$[\ce{H+}][\ce{OH-}] = 10^{-14} \to [\ce{H+}] = \frac{10^{-14}}{[\ce{OH-}]}$$ and $x = [\ce{OH-}]$. So
$$[\ce{H+}] = 5.624 \times 10^{-10}$$
Now I add the $\ce{HCl}$:
\begin{array}{cccc} &\ce{&HCl &<=> &H+ &+ &Cl-} \\ &\mathrm{E}: &\pu{0 M} & &\pu{0.15 M} & &\pu{0.15 M} \end{array}
I will ignore any common ion effect since it will be negligible (I think). The pH will be the total $\ce{H+}$ concentration, so $[\ce{H+}] = 5.624 \times 10^{-10} + 0.15$. This is a pH of about 0.823 which is totally wrong. What am I doing wrong? Also, the volumes of the buffer solution or of the acid solution have not been given.
Okay, so I now know what I was doing wrong (thanks to someone pointing it out). I was adding in $\ce{HCl}$ without thinking that it would actually react with the ammonia in equilibrium in solution (which is just me being either dumb or ignorant).
So I have the situation that:
\begin{array}{cccc} &\ce{&NH4OH &<=> &NH4+ &+ &OH-} \\ &\mathrm{E_\text{(no HCl)}}: &(0.5 - 1.778 \times 10^{-5})\,\pu{M} & &(1.778 \times 10^{-5})\,\pu{M} & &(1.778 \times 10^{-5})\,\pu{M} \\ &\mathrm{E_\text{(HCl)}}: &()\,\pu{M} & &()\,\pu{M} & &()\,\pu{M} \end{array}