How to calculate the pH of a buffer after adding HCl?


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> In a certain $\pu{1 L}$ buffer with $\pu{0.25 M}$ $\ce{NH3}$ and $\pu{0.4 M}$ $\ce{NH4Cl}$, $\pu{0.1 mol}$ of $\ce{HCl}$ has been added to the solution. What is the $\mathrm{pH}$?
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> $K\_\mathrm{a}(\ce{NH4+}) = \pu{5.56e-10}$, and $K\_\mathrm{b}(\ce{NH3}) = \pu{1.8e-5}$.
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The answer says that at last there will be $\pu{0.5 mol}$ of $\ce{NH4+}$ formed from the equation:


$$\ce{NH3 + H+ -> NH4+}$$


So $\pu{0.1 mol}$ of additional $\ce{NH4+}$ is formed and the total amount of $\ce{NH4+}$ is $\pu{0.5 mol}$.


However, I do not agree with this statement and I believe that there is still a limited power of $\ce{NH3}$ accepting $\ce{H+}$ (from $\ce{HCl}$).


My theory is that $\ce{NH3}$ accepts $x$ amount of protons and will form $x$ amount of $\ce{NH4+}$, with the original $K\_\mathrm{b}$. Meaning


$$K\_\mathrm{eq} = \frac{[\ce{NH4+}]}{[\ce{NH3}][\ce{H+}]}$$


which I thought was $1/K\_\mathrm{b}$. Is my theory correct?




I'm not sure I follow your argumentation, but I guess you don't really need $K\_\mathrm{b}$ at all since addition of strong acid $\ce{HCl}$ influences the dissociation of the one buffer component — the weak acid $\ce{NH4+}:$


$$\ce{NH4+ <<=>[$K\_\mathrm{a}$] NH3 + H+} \qquad K\_\mathrm{a} = \frac{[\ce{NH3}][\ce{H+}]}{[\ce{NH4+}]}\tag{1}$$


[Henderson–Hasselbalch equation](https://en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation) applied to this buffer system before the addition of acid allows to find initial $\mathrm{pH}$ (not required by the problem, I do this solely for demonstration):


\begin{align}
\mathrm{pH} &= \mathrm{p}K\_\mathrm{a} + \log{\frac{[\ce{NH3}]}{[\ce{NH4+}]}} \\
&= -\log{(\pu{5.56E-10})} + \log{\frac{\pu{0.25 M}}{\pu{0.40 M}}} \\
&= 9.05\tag{2}
\end{align}


Once the strong acid $(\ce{HCl},$ assuming complete dissociation) is added, the equilibrium shifts accordingly:


\begin{align}
\mathrm{pH\_1} &= \mathrm{p}K\_\mathrm{a} + \log{\frac{[\ce{NH3}] - [\ce{HCl}]}{[\ce{NH4+}] + [\ce{HCl}]}} \\
&= -\log{(\pu{5.56E-10})} + \log{\frac{\pu{0.25 M} - \pu{0.10 M}}{\pu{0.40 M} + \pu{0.10 M}}} \\
&= 8.73\tag{3}
\end{align}


You would've needed $\mathrm{p}K\_\mathrm{b}$ though when a strong base (e.g. $\ce{NaOH})$ were added.


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