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

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 $$\mathrm{pH}$$?

$$K_\mathrm{a}(\ce{NH4+}) = \pu{5.56e-10}$$, and $$K_\mathrm{b}(\ce{NH3}) = \pu{1.8e-5}$$.

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?

• Reasonably concentrated buffers like this one can be analyzed directly with Henderson-Hasselbalch assuming that we're in the buffered range (which we are).
– Zhe
Dec 11, 2017 at 19:40

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 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.