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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 $\mathrm{pH}$?

$K_\mathrm{a}(\ce{NH4+}) = \pu{5.56e-10}$, 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?

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  • $\begingroup$ Please do read How can I format math/chemistry expressions here? before you decide to revert the formatting. This network has certain guidelines, which are supported by technical means with help of Markdown and MathJax. $\endgroup$ – andselisk Dec 9 '17 at 6:20
  • $\begingroup$ Reasonably concentrated buffers like this one can be analyzed directly with Henderson-Hasselbalch assuming that we're in the buffered range (which we are). $\endgroup$ – Zhe Dec 11 '17 at 19:40
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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 ($\ce{NH4+}$):

$$\ce{NH4+ <<=>[K_\mathrm{a}] NH3 + H+} \qquad K_\mathrm{a} = \frac{[\ce{NH3}][\ce{H+}]}{[\ce{NH4+}]}$$

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 \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 \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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