# Could this solution be resolved by titration?

I have a mixture of $$\ce{NH_4NO_3}$$ and $$\ce{NH_3}$$ at $$\ce{pH}$$ of about 13.

It is possible to determine the concentration of both on them by acid-base titration?

I was told it's possible, and that you need first titrate with $$\ce{HCl}$$ and then with $$\ce{NaOH}$$.

Thanks.

Edit: In the lab were I was working, there were an auto titrator, that had a program that could tell the concentracion of $$\ce{NH_4NO_3}$$ and $$\ce{NH_3}$$ in a solution, just by titrating it and controlling the $$\ce{pH}$$ with a pH meter. I simply don't understand how it does it, because depending of the $$\ce{pH}$$, $$\ce{NH_4+}$$ and $$\ce{NH_3}$$ would be in the same form and I don't know how you could tell them apart.

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• Note that at pH=13, NH4+ ions are practically all converted to NH3 by OH- ions. Notice the pKb constant for ammonia and what NH4+/NH3 concentration ratio exists at pH 13. May 5, 2019 at 12:12

The ammonium nitrate dissociate in water:

$$\ce{NH4NO3 -> NH4+ + NO3-}$$

The $$\ce{NH4+}$$ and $$\ce{NH3}$$ form together the ammonia $$\mathrm{pH}$$ buffer with $$\ce{NH4+}$$ acidity constant $$\mathrm{p}K_\mathrm{a}=9.25$$: $$\mathrm{pH}=\mathrm{p}K_\mathrm{a}+ \log { \frac {c_{\ce{NH3}}}{c_{\ce{NH4+}}}}$$

The titration curve as $$\mathrm{pH}=f(V)$$ is the most flat at $$\mathrm{pH}=\mathrm{p}K_\mathrm{a}=9.25$$, where the buffer has the biggest capacity.

For a given solution, containing $$\ce{NH4+}$$ and $$\ce{NH3}$$, titration with $$\ce{HCl}$$ or $$\ce{NaOH}$$ causes respective acid-base reactions:

\begin{align} \ce{NH3 + H+ &-> NH4+ \\ NH4+ + OH- &-> NH3 + H2O} \\ \end{align}

The ammonia buffer causes $$\mathrm{pH}$$ being quite stable during titration, until one of the 2 ammonia forms is practically spent.

At that point, the $$\mathrm{pH}$$ titration curve gets very progressively bent up or down, respectively and the titrator stops the titration.

The content of $$\ce{NH4NO3}$$ and $$\ce{NH3}$$ is then obtained by obvious routine calculation.

\begin{align} m_{\ce{NH3}}&= M_{\ce{NH3}}\cdot c_{\ce{HCl}}\cdot V_{\ce{HCl}}\\ m_{\ce{NH4NO3}}&= M_{\ce{NH4NO3}}\cdot c_{\ce{NaOH}}\cdot V_{\ce{NaOH}}\\ \end{align}

Note that at $$\mathrm{pH}=13$$, $$\ce{NH4+}$$ ions practically do not exist, as $$\frac {c_{\ce{NH3}}}{c_{\ce{NH4+}}}=5620$$

If we consider $$\mathrm{pH}$$ of ammonia solution itself, it is \begin{align} \mathrm{pH}=14 - 0.5 \cdot ( \mathrm{p}K_\mathrm{b}-\log c) \\ 13=14 - 0.5 \cdot ( 4.75 -\log c) \\ \end{align}

$$c_{\ce{NH3}}$$ would be 563 mol/L, what is nonsense. Such $$\mathrm{pH}$$ is not achievable by ammonia alone, but a mineral hydroxide would be present.

If there was initially any ammonium nitrate, it would be converted to ammonia:

$$\ce{NH4NO3 + NaOH -> NaNO3 + NH3 + H2O}$$

In such a case, determination of both compounds by titration is not possible, as there are no ammonium ions and ammonia would be titrated together with hydroxide.

• Ok, I get it, thanks. I don't know why, but such a simple thing and I was completely stuck. And sorry, $\mathrm{pH = 13}$ was just an example, but now I see that it would make no sense with ammonia alone. So the titrator would first titrate with $\ce{HCl}$ and then with $\ce{NaOH}$, having to titrate first the $\ce{HCl}$ that it used. May 5, 2019 at 21:37