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 $\ce{NaOH}$.


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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    – Poutnik
    Commented May 5, 2019 at 12:07
  • 4
    $\begingroup$ 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. $\endgroup$
    – Poutnik
    Commented May 5, 2019 at 12:12

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – user78790
    Commented May 5, 2019 at 21:37

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