TLDR: Is there an as simple and direct as possible method to measure chloride content by titration?

I want to measure the content (mass fraction) of chloride in $\mathrm{AlCl_3}$ powder by titration. Purpose is to measure the purity of the aluminium chloride. For example assume the measured chloride content is 75 % then that means the $\mathrm{AlCl_3}$ is 94.0 % pure.

When I search for how to determine chloride by titration I get the following three methods, which appear to be more complex than needed: Mohr method, Volhard method, Fajans method. All of which use an indicator and require specific conditions and treatments or reagents. The Volhard method is also a back titration and seems to be the default method for some reason which I could not find.

I found the following procedure, which is basically the Volhard method but with an additional reagents (of which I do not know the purpose) and an additional dilution step.

1.) Prepare an flask with stopper with 50 ml water and chill in the freezer to form ice.

2.) Weigh about 1 g of sample and transfer it in to the flask along with the paper and put the stopper.

3.) Allow it to come to room temperature. Add a few drops of phenolphthalein indicator and titrate with 1N NaOH solution to the end point.

4.) Dilute the solution to 250 mL. Pipette out 20 mL in to another flask.

5.) Add 3-4 mL of conc. nitric acid , 50 mL of 0.1 N silver nitrate solution, 2 mL of nitro benzene and a few drops of ferric ammonium sulfate indicator.

6.) Titrate with 0.1 N ammonium thiocyanate solution to reddish brown end point.

No idea what the nitro benzene is for. And the additional dilution step is certainly unnecessary and lowers precision; could just weight less sample and not dilute. The NaOH step is probably to react HCl (which is formed when $\mathrm{AlCl_3}$ reacts with water and may escape and thus lower the chlorine content) to NaCl; but also not sure about that.

What speaks against just directly titrating the chloride containing sample solution with the 0.1N $\mathrm{AgNO_3}$ solution and detect the end point potentiometric? Am I missing something? Why would people do a back titration with two solutions needed, instead of just a direct titration?

  • 2
    $\begingroup$ You can consult Vogel's Quantitative Analysis. There you will find all titrimetric details for chloride and explanations. Web search will only lead to confusion. $\endgroup$
    – AChem
    Feb 11 at 16:15
  • $\begingroup$ All your answers to the question are available in Vogel. $\endgroup$
    – AChem
    Feb 11 at 16:16
  • $\begingroup$ @M.Farooq - What, no love for Skoog and West? $\endgroup$
    – Todd Minehardt
    Feb 11 at 19:33
  • 3
    $\begingroup$ @ToddMinehardt, I like classical books when it comes to titrimetric and gravimetric methods. Vogel's classical Quantitative Analysis from the 60s was a massive book with >1200 pages, all extensive details such as holding volumetric pipette. $\endgroup$
    – AChem
    Feb 11 at 22:02
  • 2
    $\begingroup$ Here is the textbook for the OP on Internet Archive. archive.org/details/textbookofquanti0006voge/page/n7/mode/… $\endgroup$
    – AChem
    Feb 11 at 22:03

1 Answer 1


This problem is not as easy as it looks. It starts from the hypothesis that $\ce{AlCl3}$ is not pure in the original flask- The impurity will be supposed not to interfere later on. Only $\ce{H+}$ and $\ce{Cl-}$ will be determined by titration.

Let's suppose that the original sample of substance weighs $m_o$ and contains $n\ce{_o mol AlCl3}$. When dissolved into water, this substance reacts according to : $$\ce{AlCl3 + 3 H2O <=> 3 H+ + 3 Cl^- + Al(OH)3} \tag{1}$$ So, if the hydrolysis is complete (and it will be during titrating by $\ce{NaOH}$), the total amount of $\ce{H+}$ and $\ce{Cl^-}$ in the solution is $3n_o$. The first titration, with $\ce{NaOH}$, determines the amount $n_1$ = $n\ce{(NaOH)}$ = $n(\ce{H+})$ = $\ce{n({Cl^-})}$ = $3n_o$ produced by the reaction $(1)$. But, apparently, the result of this titration is not reliable enough. The author of the problem is afraid of some unwanted acidic contamination. So the actual amount $n_2$ of $\ce{H+}$ and of $\ce{Cl-}$ due to ($1$) will be obtained by another way, as will be described now.

The second titration is made to determine the amount $n_2$ of $\ce{Cl-}$ ion in solution. But it cannot be done directly, as, after the first titration by $\ce{NaOH}$, the solution contains $\ce{NaCl}$ and maybe some $\ce{NaOH}$. It is first acidified by adding some drops of $\ce{HNO3}$ to destroy this unwanted $\ce{NaOH}$ . Then a known excess of $n_3$ mol silver nitrate is added. All the $n_2$ mol $\ce{Cl-}$ ions are precipitated as insoluble $\ce{AgCl}$. $$\ce{Ag^+ + Cl- -> AgCl(s)} \tag{2}$$ The solution contains now an excess of $n_3 - n_2$ mol $\ce{Ag+}$ ions. Some drops of ferric ions are added (plus nitrobenzene - I don't know why). The excess of $\ce{Ag+}$ ions is titrated by gradually adding a thiocyanate solution. In the beginning, it produces the reaction : $$\ce{Ag+ + SCN^- -> AgSCN(s)}\tag{3}$$ $\ce{AgSCN}$ is a white insoluble precipitate, which looks like $\ce{AgCl}$.

The end point can be determined with a high degree of precision. Because, when the precipitation reaction (3) is finished, the next drop of thiocyanate reacts with the ferric ions to produce a dark red color, made of the ion $\ce{Fe(SCN)^{2+}}$ which is dark red. So as soon as the last amount of $\ce{AgSCN}$ is precipitated, the solution gets red. And this point can be determined with a high precision.

If $n_4$ mol thiocyanate have been added to get to the final red color, the amount $n_2$ of $\ce{Cl-}$ ions can be determined : $n_2 = n_4 - n_3$. And the original mass of $\ce{AlCl3}$ is equal to $m = \frac{(n_4 - n_3) M(\ce{AlCl3})}{3}$, if $M(\ce{AlCl3})$ is the molar mass of $\ce{AlCl3}$. One should find that $m < m_o$

  • $\begingroup$ Maurice, I think HAlCl4 does not even exist. $\endgroup$
    – AChem
    Feb 11 at 22:46
  • $\begingroup$ @M. Farooq. You may be right, after all. I know that the chloraluminate ion $\ce{AlCl_4^-}$ does exist. But the existence of $\ce{HAlCl4}$ is not proven. So I will edit and modify my answer. $\endgroup$
    – Maurice
    Feb 12 at 20:32
  • $\begingroup$ @Maurice Thank you for the explanation. The $\ce{AlCl3}$ is supposed to be pure or almost pure (> 99.5 %) as it is purchased from a supplier, but we have to test the assay of it before using it in a production process. Now, what I really do not understand is: Why is it not common to directly titrate the chloride (be it in the form of $\ce{[Al(H2O)6Cl3]}$ or $\ce{HCl}$ or $\ce{NaCl}$) with the $\ce{AgNO3}$ solution? Then the used volume of $\ce{AgNO3}$ solution directly corresponds to the amount of chloride. Why the detour via a back titration with thiocyanate? $\endgroup$
    – Wulfram
    Feb 14 at 10:18
  • $\begingroup$ @Wolfram. Even if the product is sold $99.5$% by the supplier, you cannot avoid that some part of $\ce{AlCl3}$ has been hydrolyzed in the flask during its storage. $\endgroup$
    – Maurice
    Feb 14 at 11:16

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