Determination of chloride content in aluminium chloride by titration

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.



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> 1.) Prepare an flask with stopper with 50 ml water and chill in the freezer to form ice.
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> 2.) Weigh about 1 g of sample and transfer it in to the flask along with the paper and put the stopper.
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> 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.
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> 4.) Dilute the solution to 250 mL. Pipette out 20 mL in to another flask.
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> 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.
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> 6.) Titrate with 0.1 N ammonium thiocyanate solution to reddish brown end point.
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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?




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 $m < m\_o$


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