Home >
Community >
Determination of chloride content in aluminium chloride by titration
Upvote
VOTE
Downvote
+ Titration
+ Chemistry
Posted by
Nicholas Jonas
Determination of chloride content in aluminium chloride by 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 $m < m_o$
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 $m < m_o$
@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.More
@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?More
@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.More
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 $m < m_o$
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 $m < m_o$
More
VOTE
VOTE
VOTE
VOTE
VOTE