Analysis of mixture by precipitation and complexometric titrations

I've performed an experiment about chloride mixtures. First using $\ce{AgNO3}$ as titrant, I've titrated the mixture to determine the amount of total chloride. Then by direct titration with $\ce{EDTA}$, I wanted to determine the amount of $\ce{Mg}$ ions.


I'm confused about the amount of total chloride ions as there are two sources of chloride ions.


I'd really appreciate if someone can help me here.


Thank you so much.




Let $x$ be the amount of $\ce{MgCl2}$ in the original mixture, and $y$ be the amount of $\ce{NaCl}$.


Both of these salts dissociate completely in aqueous solution to give their constituent ions:


$$\begin{align}
\ce{MgCl2 (s) -> Mg^2+ (aq) + 2 Cl- (aq)} \\
\ce{NaCl (s) -> Na+ (aq) + Cl- (aq)}
\end{align}$$


Therefore, the total number of moles of chloride ions in the mixture is:


$$\begin{align}
\eta\_{\ce{Cl-},\text{total}} &= \eta\_{\ce{Cl-}\text{(from }\ce{MgCl2})} + \eta\_{\ce{Cl-}\text{(from }\ce{NaCl})} \\
&= 2\eta\_{\ce{MgCl2}} + \eta\_{\ce{NaCl}} \\
&= 2x + y
\end{align}$$


$\ce{Ag+}$ ions react with $\ce{Cl-}$ ions in a $1:1$ stoichiometric ratio:


$$\ce{Ag+ (aq) + Cl- (aq) -> AgCl (s)}$$


Therefore, we have $\eta\_{\ce{Ag+}} = \eta\_{\ce{Cl-},\text{total}} = 2x + y$.




---


$\ce{EDTA}$ only complexes magnesium, and again in a $1:1$ stoichiometric ratio:


$$\ce{Mg^2+ (aq) + edta^4- (aq) -> [Mg(edta)]^2- (aq)}$$


Therefore, $\eta\_{\ce{EDTA}} = \eta\_{\ce{Mg^2+}} = \eta\_{\ce{MgCl2}} = x$.




---


From your titration, you would have determined both values $\eta\_{\ce{Ag+}}$ and $\eta\_{\ce{EDTA}}$. Therefore, you have a system of two simultaneous equations in two unknowns, which is extremely simple to solve:


$$\begin{align}
\eta\_{\ce{Ag+}} &= 2x + y \\
\eta\_{\ce{EDTA}} &= x
\end{align}$$


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