# 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.

• You should specify more details - what is in the mixture, etc. In any case, why should the chloride ions from the two different sources be any different? There's no label attached to the $\ce{Cl-}$ that tells the $\ce{Ag+}$ "this ion came from the first component". As such, $\ce{Ag+}$ is going to precipitate every single chloride ion inside that mixture. Nov 22, 2015 at 14:54
• Safe to calculate like what? Regarding your "another question", there's no way to calculate the amount of $\ce{Mg^2+}$ without the EDTA titration. You will have one equation in two unknowns. Why don't you try setting up a system of equations? Let $x$ be the number of moles of $\ce{MgCl2}$ and $y$ be the number of moles of $\ce{NaCl}$. What is the total number of moles of $\ce{Cl-}$? What is the number of moles of $\ce{Mg^2+}$? Nov 22, 2015 at 15:49
• There is no guarantee that the amount of $\ce{Cl-}$ from $\ce{NaCl}$ is equal to the amount of $\ce{Cl-}$ from $\ce{MgCl2}$ - so how can you add them up to give $2n(\ce{Cl-})?$ They are independent of each other. If you use $x$ and $y$ as I suggested earlier, you might understand it better. Nov 22, 2015 at 16:26

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}

• @Mel - the answer key gives the Mg+2 as mmol/(100 ml of solution). Without knowing how large a sample was titrated you can't determine that part of the experiment. The answer key to me indicates that two different samples were taken from the solution in question. A 10 ml sample was titrated with the Ag solution and a 25 ml sample was titrated with EDTA.
– MaxW
Nov 22, 2015 at 19:33