# An anion-exchange column is lined with "OH"^(-). A solution of "MCl"_2 is passed through the column, and the result titrated to its equivalence point [ . . . ]. If "2.375 g" of "MCl"_2 was originally dissolved, determine [ . . . ] "M"?

## An anion-exchange column is lined with ${\text{OH}}^{-}$. A solution of ${\text{MCl}}_{2}$ is passed through the column, and the result titrated to its equivalence point with $\text{100.00 mL}$ of $\text{0.5 M}$ $\text{HCl}$. If $\text{2.375 g}$ of ${\text{MCl}}_{2}$ was originally dissolved, determine the likely identity of the unknown metal $\text{M}$?

Jun 9, 2017

Well, since you have an anion exchange column, presumably it, well, exchanges what anion is in that column.

The column is lined with ${\text{OH}}^{-}$.

A solution of ${\text{MCl}}_{2}$ is run through the column, so ${\text{Cl}}^{-}$ will displace ALL the ${\text{OH}}^{-}$; the ${\text{OH}}^{-}$ consequently passes through.

Since you have used

${\text{0.5 M" xx "0.100 L" = "0.05 mols H}}^{+}$

to neutralize the ${\text{0.05 mols OH}}^{-}$ that passed through, you have:

${\text{0.05 mols OH"^(-) xx "1 mol Cl"^(-)/"1 mol OH"^(-) xx "1 mol MCl"_2/"2 mol Cl"^(-) = "0.025 mols MCl}}_{2}$.

The $\text{M}$ metal had reacted to make:

${\text{M"(s) + "Cl"_2(g) -> "MCl}}_{2} \left(s\right)$

Thus, ${\text{0.025 mols MCl}}_{2}$ were made. Therefore, from the mass of the solid dissolved in water:

$\text{2.375 g MCl"_2/"0.025 mols}$

$=$ ${\text{95 g/mol MCl}}_{2}$.

${M}_{M C {l}_{2}} - {M}_{2 C l}$
= "95 g/mol" - 2("35.453 g/mol Cl")
= color(blue)(M_(M)) = "24.094 g/mol" ~~ color(blue)("24 g/mol")