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"#?

Question

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 with #"100.00 mL"# of #"0.5 M"# #"HCl"#. If #"2.375 g"# of #"MCl"_2# was originally dissolved, determine the likely identity of the unknown metal #"M"#?

2621 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-09T16:51:08

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

The column is lined with #"OH"^(-)#.

A solution of #"MCl"_2# is run through the column, so #"Cl"^(-)# will displace ALL the #"OH"^(-)#; the #"OH"^(-)# consequently passes through.

Since you have used

#"0.5 M" xx "0.100 L" = "0.05 mols H"^(+)#

to neutralize the #"0.05 mols OH"^(-)# that passed through, you have:

#"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 #"M"# metal had reacted to make:

#"M"(s) + "Cl"_2(g) -> "MCl"_2(s)#

Thus, #"0.025 mols MCl"_2# were made. Therefore, from the mass of the solid dissolved in water:

#"2.375 g MCl"_2/"0.025 mols"#

#=# #"95 g/mol MCl"_2#.

Since molar masses are additive:

#M_(MCl_2) - M_(2Cl)#

#= "95 g/mol" - 2("35.453 g/mol Cl")#

#= color(blue)(M_(M)) = "24.094 g/mol" ~~ color(blue)("24 g/mol")#

So the metal is likely magnesium.