A mass of #"18 g"# of a compound #"M"_2"CO"_3 * x"H"_2"O"# #(M_ "M metal" = "24 g/mol")# is dissolved in #"500 mL"# of water and titrated against a #"0.5-M"# #"HCl"# solution. If the volume of #"HCl"# consumed is #"500 mL"#, the value of #x# is?

1 Answer
Nov 26, 2017

Answer:

#x = 2#

Explanation:

The idea here is that you can use the number of moles of anhydrous metal carbonate that took place in the reaction to figure out the number of moles of water present in the hydrate.

You know that you have

#"M"_ 2"CO"_ (3(aq)) + 2"HCl"_ ((aq)) -> 2"MCl"_ ((aq)) + "H"_ 2"O"_ ((l)) + "CO"_ (2(g)) uarr#

The reaction consumes #2# moles of hydrochloric acid for every #1# mole of anhydrous metal carbonate.

Use the molarity and the volume of the hydrochloric acid solution to find the number of moles of acid that were consumed in the reaction.

#500 color(red)(cancel(color(black)("mL solution"))) * "0.5 moles HCl"/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.25 moles HCl"#

This implies that the reaction consumed

#0.25 color(red)(cancel(color(black)("moles HCl"))) * ("1 mole M"_2"CO"_3)/(2color(red)(cancel(color(black)("moles HCl")))) = "0.125 moles M"_2"CO"_3#

Now, use the molar mass of the metal to find the molar mass of the anhydrous metal carbonate. You will end up with

#M_ ("M M"_2"CO"_3) = 2 * "24 g mol"^(-1) + "12.011 g mol"^(-1) + 3 * "16.0 g mol"^(-1)#

#M_ ("M M"_2"CO"_3) = "108.011 g mol"^(-1)#

Use this value to find the number of grams of anhydrous metal carbonate present in the initial sample, i.e. in the hydrated metal carbonate.

#0.125 color(red)(cancel(color(black)("moles M"_2"CO"_3))) * "108.011 g"/(1color(red)(cancel(color(black)("mole M"_2"CO"_3)))) = "13.5 g"#

This implies that the hydrate contained

#m_ ("H"_ 2"O") = "18 g " - " 13.5 g"#

#m_ ("H"_ 2"O") = "4.5 g"#

of water of hydration. Use the molar mass of water to find the number of moles of water of hydration present in the hydrate.

#4.5 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "0.2498 moles H"_2"O"#

Finally, to find the value of #x#, you need to find the number of moles of water present in exactly #1# mole of hydrate. Since you know that every mole of hydrate contains #1# mole of anhydrous metal carbonate, you can say that

#"0.125 moles M"_2"CO"_3 * x"H"_2"O" " " -> " " "0.2498 moles H"_2"O"#

This implies that #1# mole of hydrate will contain

#1 color(red)(cancel(color(black)("mole hydrate"))) * ("0.2498 moles H"_2"O")/(0.125 color(red)(cancel(color(black)("moles hydrate")))) ~~ "2 moles H"_2"O"#

Therefore, you have

#x = 2#

which means that you're dealing with a dihydrate.

#"M"_2"CO"_3 * 2"H"_2"O"#