Question

please help!

A 0. 563 g sample of a pure carbonate, X_nCO_3(s) was dissolved in 50. 0 mL of 0. 2000 M HCl(aq). The excess HCl(aq) was back titrated with 24. 20 mL of 0. 0980 M NaOH(aq). How many moles of HCl react with the carbonate? What is the identity of the cation, X?

Answer 1

From the first data that we have we can see how much HCl we added.

Molarity= mol solute/Volume solution (L)

mol solute= M* V= 0,2000M* 0,050L= 0,010mol of HCl

The same with the NaOH added.

mole solute= Molarity* Volume = 0,0980M * 0,0242L = 0,0024mole NaOH

So from the reaction of HCl in excess we can calculate how much HCl reacts with the carbonate.

NaOH + HCl ------- NaCl + H2O

because 1 mol of NaOH ----------- 1 mol of HCl

then 0,0024 mole NaOH ----- 0,0024mole HCl

So we have to subtract the amount of HCl that was consumed by the NaOH and we will get the amount of HCl that react before with the carbonate.

mole HCl = mole HCl inicially - mole HCl consumed by NaOH

mole HCl = 0,010mol - 0,0024mole = 0,0076 mole

0,0076 mole this is the amount of HCl that react with the carbonate.

Now we know that we added 0,563g of the carbonate and because of the titration we know that we have 0,0076 mole of that compound.

0,563g ----------- 0,0076mole

Xg ---------------- 1 mol

X = 74,08g wiegths the XnCO3

But the anion carbonate weights: 60g/mol

So the subtract 74,08g/mol - 60g/mol= 14,08 g/mol

That means that the element X has to weight 14g/mol or have 2 atoms of a weight of 7g/mol

The only atom that weights 14g is the Nitrogen, but the valence of that is 1, 3 and 5 and because the carbonate has 2 negative charges can´t be that element.

The only element that weight 7g/mol is the lithium, and ahs a valence of +1, so we would have 2 atoms of lithium having the mass of 14g/mole.

So X is Li and n is 2.