Question

A solution contains Cr³⁺ ion and Mg²⁺ ion. The addition of 1.00 L of 1.54 M NaF solution is required to cause the complete precipitation of these ions as CrF₃ (s) and MgF₂ (s). The total mass of the precipitate is 49.6 g.

Find the mass in grams of Cr³⁺ in the original solution.

Answer 1

MW CrF3= 108.9913 g/mol

MW MgF2= 62.3018 g/mol

If X is the amount of CrF3 formed then, 49.6 -X is the mass of MgF2.

Now find moles:

mol CrF3= X/108.9913 g/mol

mol MgF2= (49.6-X)/62.3018 g/mol

mol of F-= 1.54mol/L x 1L= 1.54 mol

Then, we know that we used 1.54 mol of F- to obtain CrF3 and MgF2. Now, each mol of CrF3 needs 3 mol of F- and each mol of MgF2 needs 2 mol of F-:

3(X/108.9913g/mol) + 2[(49.6-X)/62.3018 g/mol]= 1.54 mol F-

X= 11.416384g of CrF3

mol CrF3= 11.416384g/108.9913 g/mol= 0.1047 mol= mol of Cr+3

mass Cr+3= 0.1047 mol x 51.9961g/mol= 5.446g of Cr+3 in the original solution