Question

A solution contains Ag+ and Hg2+ ions. The addition of 0.100 L of 1.24 M NaI solution is just enough to precipitate all the ions as AgI and HgI2. The total mass of the precipitate is 28.7 g Find the mass of AgI in the precipitate.

Answer 1

Ag⁺(aq) + Hg²⁺(aq) + 3I^-(aq) AgI(s) + HgI₂(s)

Note: Mercury(II) iodide is HgI₂ and not HgI.

moles I^- added = (Molarity I^-)(L of I^-) = (1.24)*(0.100) = 0.124 moles I^- added
Hg²⁺ consumes twice as much I^- in the reaction as does Ag+, since the formula HgI₂ contains 2 I^- atoms while the formula AgI contains only 1 I^- atom.

If Ag⁺ consumes x moles of I^-, then Hg²⁺ consumes 2x moles of I^-. The total moles of I^- consumed = 3x = 0.124 moles I^-.

3x = 0.124; x = 0.124 / 3 = 0.0413 moles I-
0.0413moles I^- x (126.9 g I^- / 1 mole I^-) = 5.240 g I^-

The formula AgI tells us that the mole ratio of Ag+ to I^- is 1:1.


0.0413 moles I^- x (1 mole Ag+ / 1 mole I-) = 0.0413 moles Ag+
0.0413 moles Ag+ x (107.86 g Ag / 1 mole Ag) = 4.454 g Ag+

So the mass of AgI = mass of Ag+ + mass of I- = 5.240 + 4.454 = 9.694 g AgI