The volume (in mL) of 0.125 M AgNO₃ required to quantitatively precipitate chloride ions in 0.3 g of [Co(NH₃)₆]Cl₃ is ________. (Round off to the nearest integer)
M [Co(NH₃)₆Cl₃] = 267.46 g/mol
M AgNO₃ = 169.87 g/mol

Solution

Chemistry ma’am · 3 steps

All three chlorides sit outside the bracket, so each mole of complex frees three Cl⁻.

1. Free chloride ions. Six NH₃ ligands already fill the coordination sphere of cobalt. So all three Cl⁻ are counter ions, and in water the complex gives [Co(NH₃)₆]³⁺ + 3Cl⁻. Each free Cl⁻ reacts with one Ag⁺ to form AgCl.
2. Moles of chloride. Moles of complex = 0.3/267.46 = 1.122 × 10⁻³ mol. Moles of Cl⁻ = 3 × 1.122 × 10⁻³ = 3.365 × 10⁻³ mol. This is also the number of moles of AgNO₃ needed.

   V = n/M = (3.365 × 10⁻³ mol)/(0.125 mol L⁻¹) = 0.02692 L

3. So. V = 26.92 mL, which rounds to 27. The molar mass of AgNO₃ is not needed, because the silver nitrate is given as a molarity.

The common slip: 9. Forgetting that each formula unit releases three Cl⁻ gives about 9 mL, one third of the correct volume.

The idea to keep. For precipitation with AgNO₃, count only the ionizable chlorides, the ones outside the square bracket. Then moles of Ag⁺ equal moles of free Cl⁻.

Answer from the NTA final answer key. Solution by PrepWiser.