The volume (in mL) of 0.1 M AgNO₃ required for complete precipitation of chloride ions present in 20 mL of 0.01 M solution of [Cr(H₂ O)₅ Cl] Cl₂ as silver chloride is __________.

Solution

Chemistry ma’am · 4 steps

Only the chlorides outside the square bracket are free ions, so count 2 per formula unit, not 3.

1. Free chloride only. In [Cr(H₂O)₅Cl]Cl₂ one chloride sits inside the bracket, bonded to chromium as a ligand. Ag⁺ cannot pull it out. Only the 2 chlorides outside the bracket are free Cl⁻ ions in solution.
2. Moles of chloride. Moles of complex = 20 mL × 0.01 M = 0.2 mmol. Each formula unit releases 2 Cl⁻, so free chloride = 2 × 0.2 = 0.4 mmol.

   Ag⁺ + Cl⁻ → AgCl v

3. Volume of silver nitrate. Ag⁺ and Cl⁻ react 1 : 1, so 0.4 mmol of AgNO₃ is needed. Volume = (0.4 mmol)/(0.1 M) = 4 mL.
4. So. The volume of 0.1 M AgNO₃ required is 4 mL.

The common slip: 6. Counting all three chlorides, including the one bonded to chromium, gives 0.6 mmol and 6 mL instead of 4.

The idea to keep. Silver nitrate only precipitates the counter-ions outside the coordination sphere. Ligand chlorides stay bound to the metal, so count only what is outside the bracket.

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