The volume ( in ml) of 0.1M AgNO₃ required for complete precipitation of chloride Ions present in 20ml of 0.01 M solution of [Cr(H₂O)₅ Cl]Cl₂ as silver chloride is ___

The complex [Cr(H₂O)₅Cl]Cl₂ contains:

• One chloride ion bound to chromium (coordinate bond) - NOT ionizable
• Two chloride ions outside the coordination sphere - ionizable

Only the two ionizable chloride ions will precipitate with AgNO₃.

Chemical Equation:

[Cr(H₂O)₅Cl]Cl₂ + 2AgNO₃ → [Cr(H₂O)₅Cl]⁺ + 2AgCl↓ + 2NO₃^-

Step 1: Calculate millimoles of Cl^- ions

Millimoles of Cl^- = Molarity × Volume (mL) × Number of ionizable Cl^-

= 0.01 M × 20 mL × 2 = 0.4 mmol

Step 2: Calculate millimoles of AgNO₃ required

From stoichiometry: 1 mole AgNO₃ precipitates 1 mole Cl^-

Millimoles of AgNO₃ required = 0.4 mmol

Step 3: Calculate volume of AgNO₃ solution

Volume = Millimoles required / Molarity

Volume = 0.4 mmol / 0.1 M = 4 mL

Final Answer:

The volume of 0.1 M AgNO₃ required for complete precipitation of chloride ions is 4 mL.