The volume (in mL) of 0.125 M required to quantitatively precipitate chloride ions in 0.3 g of is .

( = 267.46 g/mol, = 169.87 g/mol)

Question

The volume (in mL) of 0.125 M  required to quantitatively precipitate chloride ions in 0.3 g of  is .

( = 267.46 g/mol,  = 169.87 g/mol)

Canvas Classes · Accepted Answer

Step 1: Identify the Complex and Its Chloride Content

The compound is  — hexaamminecobalt(III) chloride.
- Molar mass = 267.46 g/mol (given)
- Formula shows: 3 chloride ions per formula unit (all three Cl are ionic, outside the coordination sphere)

Step 2: Calculate Moles of Complex

Step 3: Calculate Moles of Cl⁻

Each formula unit gives 3  ions (all are ionic, precipitable by ):

Step 4: Calculate Volume of AgNO₃ Required

The reaction:  (1:1 stoichiometry)

For 0.125 M :

Answer: 26.9

More questions from Practical Organic Chemistry