Question

What quantity of AgCl is precipitated when one mole of Pentaamminecarbonatocobalt(III) chloride is treated with excess \(\text{AgNO}_{3}\) (molar mass \(=143.3\) g \(\text{mol}^{-1}\)) solution?

• A. 143.3 g
• B. 72.0 g
• C. 286.6 g
• D. 215.0 g

Hint:

Only the chloride outside the coordination sphere precipitates with silver nitrate.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Only ionisable chloride, which lies outside the coordination sphere, reacts with \(\text{AgNO}_{3}\) to give AgCl.
Ligands inside the sphere do not precipitate.

Step 2: Write the formula
Pentaamminecarbonatocobalt(III) chloride is \([\text{Co}(\text{NH}_{3})_{5}\text{CO}_{3}]\text{Cl}\).
Co is +3 and carbonate is 2-, so the complex ion has charge +1.

Step 3: Count the ionic chloride
The complex ion \([\text{Co}(\text{NH}_{3})_{5}\text{CO}_{3}]^{+}\) is balanced by one \(\text{Cl}^{-}\).
So one mole of complex gives one mole of AgCl.

Step 4: Calculation
Mass of AgCl = 1 mol \(\times\) 143.3 g \(\text{mol}^{-1}\) = 143.3 g

Final Answer:
143.3 g of AgCl is precipitated. \[ \boxed{\text{(A) }\text{143.3 g}} \]