Werner's Theory: AgNO3 Precipitation Calculations
One of the most frequently asked question formats in the Coordination Compounds chapter combines basic stoichiometry with Werner's Theory. These questions test whether you understand the physical difference between the coordination sphere and the ionization sphere.
The Question (JEE 2023): The volume (in mL) of 0.1M AgNO3 required for complete precipitation of chloride ions present in 20 mL of a 0.01 M solution of [Cr(H2O)5Cl]Cl2 as silver chloride is...
Let's break down the logic exactly as an examiner would expect you to solve it.
Video Tutorial: Solving the Precipitation Puzzle
Watch Abhishek Sengar sir from CHEMCA expertly decode the complex's formula, calculate the ionizable molarity, and apply the classic M1V1 = M2V2 rule.
Step 1: Identifying Ionizable Chlorides (Werner's Theory)
Before doing any math, we must inspect the given complex: [Cr(H2O)5Cl]Cl2.
- Anything inside the square brackets `[ ]` is part of the coordination sphere. These are non-ionizable (Secondary Valency). The single Cl inside the bracket is trapped and will NOT react with Silver Nitrate.
- Anything outside the square brackets is part of the ionization sphere (Primary Valency). The two Cl atoms outside the bracket will freely ionize in water and form a precipitate!
When this complex dissolves in water, it ionizes like this:
Step 2: Calculating Molarity of Cl-
Because 1 molecule of the complex releases exactly 2 freely roaming Chloride ions, the molarity of the Chloride ions in the solution will be double the molarity of the complex itself.
- Molarity of Complex = 0.01 M
- Number of ionizable Cl- = 2
- Molarity of Cl- (MCl) = 2 × 0.01 = 0.02 M
- Volume of Cl- solution (VCl) = 20 mL
Fig: Only the counter ions residing outside the square brackets will react with the Silver Nitrate to form the white AgCl precipitate.
Step 3: The Stoichiometric Calculation
The precipitation reaction between Silver ions (Ag+) and Chloride ions (Cl-) is a simple 1:1 ratio:
Because it is a 1:1 ratio, the number of moles of Ag+ required must exactly equal the number of moles of Cl- present in the beaker.
Moles of Ag+ = Moles of Cl-
Since Moles = Molarity × Volume, we can set up the classic equation:
Now, we simply plug in all our known values to solve for the missing volume (VAg).
- MAg = 0.1 M (Given in the question)
- VAg = ? (This is what we need to find)
- MCl = 0.02 M (Calculated in Step 2)
- VCl = 20 mL (Given in the question)
0.1 × VAg = 0.02 × 20
0.1 × VAg = 0.40
VAg = 0.40 / 0.1
VAg = 4 mL
Final Answer: 4 mL of the 0.1M AgNO3 solution is required for complete precipitation!
Practice Questions for JEE & NEET
Let's test your ability to apply Werner's Theory to variations of this classic problem!
Question 1: Using the exact same concentrations and volumes as the problem above (20 mL of 0.01M complex, titrated with 0.1M AgNO3), what would the required volume of AgNO3 be if the complex was [Cr(H2O)6]Cl3?
Answer: 6 mL
Reasoning:
1. Look at the formula: [Cr(H2O)6]Cl3. There are 3 ionizable chloride ions outside the bracket.
2. Molarity of Cl- = 3 × Molarity of Complex = 3 × 0.01 M = 0.03 M.
3. Apply the formula: MAg × VAg = MCl × VCl
4. 0.1 × VAg = 0.03 × 20
5. 0.1 × VAg = 0.60
6. VAg = 0.60 / 0.1 = 6 mL.
Question 2: According to Werner's Theory, why doesn't the Chlorine atom residing inside the square bracket precipitate with Silver Nitrate?
Answer: Because it satisfies the Secondary Valency and forms a non-ionizable Coordinate Covalent bond.
Reasoning:
Werner proposed two types of valencies. The Primary Valency (outer sphere) is ionizable and represents the oxidation state. The Secondary Valency (inner coordination sphere) represents the coordination number.
Ligands inside the coordination sphere are bound directly to the central metal atom via strong, directional Coordinate Covalent Bonds. These bonds do not break apart when dissolved in water, meaning the inner Chlorine is "trapped" and completely unavailable to react with the Ag+ ions!
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