Percentage Dissociation in Isotonic Solutions: K2SO4 and Glucose
In the Solutions chapter, JEE and NEET examiners love to test two concepts simultaneously. A classic example is combining Osmotic Pressure with the Van't Hoff Factor to find the percentage dissociation of a salt.
Let's break down a highly repetitive Previous Year Question (PYQ) step-by-step.
Video Tutorial: Solving the Isotonic Numericals
Watch Abhishek Sengar sir from CHEMCA demonstrate the mathematical workflow to effortlessly arrive at the final percentage.
Step-by-Step Problem Breakdown
Problem Statement:
A 0.004 M K2SO4 solution is isotonic with a 0.01 M Glucose solution. Find the percentage dissociation of K2SO4.
Two solutions are isotonic if they exert the exact same Osmotic Pressure (π) across a semi-permeable membrane at the same temperature. Therefore:
π1 = π2
-
Set up the Osmotic Pressure Equation:
The formula for osmotic pressure is π = iCRT. (Where i is the Van't Hoff factor, C is concentration, R is the gas constant, and T is temperature).
π(K2SO4) = π(Glucose)Since R and T are constants, they cancel out, leaving us with:
i1 · C1 · RT = i2 · C2 · RT
i1 · C1 = i2 · C2 -
Find the Van't Hoff factor (i) for K2SO4:
Glucose is a non-electrolyte. It does not dissociate in water, so its Van't Hoff factor (i2) is exactly 1. Plug in the given concentrations:
i1 × 0.004 = 1 × 0.01
i1 = 0.01 / 0.004 = 10 / 4 = 2.5 -
Ionize K2SO4 to find 'n':
To use the dissociation formula, we need to know how many maximum ions one molecule of Potassium Sulfate can produce (n).
K2SO4 → 2 K+ + SO42-It produces two K+ ions and one SO42- ion. Total ions: n = 3. -
Calculate the Degree of Dissociation (α):
We apply the master formula for dissociation: α = (i - 1) / (n - 1)
α = (2.5 - 1) / (3 - 1)
α = 1.5 / 2 = 0.75
Fig: Equating the effective concentration (i × C) of two Isotonic solutions.
Practice Questions for JEE & NEET
Verify your grasp on isotonic solutions and osmotic pressure with these conceptual checks.
Question 1: A 0.1 M solution of Sodium Chloride (NaCl) is found to be isotonic with a 0.2 M solution of Urea at the same temperature. Assuming NaCl undergoes complete dissociation, is this data experimentally correct? Show mathematically.
Answer: Yes, the data is correct.
Reasoning:
- For solutions to be isotonic, their effective concentrations (i \cdot C) must be equal.
- For Urea: It is a non-electrolyte, so i = 1.
Effective concentration = 1 \times 0.2 = 0.2 \text{ M}. - For NaCl: It undergoes complete dissociation into Na+ and Cl-, so i = n = 2.
Effective concentration = 2 \times 0.1 = 0.2 \text{ M}. - Since 0.2 \text{ M} = 0.2 \text{ M}, their osmotic pressures are equal, proving they are isotonic.
Question 2: What happens to a human red blood cell (RBC) when it is placed into a solution that is Hypertonic compared to the intracellular fluid?
Answer: The cell will shrink (Plasmolysis / Crenation).
Reasoning:
A "Hypertonic" solution has a higher solute concentration (and thus higher osmotic pressure) than the fluid inside the red blood cell. Due to Osmosis, solvent (water) always flows through a semi-permeable membrane from the region of lower solute concentration to higher solute concentration. Thus, water will flow out of the RBC, causing it to shrivel and collapse.
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