Q.2
Calculate the ionic strength of a solution containing mixture of 0.02 M BaClz and 0.1 M HCI
NOTE: This question was officially DROPPED by NTA in the Final Answer Key. Full marks awarded to all candidates.
- ⚠️ Note: This question was officially DROPPED by NTA/ICAR in the final answer key because the mathematically calculated ionic strength ($0.16\text{ M}$) was omitted from the answer options. Full marks were awarded to all candidates.
- Calculation of Ionic Strength ($I$) of Electrolyte Solutions:
- The Lewis-Randall formula for ionic strength is:
- For $0.02\text{ M } \text{BaCl}_2$:
- $\text{Ba}^{2+}$: $c = 0.02\text{ M},\ z = +2 \implies c z^2 = 0.02 \times 4 = 0.08$
- $\text{Cl}^-$: $c = 0.04\text{ M},\ z = -1 \implies c z^2 = 0.04 \times 1 = 0.04$
- Subtotal for $\text{BaCl}_2 = \frac{1}{2}(0.08 + 0.04) = 0.06\text{ M}$
- For $0.1\text{ M } \text{HCl}$:
- $\text{H}^+$: $c = 0.1\text{ M},\ z = +1 \implies c z^2 = 0.1 \times 1 = 0.10$
- $\text{Cl}^-$: $c = 0.1\text{ M},\ z = -1 \implies c z^2 = 0.1 \times 1 = 0.10$
- Subtotal for $\text{HCl} = \frac{1}{2}(0.10 + 0.10) = 0.10\text{ M}$
- Total Ionic Strength:
- Discrepancy Note: The examination options listed 0.80, 0.70, 0.65, and 0.60. Since the correct value $0.16\text{ M}$ was not available, NTA dropped the question.
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