Question

The ionic strength of a solution which is 0.1 molal in NaCl and 0.01 molal in calcium chloride is (assuming complete dissociation)

• A. 0.13 molal
• B. 0.26 molal
• C. 0.11 molal
• D. 0.056 molal

Hint:

To calculate ionic strength, identify all ions in the solution along with their concentrations and charges

Answer 1

The Correct Option is A

The ionic strength (\( I \)) of a solution is a measure of the total concentration of ions in that solution. It is given by the formula: \[ I = \frac{1}{2} \sum_i c_i z_i^2 \] where:

• \( c_i \): concentration of the \( i^{\text{th}} \) ion (in molal units)
• \( z_i \): charge of the \( i^{\text{th}} \) ion

For the given solution, we have two solutes: NaCl and CaCl\(_2\). Assuming complete dissociation, we can determine the concentration of each ion in the solution.

1. Dissociation of Compounds:
   NaCl → Na\(^+\) + Cl\(^-\)
   CaCl\(_2\) → Ca\(^{2+}\) + 2Cl\(^-\)
2. Determining Ion Concentrations:
   
   • From 0.1 molal NaCl:
     Na\(^+\) concentration = 0.1 molal
     Cl\(^-\) concentration from NaCl = 0.1 molal
   • From 0.01 molal CaCl\(_2\):
     Ca\(^{2+}\) concentration = 0.01 molal
     Cl\(^-\) concentration from CaCl\(_2\) = \( 2 \times 0.01 = 0.02 \, \text{molal} \)
3. Total Ion Concentrations:
   Na\(^+\): 0.1 molal
   Cl\(^-\): \( 0.1 + 0.02 = 0.12 \, \text{molal} \)
   Ca\(^{2+}\): 0.01 molal
4. Calculating Ionic Strength:
   \[ I = \frac{1}{2} \left[ (0.1)(+1)^2 + (0.12)(-1)^2 + (0.01)(+2)^2 \right] \] \[ I = \frac{1}{2} \left[ 0.1 \times 1 + 0.12 \times 1 + 0.01 \times 4 \right] \] \[ I = \frac{1}{2} \left[ 0.1 + 0.12 + 0.04 \right] = \frac{1}{2} \times 0.26 = 0.13 \, \text{molal} \]

Therefore, the ionic strength of the solution is 0.13 molal.