Question

The molar concentrations (M, i.e. mol/L) of some ionic species in a water sample were estimated as follows: \[ \text{Na}^+ = 0.25 \, \text{M}; \, \text{Ca}^{2+} = 0.12 \, \text{M}; \, \text{Cl}^- = 0.32 \, \text{M}; \, \text{HCO}_3^- = 0.05 \, \text{M}. \] The ionic strength of this water sample is _______ M (correct up to two decimal places).

Hint:

The ionic strength depends on the concentration and charge of each ion in the solution.

Answer 1

Correct Answer: 0.5

The ionic strength \( I \) of a solution is given by: \[ I = \frac{1}{2} \sum_i c_i z_i^2, \] where \( c_i \) is the concentration of the \( i \)-th ion and \( z_i \) is the charge of the \( i \)-th ion. Given:
- \( \text{Na}^+ \) has a concentration of 0.25 M and a charge of 1,
- \( \text{Ca}^{2+} \) has a concentration of 0.12 M and a charge of 2,
- \( \text{Cl}^- \) has a concentration of 0.32 M and a charge of -1,
- \( \text{HCO}_3^- \) has a concentration of 0.05 M and a charge of -1.
Substitute the values: \[ I = \frac{1}{2} \left( 0.25 \times 1^2 + 0.12 \times 2^2 + 0.32 \times 1^2 + 0.05 \times 1^2 \right) = \frac{1}{2} \left( 0.25 + 0.48 + 0.32 + 0.05 \right) = \frac{1}{2} \times 1.10 = 0.55 \, \text{M}. \] Thus, the ionic strength of the water sample is \( 0.55 \, \text{M} \).