Solution Exercise 22
For column D: Divide the concentration in mg/L by the atomic weight of each ion to convert the concentration from mg/L to molarity (mmol/L).
For column F: Multiply the molarity by the valence charge to calculate the concentration in milliequivalents per liter (meq/L).
| A | B | C | D | E | F |
| Solute | Measured Concentration (mg/L) | Atomic weight(g) | Molarity (mmol/L) | Valence (charge) | meq/L |
| [latex]{Ca}_{}^{2 +}[/latex] | 75 | 40.08 | 1.87 | 2 | 3.74 |
| [latex]{Mg}_{}^{2 +}[/latex] | 25 | 24.31 | 1.03 | 2 | 2.06 |
| [latex]{Na}_{}^{+}[/latex] | 5.6 | 23 | 0.24 | 1 | 0.24 |
| [latex]K_{}^{+}[/latex] | 2.1 | 39.1 | 0.05 | 1 | 0.054 |
| [latex]{Fe(III)}_{}^{}[/latex] | 0.2 | 55.8 | 0.004 | 3 | 0.011 |
| [latex]{HCO}_{3}^{-}[/latex] | 109 | 61 | 1.79 | 1 | 1.79 |
| [latex]{SO}_{4}^{2 -}[/latex] | 242 | 96 | 2.52 | 2 | 5.04 |
| [latex]{Cl}_{}^{-}[/latex] | 9.3 | 35.5 | 0.26 | 1 | 0.26 |
| [latex]{NO}_{3}^{-}[/latex] | 15 | 62 | 0.24 | 1 | 0.24 |
The sum of the cation milliequivalents (i.e., positively charged ions) is 6.11 meq/L. The sum of the anion milliequivalents (i.e., negatively charged ions) is 7.33 meq/L. The charge balance is calculated as:
[latex]C.B. = \frac{\left( \Sigma_{cations} - \ \Sigma_{anions} \right)}{\left( \Sigma_{cations} + \ \Sigma_{anions} \right)}(100)[/latex]
[latex]\frac{\left( 6.11\frac{meq}{L} - 7.33\frac{meq}{L} \right)}{\left( 6.11\frac{meq}{L} + 7.33\frac{meq}{L} \right)}(100) = - 0.091(100) = - 9.1\%[/latex]
The absolute value of the charge balance is greater than 5%, which means that the charge balance is unacceptable. The calculated value is negative, which means that there are excess anions (or insufficient cations) represented in the analytical results.