Question

Determine how many grams of resin are needed to remove the calcium ions from 200 mL of water that has 15.4 grains/gallon of hardness, if the resin exchange capacity is 2.78 millimoles of divalent cation per gram of resin.

Answer 1

Task

Determine how many grams of resin are needed to remove the calcium ions from 200 mL of water that has 15.4 grains/gallon of hardness, if the resin exchange capacity is 2.78 millimoles of divalent cation per gram of resin.

Given

Water volume: $V = 200 \, \text{mL} = 0.2 \, \text{L}$

Water hardness: $H = 15.4$ grains/gallon

Cation exchange capacity of resin: $CEC = 2.78 \, \text{mmol} \, (\text{Ca}^{2+})/\text{g}$

Mass of resin: $m = ?$

Solution

1. Conversion of hardness from grains/gallon to mg/L (1 grains/gallon = 17.118061 mg/L):

$$H = 15.4 \cdot 17.118061 = 263.6181 \, \text{mg/L}$$

2. Calculation of mass of calcium carbonate in given volume of water:

$$m(\text{CaCO}_3) = H \cdot V = 263.6181 \cdot 0.2 = 52.7236 \, \text{mg} = 0.052736 \, \text{g}$$

3. Calculation of number of moles of $\text{Ca}^{2+}$ in given volume of water (molar weight of calcium carbonate $M(\text{CaCO}_3) = 100.0869 \, \text{g/mol}$):

$$n(\text{Ca}^{2+}) = n(\text{CaCO}_3) = m(\text{CaCO}_3)/M(\text{CaCO}_3) = 0.052736 / 100.0869 = 0.0005268 \, \text{mol} = 0.5269 \, \text{mmol}$$

4. Calculation of needed mass of the resin:

$$m = n(\text{Ca}^{2+})/\text{CEC} = 0.5269 / 2.78 = 0.1895 \, \text{g}$$

Answer: 0.1895 g