Q. 13.1.3

The U.S. Environmental Protection Agency’s (EPA’s) secondary standards for contaminants that may cause cosmetic or aesthetic effects in drinking water suggest an upper limit of 250 mg/L for chloride ion. If 3.88 × 10$^{4}$ L of water in a storage tank contains 4.14 g of Cl, what is the contaminant level in ppm? Is this level acceptable based on EPA guidelines?

Verified Solution

You are asked to calculate the concentration of a dilute aqueous solution and to compare it with an EPA contaminant concentration.
You are given the mass of solute and solvent in the solution and EPA secondary standard for the contaminant.
Because this is a dilute aqueous solution, we can assume a density of 1 g/mL (ppm 5 mg solute/L solution).

ppm Cl$^{-}=\frac{\left(4.14 \text{ g Cl}^{-}\right)\left(\frac{10^{3}\text{ mg}}{1\text{ g}} \right) }{3.88\times 10^{4}\text{ L solution}}$ = 0.107 ppm

The level of the contaminant is acceptable by EPA standards.