Question

Too much fluoride ion in drinking water can cause the following undesireable effects: tooth discoloration. The EPA's Secondary Maximum Contaminant Level for fluoride ion in drinking water is 2.00 mg/L.(1) If 1.37×104 liters of water in a storage tank contains 20.8 grams of F-, what is the contaminant level in ppm? ppm Assume that the density of the impure water is 1.00 g/mL.(2) Is this level acceptable based on EPA guidelines?

Answer 1

Let's start with calculating the concentration of the fluoride ion with the provided data:

`\%m/v\text{ F=}\frac{20.8\text{ g}}{1.37*10^4L}=1.5182\text{ *10}^(-3)\text{ g/L}`

To convert to ppm, we have to remember the formula:

`ppm=\frac{mass\text{ of substrate \lparen mg\rparen}}{mass\text{ of the solution \lparen kg\rparen}}`

But, as the dissolvent is water and we can assume that the density is 1 g/ml, the same as 1 kg/L, we can calculate the ppm by dividing the mg of F- into the provided volume of water:

`ppm=\frac{(20.8*1000)\text{ mg F}}{1.37*10^4\text{ L sln}}=1.5182\text{ ppm F}^-\text{ }`

So, the answer is 1.5182 ppm F-.

For the second part, we only have to notice that ppm is the same (in this case) that mg/L. So, it means that the level is acceptable as it is 1.5182 mg/L, which is lower than 2 mg/L.