Expert: Steve Holleran - 8/3/2008

Question
The fluoridation of city water supplies has been practiced in the United States for several decades. It is done by continuously adding sodium fluoride to water as it comes from a reservoir.



Assume you live in a medium-sized city of 150,000 people and that 660 L (170 gal) of water is consumed per person per day.



What mass of sodium fluoride (in kilograms) must be added to the water supply each year (365 days) to have the required fluoride concentration of 1 ppm (part per million)--that is, 1 kilogram of fluoride is per 1 million kilograms of water?



(Sodium fluoride is 45.0% fluoride, and water has a density of 1.00g/cm^3).



Please answer this question if you know how and please show or explain how you got your answer. I tried to answer this but I think my answer is so wrong. Thank you so much! I really appreciate it!


Answer
Hi Bri,

Okay, this is not my specialty, but let's see if this makes sense:

150,000 people * 660 L / day = 99,000,000 L / day

99,000,000 * 365 days/ yr = 3.6135 x 10^10 L / year.

Since 1 L water = 1 Kg water, then

3.6135 x 10^10 L = 3.6135 x 10^10 Kg

Since you want 1 Kg flouride per 10^6 Kg water (and 1L = 1 Kg water),

3.6135 x 10^10 / 10^6 = 36,165 Kg flouride needed

Since sodium fluoride is 45% fluoride, you need

      36,165 / .45 = 80,300 Kg sodium fluoride.


The conversions of weight to volume bother me sometimes, but since 1 liter water = 1 kilogram water, that makes it easier here.

Hope this helps,
Steve