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Chapter 2: Q143CP (page 87)

Fluoride ion is poisonous in relatively low amounts: 0.2 g of F- per 70 kg of body weight can cause death. Nevertheless, in order to prevent tooth decay, F- ions are added to drinking water at a concentration of 1 mg of F- ion per L of water. How many liters of fluoridated drinking water would a 70-kg person have to consume in one day to reach this toxic level? How many kilograms of sodium fluoride would be needed to treat a 8.50脳107 -gal reservoir?

Short Answer

Expert verified

(a) The person need to drink 200 L of water to reach toxic level.

(b) There is a need of 3.2172脳102Kgof fluoride for 8.50脳107galof water.

Step by step solution

01

Calculating the toxic level

The concentration and toxic amount can be 1 mg/L and 0.2 g respectively.

V(water,toxic)=toxicamountconcentration=0.2g10鈥�3g/L=200L

02

Calculating the amount of sodium fluoride would be needed to treat a 8.50×107 -gal reservoir

It can be calculated as,

1gallon=3.785litersV=8.50脳107gal=8.50脳107gal脳3.785liters1gallon=3.2172脳108Lm=volume脳density=3.2172脳108L脳1mg/L=3.2172脳108mgm=3.2172脳102kg

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