Chapter 1, Problem 115AP

Interpretation Introduction

Interpretation:

For a city of 50,000 people, the quantity of sodium fluoride that would be needed per yeartoadd in drinking water and the percentage ofsodium fluoride wastedto be calculated.

Concept Introduction:

Conversion factor is a fraction that is used to convert one unit to another. Use of more than one factor to find a solution is called dimensional analysis.

$\text{1 gal}=\text{3}\text{.79 L}$ the conversion factorsfor conversion of gal to liters $\left(\frac{\text{3}\text{.79L}}{\text{1gal}}\right)$

$\text{1 L}=\text{1000 mL}$ the conversion factorsfor conversion of gal to liters $\left(\frac{\text{1000\#x2009;mL}}{\text{1\#x2009;L}}\right)$,

$\text{1 kg}=\text{1000 g}$ the conversion factors for conversion of gal to liters $\left(\frac{\text{1000 g}}{\text{1 kg}}\right)$,

Answer to Problem 115AP

Solution: $2.3\times {10}^4\text{ kg NaF}$, $99\%$

Explanation of Solution

Given information:

The number of people in the city is 50,000.

The daily consumption of water per person is $\text{150 gal}$.

The concentration of fluorine in water is $1\text{ ppm}$, or $1\text{ g}$, in 196 of water.

Sodium fluoride contains 45.0% fluorine by mass.

Water used for drinking and cooking per day is 6 L per person.

There are 50,000 people in the city, and daily consumption of water, per person, is $\text{150 gal}$.

Multiply 50,000 with $\frac{\text{150}\text{gal}}{\text{1}}$ to find the total daily consumption of water.

$50,000\times \frac{\text{150 gal}}{\text{1}}=75,00,000\text{ gal}$

The concentration of fluorine in water is $1\text{ ppm}$, or $1\text{ g}$, in 10⁶g of water. It can be expressed as:

$\text{1g\#x2009;F}={10}^6{\text{g\#x2009;H}}_{\text{2}}\text{O}$

The conversion factor to calculate the amount of fluorine needed is $\left(\frac{\text{1g\#x2009;F}}{{10}^6{\text{g\#x2009;H}}_{\text{2}}\text{O}}\right).\; Convert\; the\; calculated\; daily\; water\; requirement,\; in\; gallons,\; by\; the\; conversion\; factors$ \left(\frac{\text{3}\text{.79}\text{L}}{\text{1}\text{gal}}\right)$,$ \left(\frac{\text{1000\#x2009;mL}}{\text{1\#x2009;L}}\right)$,\; and$ \left|\frac{\text{1}\text{.0\#x2009;g}}{\text{1\#x2009;mL}}\right|$to\; grams.$

$75,00,000\cancel{\text{gal}}\times \frac{\text{3}\text{.79}\cancel{\text{L}}}{\text{1}\cancel{\text{gal}}}\times \frac{\text{1000\#x2009;}\cancel{\text{mL}}}{\text{1\#x2009;}\cancel{\text{L}}}\times \frac{\text{1}\text{.0\#x2009;g}}{\text{1\#x2009;}\cancel{\text{mL}}}=284.25\times {10}^8\text{g}$

Multiply the amount of water, in grams, by the conversion factor $\left(\frac{\text{1g\#x2009;F}}{{10}^6{\text{g\#x2009;H}}_{\text{2}}\text{O}}\right)$.

$\left(284.25\times {10}^8\cancel{{\text{g\#x2009;H}}_{\text{2}}\text{O}}\right)\left(\frac{\text{1g\#x2009;F}}{{10}^6\cancel{{\text{g\#x2009;H}}_{\text{2}}\text{O}}}\right)=284.25\times {10}^2\text{g\#x2009;F}$

Sodium fluoride contains 45.0% fluorine by mass. It can be expressed as:

$\text{100 g NaF}=\text{45 g F}$

Multiply the amount of fluorine by theconversion factor $\left(\frac{\text{100\#x2009;g\#x2009;NaF}}{\text{45\#x2009;g\#x2009;F}}\right)$ to get the amount of sodium fluoride needed.

$\left(284.25\times {10}^2\cancel{\text{g\#x2009;F}}\right)\left(\frac{\text{100\#x2009;g\#x2009;NaF}}{\text{45\#x2009;}\cancel{\text{g\#x2009;F}}}\right)=6.32\times {10}^2\text{g\#x2009;NaF}$

Convert the amount of sodium fluoride to kilograms using theconversion factor $\left(\frac{\text{1\#x2009;kg}}{\text{1000\#x2009;g}}\right)$.

$\left(6.32\times {10}^2\cancel{\text{g }}\text{NaF}\right)\left(\frac{\text{1 kg}}{\text{1000 }\cancel{\text{g}}}\right)=0.632\text{ kg}$

The amount of sodium fluoride required per day is 0.632 kg/day.

$\text{1yr}=\text{365 day}$

Multiply 0.632 kg/day with theconversion factor $\frac{\text{365\#x2009;day}}{\text{1yr}}$ to get the amount of sodium fluoride required per year.

$\frac{0.632\text{kg}}{\text{1 }\cancel{\text{day}}}\times \frac{\text{365 }\cancel{\text{day}}}{\text{1yr}}=2.3\times {10}^4\text{kg}/\text{yr}$

Hence, $2.3\times {10}^4\text{kg}/\text{yr}$ of sodium fluoride is needed per year.

If the daily usage of water for drinking and cooking is 6 L per person, then, for 50,000 people, the water used is as:

$50,000\times 6\text{L}=3\times {10}^5\text{L}$.

Convert this to gallons.

$\left(3\times {10}^5\cancel{\text{L}}\right)\left(\frac{\text{1gal}}{\text{3}\text{.79}\cancel{\text{L}}}\right)=0.79\times {10}^5\text{gal}$

The amount of sodium fluoride required for $75,00,000\text{ gal}$ of water is 9.632 kg/day.

Thus, for $0.79\times {10}^5\text{gal}$ of water, the required amount of sodium fluoride is as: $\frac{0.632\text{kg}}{75,00,000\cancel{\text{gal}}}\times 0.79\times {10}^5\cancel{\text{gal}}=\text{0}{\text{.6657×10}}^{-\text{2}}\text{kg}$

Thus, the amount of sodium fluoride wasted is calculated as:

$\left(0.632-\text{0}{\text{.6657×10}}^{-\text{2}}\right)\text{kg}=\text{0}\text{.625 kg }$

Hence, the percentage wastage is as:

$\begin{array}{c}\frac{\text{0}\text{.625}\cancel{\text{kg}}}{\text{0}\text{.632 }\cancel{\text{kg}}}\times 100=98.9\%\\ \simeq 99\%\end{array}$

Hence, $2.3\times {10}^4\text{kg}/\text{yr}$ of sodium fluoride is needed per yearand the percentage of sodium fluoride wasted is $99\%$.