Q127P - Chemistry: Molecular Nature Of Matter And Change

Question

A water treatment plant needs to attain a fluoride concentration of $\mathbf{4}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{-5}\mathbf{\text{ M}}$.

(a) What mass of $\mathbf{NaF}$ must be added to a $\mathbf{5000}\mathbf{\text{ L}}$ blending tank of water?

(b) What mass per day of fluoride is ingested by a person who drinks $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{L}$ of this water?

Step-by-Step Solution

Answer

1. The mass of is $9.45\text{ g}$ needs to attain a fluoride concentration of $4.50\times {10}^{-5} M$.
2. The mass of fluoride is $1.70\text{ g}$ per day ingested by a person who drink $2.0\text{ L}$ of this water.

A concept:

Molarity is defined as the number of moles of solute per liter solution.

$$\begin{gathered} \mathbf{\text{molarity}}\mathbf{=}\frac{\mathbf{\text{moles\hspace{0.17em}of\hspace{0.17em}solute}}}{\mathbf{\text{volume\hspace{0.17em}of\hspace{0.17em}solution\hspace{0.17em}}}\mathbf{\left(}\mathbf{L}\mathbf{\right)}} \\ \mathbf{\text{moles\hspace{0.17em}of\hspace{0.17em}solute}}\mathbf{=}\frac{\mathbf{\text{mass\hspace{0.17em}of\hspace{0.17em}solute}}}{\mathbf{\text{molar\hspace{0.17em}mass\hspace{0.17em}of\hspace{0.17em}solute}}} \end{gathered}$$

Given that, $4.50\times {10}^{-5}M$ is the fluoride molarity or concentration to be obtained and volume of blending tank of water is $5000\text{ L}$.

(a) Mass of NaF :

Molar mass of $\text{NaF}$ is defined as below.

Molarity is defined as the number of moles of solute per liter solution.

 

One mole of fluoride is given by one mole of $\text{NaF}$.

Thus, Fluoride concentration is equivalent to the concentration of  $\text{NaF}$.

 

Hence, the mass of $\text{NaF}$ can be calculated as,

$$\begin{gathered} \text{molarity}=\frac{\text{mass\hspace{0.17em}of\hspace{0.17em}NaF}}{\text{molar\hspace{0.17em}mass\hspace{0.17em}of\hspace{0.17em}NaF}}\times \frac{\text{1}}{\text{volume }\left(L\right)} \\ 4.50\times {10}^{-5}M=\frac{\text{mass\hspace{0.17em}of\hspace{0.17em}NaF}}{\text{41.99}{\displaystyle \raisebox{1ex}{$\text{\hspace{0.17em}g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.}}\times \frac{\text{1}}{\text{5000 L}} \\ \frac{\text{mass\hspace{0.17em}of\hspace{0.17em}NaF}}{\text{41.99}{\displaystyle \raisebox{1ex}{$\text{\hspace{0.17em}g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.}}=4.50\times {10}^{-5}M\times \text{5000 L} \\ \text{mass\hspace{0.17em}of\hspace{0.17em}NaF}=\text{9.45 g} \end{gathered}$$

Therefore, $9.45\text{ g}$ of $\text{NaF}$ needs to attain a fluoride concentration of $4.50\times {10}^{-5}M$.

(b) Mass of fluoride:

Molar mass of fluoride is $\text{18.99 }\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$ and volume is $\text{2.0 L}$.

Fluoride concentration is equal to $\text{NaF}$ concentration.

 

Thus, the mass of fluoride can be calculated as follow.

 $$\begin{gathered} \text{molarity}=\frac{\text{mass\hspace{0.17em}of\hspace{0.17em}F}}{\text{molar\hspace{0.17em}mass\hspace{0.17em}of\hspace{0.17em}F}}\times \frac{\text{1}}{\text{volume }\left(L\right)} \\ 4.50\times {10}^{-5}M=\frac{\text{mass\hspace{0.17em}of\hspace{0.17em}F}}{\text{18.99\hspace{0.17em}}{\displaystyle \raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.}}\times \frac{\text{1}}{\text{2.0 L}} \\ \frac{{\displaystyle \text{mass\hspace{0.17em}of\hspace{0.17em}F}}}{{\displaystyle \text{18.99\hspace{0.17em}}\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.}}=4.50\times {10}^{-5}M\times 2.0\text{ L} \\ \text{mass\hspace{0.17em}of\hspace{0.17em}F}=1.70\text{ g} \end{gathered}$$

 

Hence, the mass of fluoride is $1.70\text{ g}$ per day ingested by a person who drink $2.0\text{ L}$ of this water.