Question

The use of silica to form slag in the production of phosphorus from phosphate rock was introduced by Robert Boyle more than years ago. When fluorapatite $\mathbf{[}{\mathbf{Ca}}_{\mathbf{5}}{\mathbf{\left(}{\mathbf{PO}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{3}}\mathbf{}\mathbf{F}\mathbf{]}$is used in phosphorus production, most of the fluorine atoms appear in the slag, but some end up in toxic and corrosive ${\mathbf{SiF}}_{\mathbf{4}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$.

(a) If by mass of the fluorine in $\mathbf{100}\mathbf{.}\mathbf{kg}$of ${\mathbf{Ca}}_{\mathbf{5}}{\mathbf{\left(}{\mathbf{PO}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{3}}\mathbf{F}$forms ${\mathbf{SiF}}_{\mathbf{4}}$, what volume of this gas is collected at $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{atm}$and the industrial furnace temperature of $\mathbf{1450}\mathbf{°}\mathbf{C}$?

(b) In some facilities, the ${\mathbf{SiF}}_{\mathbf{4}}$ is used to produce sodium hexafluoro silicate $\left({\mathrm{Na}}_2{\mathrm{SiF}}_6\right)$which is sold for water fluoridation:

$\mathbf{2}{\mathbf{SiF}}_{\mathbf{4}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{N}{\mathbf{a}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{\to }{\mathbf{Na}}_{\mathbf{2}}{\mathbf{SiF}}_{\mathbf{6}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{Si}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{C}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{HF}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$

How many cubic meters of drinking water can be fluoridated to a level of $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{ppm}$

of ${\mathbf{F}}^{\mathbf{-}}$

using the

${\mathbf{SiF}}_{\mathbf{4}}$

produced in part (a)?

Short answer

(a) The value is $105\times {10}^3\mathrm{L}$

(b) The value is$424.0\times {10}^3 {\mathrm{m}}^3$.

Step-by-step solution

Volume

A substance's volume is the amount of space it takes up, whereas its mass is the amount of stuff it contains. The density of a sample is the amount of mass per unit volume.

Subpart (a)

First, we must determine the molar mass of $\left[{\mathrm{Ca}}_5{\left({\mathrm{PO}}_4\right)}_3\mathrm{F}\right]$by multiplying the atomic masses of the elements listed in the periodic table.

role="math" localid="1663615924375" $\begin{array}{c}\mathrm{M}\left(\left[{\mathrm{Ca}}_5{\left({\mathrm{PO}}_4\right)}_3\mathrm{F}\right]\right)=5\mathrm{A}\left(\mathrm{Ca}\right)+3\mathrm{A}\left(\mathrm{P}\right)+12\mathrm{A}\left(\mathrm{O}\right)+\mathrm{A}\left(\mathrm{F}\right)\\ \mathrm{M}\left(\left[{\mathrm{Ca}}_5{\left({\mathrm{PO}}_4\right)}_3\mathrm{F}\right]\right)=(5\times 40,08+3\times 30,974+12\times 15.999+18.998)\mathrm{gmo}{\mathrm{l}}^{-1}\\ \mathrm{M}\left(\left[{\mathrm{Ca}}_5{\left({\mathrm{PO}}_4\right)}_3\mathrm{F}\right]\right)=504,308{\mathrm{gmol}}^{-1}\end{array}$

The amount of $\left[{\mathrm{Ca}}_5{\left({\mathrm{PO}}_4\right)}_3\mathrm{F}\right]$from which we may compute the amount of fluorine is then calculated. We must also take into account the fact that only $15\%$of fluorine by mass generates ${\mathrm{SiF}}_4$.

$\begin{array}{c}\mathrm{m}{\left(\left[{\mathrm{Ca}}_5{\left({\mathrm{PO}}_4\right)}_3\mathrm{F}\right]\right)}_{15\%}=\frac{15\%}{100\%}\times 100\times {10}^3\mathrm{g}\\ \mathrm{m}{\left(\left[{\mathrm{Ca}}_5{\left({\mathrm{PO}}_4\right)}_3\mathrm{F}\right]\right)}_{15\%}=15\times {10}^3\mathrm{g}\\ \mathrm{n}\left(\left[{\mathrm{Ca}}_5{\left({\mathrm{PO}}_4\right)}_3\mathrm{F}\right]\right)=\frac{\mathrm{m}\left(\left[{\mathrm{Ca}}_5{\left({\mathrm{PO}}_4\right)}_3\mathrm{F}\right]\right)}{\mathrm{M}\left(\left[{\mathrm{Ca}}_5{\left({\mathrm{PO}}_4\right)}_3\mathrm{F}\right]\right)}\\ \mathrm{n}\left(\left[{\mathrm{Ca}}_5{\left({\mathrm{PO}}_4\right)}_3\mathrm{F}\right]\right)=\frac{15\times {10}^3\mathrm{g}}{504.308{\mathrm{gmol}}^{-1}}\\ \mathrm{n}\left(\left[{\mathrm{Ca}}_5{\left({\mathrm{PO}}_4\right)}_3\mathrm{F}\right]\right)=29.76\mathrm{mol}\end{array}$

$\begin{array}{c}\mathrm{n}\left(\mathrm{F}\right)=\mathrm{n}\left(\left[{\mathrm{Ca}}_5{\left({\mathrm{PO}}_4\right)}_3\mathrm{F}\right]\right)\\ =29.76\mathrm{mol}\end{array}$

Evaluating the values

We need four atoms of fluorine to make one molecule of ${\mathrm{SiF}}_4$thus we can compute the amount of ${\mathrm{SiF}}_4$by dividing the amount of fluorine by $4$.

role="math" localid="1663616607318" $\begin{array}{c}\mathrm{n}\left({\mathrm{SiF}}_4\right)=\frac{\mathrm{n}\left(\mathrm{F}\right)}{4}\\ \mathrm{n}\left({\mathrm{SiF}}_4\right)=\frac{29,74\mathrm{mol}}{4}\\ \mathrm{n}\left({\mathrm{SiF}}_4\right)=743.5\mathrm{mol}\end{array}$

We must apply the ideal gas law to compute the volume of the ${\mathrm{SiF}}_4\left(\mathrm{g}\right)$:

$\begin{array}{c}\mathrm{pV}=\mathrm{nRT}\\ \mathrm{p}=\mathrm{pressure}\left(\mathrm{Pa}\right)\\ \mathrm{V}=\mathrm{volume}\left({\mathrm{m}}^3\right)\\ \mathrm{n}=\mathrm{amountofsubstance}\left(\mathrm{mol}\right)\\ \mathrm{R}=\mathrm{gasconstant}\left(\left(\mathrm{Pa}\times {\mathrm{m}}^3\times {\mathrm{K}}^{-1}\times {\mathrm{mol}}^{-1}\right.\right)\\ \mathrm{T}=\mathrm{temperature}\left(\mathrm{K}\right)\end{array}$

We must first compute pressure in Pascals (Pa) and temperature in Kelvins (K) before determining the volume:

$\begin{array}{c}\mathrm{p}=1.00\mathrm{atm}\times 101325\frac{\mathrm{Pa}}{\mathrm{atm}}\\ \mathrm{p}=101325\mathrm{Pa}\\ \mathrm{T}=(1450+273.15)\mathrm{K}\\ \mathrm{T}=1723.15\mathrm{K}\end{array}$

We may get the volume of the gas by multiplying the equation with pressure:

$\begin{array}{c}\mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{V}}\\ \mathrm{V}=\frac{743.5\mathrm{mol}\times 8.314\mathrm{Pa}\times {\mathrm{m}}^3\times {\mathrm{K}}^{-1}\times {\mathrm{mol}}^{-1}\times 1723.15\mathrm{K}}{101325\mathrm{Pa}}\\ \mathrm{V}=105{\mathrm{m}}^3\\ =105\times {10}^3\mathrm{L}\end{array}$

Therefore, the value is$105\times {10}^3\mathrm{L}$.

Subpart (b)

The mass concentration $\left({\mathrm{\gamma }}_{\mathrm{i}}\right)$is calculated by dividing the mass of the ingredient $\left({\mathrm{m}}_{\mathrm{i}}\right)$by the volume of the mixture $\left(\mathrm{V}\right)$. The unit of mass concentration is ppm, which equals $\frac{\mathrm{mg}}{\mathrm{L}}$.

$\left({\mathrm{\gamma }}_{\mathrm{i}}\right)=\frac{m}{v}$

To determine the volume of water that can be fluoridated, we must first determine the quantity of $\mathrm{HF}\left({\mathrm{F}}^{-}\right)$produced in the following reaction:

$2{\mathrm{SiF}}_4\left(\mathrm{g}\right)+\mathrm{N}{\mathrm{a}}_2{\mathrm{CO}}_3\left(\mathrm{s}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to {\mathrm{Na}}_2{\mathrm{SiF}}_6\left(\mathrm{aq}\right)+\mathrm{Si}{\mathrm{O}}_2\left(\mathrm{s}\right)+\mathrm{C}{\mathrm{O}}_2\left(\mathrm{g}\right)+2\mathrm{HF}\left(\mathrm{aq}\right)$

The amount of produced in the process can be calculated as follows:

$\begin{array}{c}\mathrm{n}\left({\mathrm{Na}}_2{\mathrm{SiF}}_6\right)=\frac{1}{2}\times \mathrm{n}\left({\mathrm{SiF}}_4\right)\\ \mathrm{n}\left({\mathrm{Na}}_2{\mathrm{SiF}}_6\right)=372\mathrm{mol}\end{array}$

Each molecule of ${\mathrm{Na}}_2{\mathrm{SiF}}_6$has six fluorine atoms, implying that one molecule of ${\mathrm{Na}}_2{\mathrm{SiF}}_6$can yield six fluorine ions during the fluoridation process, allowing us to determine the amount of fluorine ions:

$\begin{array}{c}\mathrm{n}\left({\mathrm{F}}^{-}\right)=6\times \mathrm{n}\left({\mathrm{Na}}_2{\mathrm{SiF}}_6\right)\\ \mathrm{n}\left({\mathrm{F}}^{-}\right)=6\times 372\mathrm{mol}\\ \mathrm{n}\left({\mathrm{F}}^{-}\right)=2,232\mathrm{mol}\end{array}$

We can compute the mass from the number of fluorine ions. Finally, we can compute the volume of water that can be fluoridated using the mass concentration equation:

$\begin{array}{c}\mathrm{m}\left({\mathrm{F}}^{-}\right)=\mathrm{n}\left({\mathrm{F}}^{-}\right)\times \mathrm{M}\left({\mathrm{F}}^{-}\right)\\ \mathrm{m}\left({\mathrm{F}}^{-}\right)=22.32\mathrm{mol}\times 18,998{\mathrm{gmol}}^{-1}\\ \mathrm{m}\left({\mathrm{F}}^{-}\right)=424035\mathrm{g}=424.0\times {10}^6\mathrm{mg}\end{array}$

$\begin{array}{c}\mathrm{\gamma }\left({\mathrm{F}}^{-}\right)=\frac{\mathrm{m}\left({\mathrm{F}}^{-}\right)}{\mathrm{V}\left({\mathrm{H}}_2\mathrm{O}\right)}\\ \mathrm{V}\left({\mathrm{H}}_2\mathrm{O}\right)=\frac{\mathrm{m}\left({\mathrm{F}}^{-}\right.}{\mathrm{\gamma }\left({\mathrm{F}}^{-}\right)}\\ \mathrm{V}\left({\mathrm{H}}_2\mathrm{O}\right)=\frac{424.0\times {10}^6\mathrm{mg}}{1{\mathrm{mgL}}^{-1}}\\ \mathrm{V}\left({\mathrm{H}}_2\mathrm{O}\right)=424.0\times {10}^6 \mathrm{L}\\ =424.0\times {10}^3 m^3\end{array}$

Therefore, the value is$424.0\times {10}^3 {\text{m}}^3$