Chapter 5, Problem 5.109P

Interpretation Introduction

Interpretation:

The percentage of ${\text{SO}}_{\text{2}}$ in the air sample is to be calculated.

Concept introduction:

The expression to calculate the molarity is as follows:

$\text{Molarity of solute}=\frac{\text{Moles of solute}}{\text{Volume of solute}}$

The ideal gas equation can be expressed as follows,

$PV=nRT$

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of the gas and $R$ is the gas constant.

Answer to Problem 5.109P

The percentage of ${\text{SO}}_{\text{2}}$ in the air sample is $0.797\%$.

Explanation of Solution

The equation for the reaction of ${\text{I}}_{\text{2}}\left(aq\right)$ with ${\text{S}}_2{\text{O}}_3^{2-}\left(aq\right)$ is as follows:

${\text{2S}}_{\text{2}}{\text{O}}_3^{2-}\left(aq\right){\text{ + I}}_{\text{2}}\left(aq\right)\to {\text{S}}_4{\text{O}}_6^{2-}\left(s\right)+2{\text{I}}^{-}\left(aq\right)$        (1)

The expression to calculate the initial mole ${\text{I}}_{\text{2}}$ is as follows:

${\text{Moles of I}}_{\text{2}}=\left({\text{Molarity of I}}_{\text{2}}\right)\left({\text{Volume of I}}_{\text{2}}\right)$        (2)

Substitute the value $200\text{ mL}$ for volume of ${\text{I}}_{\text{2}}\left(aq\right)$ and $0.01017\text{ mol/L}$ for molarity of ${\text{I}}_{\text{2}}\left(aq\right)$ in the equation (2).

$\begin{array}{c}{\text{Moles of I}}_{\text{2}}=\left(0.01017\text{ mol/L}\right)\left(200\text{ mL}\right)\left(\frac{1\text{ l}}{1000\text{ mL}}\right)\\ =2.034\times {10}^{-4}\text{ mol}\end{array}$

The expression to calculate the mole ${\text{S}}_2{\text{O}}_3^{2-}\left(aq\right)$ is as follows:

${\text{Moles of S}}_2{\text{O}}_3^{2-}\left(aq\right)=\left({\text{Molarity of S}}_2{\text{O}}_3^{2-}\left(aq\right)\right)\left({\text{Volume of S}}_2{\text{O}}_3^{2-}\left(aq\right)\right)$        (3)

Substitute the value $\text{11}\text{.37 mL}$ for volume of ${\text{S}}_2{\text{O}}_3^{2-}\left(aq\right)$ and $0.0105\text{ mol/L}$ for molarity of ${\text{S}}_2{\text{O}}_3^{2-}\left(aq\right)$ in the equation (3).

$\begin{array}{c}{\text{Moles of S}}_2{\text{O}}_3^{2-}\left(aq\right)=\left(0.0105\text{ mol/L}\right)\left(\text{11}\text{.37 mL}\right)\left(\frac{1\text{ l}}{1000\text{ mL}}\right)\\ =1.19\times {10}^{-4}\text{ mol}\end{array}$

According to the balanced chemical equation, two moles of the ${\text{S}}_2{\text{O}}_3^{2-}$ reacts with one mole of ${\text{I}}_{\text{2}}$. Thus, the mole of ${\text{I}}_{\text{2}}\left(aq\right)$ that react with ${\text{S}}_2{\text{O}}_3^{2-}\left(aq\right)$ is calculated as follows:

${\text{Moles of I}}_{\text{2}}\left(aq\right)=\left[\left({\text{Moles of S}}_2{\text{O}}_3^{2-}\left(aq\right)\right)\left(\frac{{\text{1 mol I}}_{\text{2}}\left(aq\right)}{{\text{2 mol S}}_2{\text{O}}_3^{2-}\left(aq\right)}\right)\right]$        (4)

Substitute the value $1.19\times {10}^{-4}\text{ mol}$ for moles of ${\text{S}}_2{\text{O}}_3^{2-}\left(aq\right)$ in the equation (4).

$\begin{array}{c}{\text{Moles of I}}_{\text{2}}\left(aq\right)=\left[\left(1.19\times {10}^{-4}\text{ mol}\right)\left(\frac{{\text{1 mol I}}_{\text{2}}\left(aq\right)}{{\text{2 mol S}}_2{\text{O}}_3^{2-}\left(aq\right)}\right)\right]\\ =5.96925\times {10}^{-5}\text{ mol}\end{array}$

The expression to calculate the mole ${\text{I}}_{\text{2}}\left(aq\right)$ left after the reaction is as follows:

${\text{Moles of I}}_{\text{2}}=\left[\left({\text{initial moles of I}}_{\text{2}}\right)-\left({\text{moles of I}}_{\text{2}}{\text{ react with S}}_{\text{2}}{\text{O}}_3^{-}\right)\right]$        (5)

Substitute the value $2.034\times {10}^{-4}\text{ mol}$ for initial moles of ${\text{I}}_{\text{2}}\left(aq\right)$ and $5.96925\times {10}^{-5}\text{ mol}$ for moles of ${\text{I}}_{\text{2}}\left(aq\right)$ reacted with ${\text{S}}_2{\text{O}}_3^{2-}\left(aq\right)$ in the equation (5).

$\begin{array}{c}{\text{Moles of I}}_{\text{2}}=\left[\left(2.034\times {10}^{-4}\text{ mol}\right)-\left(5.96925\times {10}^{-5}\text{ mol}\right)\right]\\ =1.437075\times {10}^{-5}\text{ mol}\end{array}$

The equation for the reaction of ${\text{I}}_{\text{2}}\left(aq\right)$, ${\text{H}}_{\text{2}}\text{O}\left(l\right)$ with ${\text{SO}}_2\left(aq\right)$ is as follows:

${\text{SO}}_2\left(aq\right){\text{ + I}}_{\text{2}}\left(aq\right){\text{ + 2H}}_{\text{2}}\text{O}\left(l\right)\to {\text{HSO}}_4^{2-}\left(aq\right)+2{\text{I}}^{-}\left(aq\right)+3{\text{H}}^{+}\left(aq\right)$        (6)

From the equation (6), one mole of the ${\text{I}}_{\text{2}}\left(aq\right)$ reacts with one mole of ${\text{SO}}_2\left(aq\right)$. Thus, the moles of ${\text{SO}}_2\left(aq\right)$ is calculated from ${\text{I}}_{\text{2}}\left(aq\right)$ as follows:

${\text{Moles of SO}}_{\text{2}}\left(aq\right)=\left[\left({\text{Moles of I}}_{\text{2}}\left(aq\right)\right)\left(\frac{{\text{1 mol SO}}_{\text{2}}\left(aq\right)}{{\text{1 mol I}}_{\text{2}}\left(aq\right)}\right)\right]$        (7)

Substitute the value $1.437075\times {10}^{-5}\text{ mol}$ for moles of ${\text{I}}_{\text{2}}\left(aq\right)$ in the equation (7).

$\begin{array}{c}{\text{Moles of SO}}_{\text{2}}\left(aq\right)=\left[\left(1.437075\times {10}^{-5}\text{ mol}\right)\left(\frac{{\text{1 mol SO}}_{\text{2}}\left(aq\right)}{{\text{1 mol I}}_{\text{2}}\left(aq\right)}\right)\right]\\ =1.437075\times {10}^{-5}\text{ mol}\end{array}$

The formula to convert $°\text{C}$ to Kelvin is:

$T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15$        (8)

Substitute $\text{38 }°\text{C}$ for $T\left(°\text{C}\right)$ in equation (8).

$\begin{array}{c}T\left(\text{K}\right)=38°\text{C}+273.15\\ =311.15\text{ K}\end{array}$

The expression to convert pressure from $\text{torr}$ into atm is as follows,

$\begin{array}{c}P=\text{ 700 torr}\left(\frac{1\text{ atm}}{760\text{ torr}}\right)\\ =0.92105631\text{ atm}\end{array}$

The ideal gas equation is as follows,

$PV=nRT$        (9)

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of air and $R$ is the gas constant.

Rearrange equation (9) for $n$ as follows:

$n=\frac{PV}{RT}$        (10)

Substitute the value $\text{0}\text{.921052631 atm}$ for $P$, $311.15\text{ K}$ for $T$, $\text{500 mL}$ for $V$ and $0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}$ for $R$ in the equation (10).

$\begin{array}{c}n=\frac{\left(\text{0}\text{.921052632 atm}\right)\left(\text{500 mL}\right)\left(\frac{1\text{ L}}{1000\text{ mL}}\right)}{\left(0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}\right)\left(298.15\text{ K}\right)}\\ =0.018028\text{ mol}\end{array}$

The percentage volume equals to the percentage mole of ${\text{SO}}_2\left(aq\right)$. The percentage volume is calculted as follows:

$\text{Volume }\%{\text{ SO}}_{\text{2}}=\left(\frac{{\text{Moles of SO}}_{\text{2}}}{\text{Moles of air}}\right)\left(100\text{}\text{ \%}\right)$        (11)

Substitute $0.018028\text{ mol}$ for moles of air and $1.437075\times {10}^{-5}\text{ mol}$ for moles of ${\text{SO}}_2\left(aq\right)$ in equation (11).

$\begin{array}{c}\text{Volume }\%{\text{ SO}}_{\text{2}}=\left(\frac{1.437075\times {10}^{-5}\text{ mol}}{0.018028\text{ mol}}\right)\left(100\text{ \%}\right)\\ =0.797146\text{ \%}\\ =0.797\%\end{array}$

Conclusion

The percentage of ${\text{SO}}_{\text{2}}$ in the air sample is $0.797\%$.