Chapter 19, Problem 109IP

(a)

Interpretation Introduction

Interpretation: The answers are to be stated for the given options.

Concept introduction: Oxidation state is a number that is assigned to each element in a compound. This number describes the ability to lose or gain electrons.

The $\text{pH}$ is calculated using the formula,

$\text{pH}=-{\log }_{\text{10}}\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]$

To determine: The oxidation state of tellurium in ${\text{Te(OH)}}_{\text{6}}$.

Explanation of Solution

The oxidation state of tellurium in ${\text{Te(OH)}}_{\text{6}}$ is $\underset{\_}{+6}$.

Oxidation state of oxygen is $-2$.

Oxidation state of hydrogen is $+1$.

It is assumed that oxidation state of tellurium in ${\text{Te(OH)}}_{\text{6}}$ is $x$.

The compound ${\text{Te(OH)}}_{\text{6}}$ contains six oxygen and hydrogen atoms and single tellurium atom. Since, stable molecule has a zero overall charge. Therefore,

$\begin{array}{c}6\left(+1\right)+x+6\left(-2\right)=0\\ x=+6\end{array}$

Therefore, the oxidation state of tellurium in ${\text{Te(OH)}}_{\text{6}}$ is $\underset{\_}{+6}$.

(b)

Interpretation Introduction

Interpretation: The answers are to be stated for the given options.

Concept introduction: Oxidation state is a number that is assigned to each element in a compound. This number describes the ability to lose or gain electrons.

The $\text{pH}$ is calculated using the formula,

$\text{pH}=-{\log }_{\text{10}}\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]$

To determine: The $\text{pH}$ value of a solution of ${\text{Te(OH)}}_{\text{6}}$ formed by dissolving the isolated ${\text{TeF}}_{\text{6}}\left(g\right)$ in $\text{115 mL}$ solution.

Explanation of Solution

Given

Density of tellurium $(\text{Te})$  is $6.24{\text{g/cm}}^{\text{3}}$.

Length of tellurium edge is $0.545\text{cm}$.

Since, the length of tellurium edge is $0.545\text{cm}$. The volume of tellurium is,

${\left(0.545\text{cm}\right)}^3=0.1619{\text{cm}}^3$

Formula

Mass of $\text{Te}$ is calculated using the formula,

$\text{Mass}\text{of}\text{Te}=\text{Density}\times \text{Volume}$

Substitute the values of density and volume of tellurium in the above equation.

$\begin{array}{c}\text{Mass}\text{of}\text{Te}=\text{Density}\times \text{Volume}\\ =\text{6}\text{.24}{\text{g/cm}}^3\times 0.1619{\text{cm}}^3\\ =\underset{\_}{1.01g}\end{array}$

Moles of tellurium is $\underset{\_}{0.00791mol}$.

Mass of tellurium is $1.01\text{g}$.

Atomic mass of tellurium is $\text{127}\text{.6}\text{g/mol}$.

Formula

The number of moles of $\text{Te}$ is calculated using the formula,

$\text{Moles}\text{of}\text{Te}=\frac{\text{Mass}\text{of}\text{Te}}{\text{Atomic}\text{mass}\text{of}\text{Te}}$

Substitute the values of mass and atomic mass of tellurium in the above equation.

$\begin{array}{c}\text{Moles}\text{of}\text{Te}=\frac{\text{Mass}\text{of}\text{Te}}{\text{Atomic}\text{mass}\text{of}\text{Te}}\\ =\frac{\text{1}\text{.01}\text{g}}{\text{127}\text{.6}\text{g/mol}}\\ =\underset{\_}{0.00791mol}\end{array}$

Moles of fluorine is $\underset{\_}{0.101mol}$.

Given

Volume of fluorine gas is $2.34\text{L}$.

Pressure is $1.06\text{atm}$.

Temperature is $25°\text{C}$.

The conversion of degree Celsius $\left(°\text{C}\right)$ into Kelvin $\left(\text{K}\right)$ is done as,

$T\left(\text{K}\right)=T\left(°\text{C}\right)+\text{273}$

Hence,

The conversion of $\text{25}°\text{C}$ into Kelvin is,

$\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+\text{273}\\ T\left(\text{K}\right)=\left(\text{25}+\text{273}\right)\text{K}\\ =\text{298}\text{K}\end{array}$

Formula

Number of moles is calculated as,

$\begin{array}{c}PV=nRT\\ n=\frac{PV}{RT}\end{array}$

Where,

• $P$ is the total pressure.
• $V$ is the volume.
• $n$ is the total moles.
• $R$ is the universal gas constant $(\text{0}\text{.08206}\text{L}\cdot \text{atm}/\text{K}\cdot \text{mol})$.
• $T$ is the absolute temperature.

Substitute the values of $P,V,R$ and $T$ in the above equation.

$\begin{array}{c}n=\frac{PV}{RT}\\ =\frac{\text{1}\text{.06}\text{atm}\times \text{2}\text{.34}\text{L}}{(\text{0}\text{.08206}\text{L}\cdot \text{atm}/\text{K}\cdot \text{mol})\left(\text{298}\text{K}\right)}\\ =\underset{\_}{0.101mol}\end{array}$

Molarity of ${\text{Te(OH)}}_{\text{6}}$ is $\underset{\_}{0.06878M}$.

The reaction of formation of ${\text{TeF}}_{\text{6}}$ is,

$\text{Te}\left(s\right)+3{\text{F}}_2\left(g\right)\stackrel{}{\to }{\text{TeF}}_6\left(g\right)$

It is clear from the above equation that, one mole of tellurium reacts with three moles of fluorine to yield one mole of ${\text{TeF}}_{\text{6}}$. In this reaction, tellurium is limited, whereas fluorine gas is in excess. Hence, the chemical equivalence is,

Number of moles of tellurium is equal to the number of moles of ${\text{TeF}}_{\text{6}}$ formed. The number of moles of ${\text{TeF}}_{\text{6}}$ is $0.00791\text{mol}$.

Since one mole of ${\text{TeF}}_{\text{6}}$ leads to the formation of one mole of ${\text{Te(OH)}}_{\text{6}}$. Therefore, $0.00791\text{mol}$ mole of ${\text{TeF}}_{\text{6}}$ leads to the formation of,

$0.00791\text{mol}\times \text{1}=0.00791\text{mol}{\text{of Te(OH)}}_{\text{6}}$

Formula

The molarity of compound in a solution is calculated using the formula,

$\text{Molarity}\text{of}\text{compound}=\frac{\text{Moles}\text{of}\text{compound}}{\text{Volume}\text{of}\text{solution}}$

Substitute the values of number of moles of ${\text{Te(OH)}}_{\text{6}}$ and volume of solution in the above equation.

$\begin{array}{c}\text{Molarity}\text{of}\text{compound}=\frac{\text{Moles}\text{of}\text{compound}}{\text{Volume}\text{of}\text{solution}}\\ =\frac{\text{0}\text{.00791}\text{mol}}{\text{0}\text{.115}\text{L}}\\ =\underset{\_}{0.06878M}\end{array}$

Concentration of $\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]$ is $\underset{\_}{6.74\times 10^{-3}M}$.

Given

The $\text{p}{K}_{\text{a1}}$ value of ${\text{Te(OH)}}_{\text{6}}$ is $7.68$.

The $\text{p}{K}_{\text{a2}}$ value of ${\text{Te(OH)}}_{\text{6}}$ is $11.29$.

Molarity of ${\text{Te(OH)}}_{\text{6}}$ $\text{0}\text{.06878}\text{M}$.

The $K_{\text{a}}$ value of $\text{HF}$ is $6.6\times {10}^{-4}$. Since, $\text{HF}$ is a stronger acid than ${\text{Te(OH)}}_{\text{6}}$, only $\text{HF}$ will be dissociated into protons, whereas dissociation of ${\text{Te(OH)}}_{\text{6}}$ will be suppressed.

The formula to calculate the concentration of proton $\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]$ of $\text{HF}$ is,

$\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]=\sqrt{K_{\text{a}}\left(\text{HF}\right)\times \text{Molarity}\text{of HF}}$

Substitute the values of $K_{\text{a}}$ and molarity of $\text{HF}$ in the above equation.

$\begin{array}{c}\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]=\sqrt{K_{\text{a}}\left(\text{HF}\right)\times \text{Molarity}\text{of HF}}\\ =\sqrt{6.6\times {10}^{-4}\times \text{0}\text{.06878}\text{M}}\\ =\underset{\_}{6.74\times 10^{-3}M}\end{array}$

Required $\text{pH}$ value is $\underset{\_}{2.17}$.

Concentration of $\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]$ is $\text{6}\text{.74}\times {\text{10}}^{-\text{3}}\text{M}$.

Formula

The $\text{pH}$ is calculated using the formula,

$\text{pH}=-{\log }_{\text{10}}\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]$

Substitute the value of concentration of $\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]$ in the above equation.

$\begin{array}{c}\text{pH}=-{\log }_{\text{10}}\left[{\text{H}}^{+}\right]\\ =-{\log }_{\text{10}}\left[\text{6}\text{.74}\times {\text{10}}^{-\text{3}}\right]\\ =\underset{\_}{2.17}\end{array}$