Chapter 18, Problem 124CP

(a)

Interpretation Introduction

Interpretation : The oxidation state of tellurium in ${\text{Te(OH)}}_{\text{6}}$ is calculated.

Concept Introduction :

The oxidation state is also known as oxidation number that defines the degree of oxidation. This number represents the number of electrons that an atom can lose or gain or share due to the bonding with another element atom.

Answer to Problem 124CP

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

Explanation of Solution

Given that

The molecular formula for telluric acid is ${\text{H}}_{\text{6}}{\text{TeO}}_{\text{6}}$

Oxidation state of O is -2 and H is +1. Let x be the oxidation state of $\text{Te}$ then

  $\begin{array}{l}6\left(+1\right)+x+6\left(-2\right)=0\\ 6+x-12=0\\ x=+6\end{array}$

Hence, the oxidation state of tellurium in ${\text{Te(OH)}}_{\text{6}}$ is $x=+6$

(b)

Interpretation Introduction

Interpretation : The pH of the solution of ${\text{Te(OH)}}_{\text{6}}$ formed by dissolving the isolated ${\text{TeF}}_{\text{6}}\text{(g)}$ in 115 mL solution should be calculated.

Concept Introduction :

The tellurium hexafluoride can be prepared by the reaction of elemental tellurium with the fluorine gas. The pH is defined as the measure of acidity or alkalinity of the solution, hydrogen ion concentration. When the pH value is less than 7 then it is acidic and the pH value is greater than 7 then it is base or alkaline solution.

Answer to Problem 124CP

The pH of the solution of ${\text{Te(OH)}}_{\text{6}}$ formed by dissolving the isolated ${\text{TeF}}_{\text{6}}\text{(g)}$ in 115 mL solutionis $pH=2.1526$.

Explanation of Solution

The Structures of selenic acid and telluric acid is given below as,

   Sulfuric and selenic acids have tetrahedral structure where as telluric acid has octahedral structure.

Given data is

Density of $\text{Te}=6.24{\text{g/cm}}^{\text{3}}$

Volume of $\text{Te}=0.545{\text{cm}}^{\text{3}}$

  $=0.1619{\text{cm}}^3$

Mass of $\text{Te}=\text{density}\times \text{volume}$

  $\begin{array}{l}=6.24{\text{g/cm}}^{\text{2}}\times 0.1619{\text{cm}}^{\text{3}}\\ \text{Mass of}\text{Te}=1.01\text{g}\end{array}$

Moles of $\text{Te}=\frac{\text{mass}}{\text{atomic}\text{mass}}$

  $\begin{array}{l}=\frac{1.01\text{g}}{127.6\text{g/mol}}\\ \text{Moles}\text{of}\text{Te}=0.00791\text{mol}\end{array}$

Also given that

Volume of $F_2=2.34\text{L}$

Pressure of $F_2=1.06\text{atm}$

Temperature of $F_2={25}^{\circ }C=298\text{k}$

  $\begin{array}{l}\text{Moles}\text{of}{\text{F}}_{\text{2}}\text{,}\text{n}=\frac{\text{PV}}{\text{RT}}\\ =\frac{\left(1.06\text{atm}\right)\left(2.34\text{L}\right)}{\left(0.0821\text{L}\text{atm/mol}\text{K}\right)\left(298\text{k}\right)}\\ \text{Moles}\text{of}{\text{F}}_{\text{2}}\text{,}\text{n}=0.101\text{mol}\end{array}$

According to the stoichiometric equation one mole Te reacts with 3 moles of $F_2$

  $0.00791$ a mole of Te reacts with $0.02373\text{mol}$

  $F_2$ . But there are $0.01$ moles of $F_2$ . Hence Te is the limiting reactant.

Number of moles $Te{F}_6$ formed = moles of Te= $0.00791\text{mol}$

Number of moles $Te{\left(OH\right)}_6=\frac{\text{moles}}{\text{volume}\text{in}\text{liters}}$

  $=\frac{0.00791\text{mol}}{0.115\text{L}}$

Number of moles $=0.06878\text{M}$

  $Te{\left(OH\right)}_6$ is the weak acid and it has $pKa1=7.68\text{and}pKa2=11.29$

Molarity of HF = 0.06878 M

HF has $K_a=7.2\times {10}^{-4}$ so HF is strong acid than $Te{\left(OH\right)}_6$ acid. Hence the proton coming from HF only and the dissociation of $Te{\left(OH\right)}_6$ is suppressed by HF.

  $\begin{array}{l}\left[H_3{O}^{+}\right]=\sqrt{K_a\left(HF\right)\times \text{molarity}\text{of}\text{HF}}\\ =\sqrt{7.2\times {10}^{-4}\times 0.06878\text{M}}\\ \left[H_3{O}^{+}\right]=7.04\times {10}^{-3}\text{M}\end{array}$

  $\begin{array}{l}pH=-\log \left[7.04\times {10}^{-3}\right]\\ pH=2.1526\end{array}$

Thus, the pH of the solution of ${\text{Te(OH)}}_{\text{6}}$ formed by dissolving the isolated ${\text{TeF}}_{\text{6}}\text{(g)}$ in 115 mL solution is $pH=2.1526$.