Question

While selenic acid has the formula \(\mathrm{H}_{2} \mathrm{SeO}_{4}\) and thus is directly related to sulfuric acid, telluric acid is best visualized as \(\mathrm{H}_{6} \mathrm{TeO}_{6}\) or \(\mathrm{Te}(\mathrm{OH})_{6}\) a. What is the oxidation state of tellurium in \(\mathrm{Te}(\mathrm{OH})_{6}\) ? b. Despite its structural differences with sulfuric and selenic acid, telluric acid is a diprotic acid with \(\mathrm{p} K_{a_{1}}=7.68\) and \(\mathrm{p} K_{\mathrm{a}_{2}}=11.29 .\) Telluric acid can be prepared by hydrolysis of tellurium hexafluoride according to the equation $$\mathrm{TeF}_{6}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Te}(\mathrm{OH})_{6}(a q)+6 \mathrm{HF}(a q)$$ Tellurium hexafluoride can be prepared by the reaction of elemental tellurium with fluorine gas: $$\mathrm{Te}(s)+3 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{TeF}_{6}(g)$$ If a cubic block of tellurium (density \(=6.240 \mathrm{~g} / \mathrm{cm}^{3}\) ) measuring \(0.545 \mathrm{~cm}\) on edge is allowed to react with \(2.34 \mathrm{~L}\) fluorine gas at \(1.06\) atm and \(25^{\circ} \mathrm{C}\), what is the \(\mathrm{pH}\) of a solution of \(\mathrm{Te}(\mathrm{OH})_{6}\) formed by dissolving the isolated \(\mathrm{TeF}_{6}(g)\) in \(115 \mathrm{~mL}\) water?

Short answer

The oxidation state of tellurium in \(\mathrm{Te}(\mathrm{OH})_{6}\) is +6. After conducting the given reactions and dissolving the formed \(\mathrm{Te}(\mathrm{OH})_{6}\) in water, the pH of the solution is approximately 7.68.

Step-by-step solution

Calculate the oxidation state of tellurium in \(\mathrm{Te}(\mathrm{OH})_{6}\)

To determine the oxidation state of tellurium, we have to distribute the oxidation states of the other atoms in the compound: The oxidation state of hydrogen in \(\mathrm{OH}\) is +1 and that of oxygen is -2. Since there are 6 \(\mathrm{OH}\) groups, the total combined oxidation state of these groups is: \(6 \times (-2 + 1) = -6\) The oxidation state of tellurium is thus +6, because the compound is neutral: \(\mathrm{Te}(\mathrm{OH})_{6}: \; (+6) + 6 \times (-2 + 1) = 0\)

Calculate the moles of tellurium in the cubic block

Determine the mass of tellurium based on its given density and volume: $$\text{Mass} = \text{Density} \times \text{Volume}$$ $$\text{Mass} = 6.240 (g/cm^3) \times (0.545\, cm)^3= 0.906\, g$$ Then, calculate the moles of tellurium using its molar mass (127.6 g/mol) $$\text{Moles of Te} = \frac{\text{Mass}}{\text{Molar Mass}}$$ $$\text{Moles of Te} =\frac{0.906\, g}{127.6\, g/mol} = 0.0071\, mol$$

Determine the moles of fluorine gas

First, convert the given conditions of fluorine gas to moles using the ideal gas law formula: $$\text{PV} = \text{nRT}$$ $$ \text{n} = \frac{\text{PV}}{\text{RT}}$$ $$ \text{n} = \frac{(1.06\, \text{atm})(2.34\, L)}{(0.0821\, \frac{L\cdot\text{atm}}{mol\cdot K})(298\, K)}=0.099\, \text{mol}\, \ce{F2}$$

Calculate the moles of \(\mathrm{TeF}_{6}(g)\) formed

Determine the limiting reactant in the reaction to produce \(\mathrm{TeF}_{6}(g)\). The mole ratio of \(\mathrm{Te}\) to \(\ce{F2}\) is 1:3. $$0.0071\, \text{mol Te} \times3= 0.0213\, \text{mol } \ce{F2}$$ Since we have 0.099 mol of \(\ce{F2}\) and only need 0.0213 mol for the reaction, the limiting reactant is tellurium. The moles of \(\mathrm{TeF}_{6}(g)\) formed will be equal to the moles of Te: $$0.0071\, \text{mol}$$

Calculate the moles of \(\mathrm{Te}(\mathrm{OH})_{6}\) formed during hydrolysis

The hydrolysis reaction of \(\mathrm{TeF}_{6}(g)\) to \(\mathrm{Te}(\mathrm{OH})_6\) has a mole ratio of 1:1. Therefore, the moles of \(\mathrm{Te}(\mathrm{OH})_6\) formed are equal to the moles of \(\mathrm{TeF}_{6}(g)\): $$0.0071\, \text{mol}$$

Determine the concentration of \(\mathrm{Te}(\mathrm{OH})_{6}\) in the final solution

To find the concentration of \(\mathrm{Te}(\mathrm{OH})_6\) in the solution, divide the moles of \(\mathrm{Te}(\mathrm{OH})_6\) by the volume of the solution in liters: $$\text{Concentration} = \frac{\text{Moles}}{\text{Volume}}$$ $$\text{Concentration} = \frac{0.0071 \,\text{mol}}{0.115\, L}= 0.062\, M$$

Calculate the pH of the solution using \(\mathrm{p} K_{a_{1}}\)

To approximate the pH of the solution, we can use the first \(\mathrm{p} K_a\) because the second \(\mathrm{p} K_a\) is beyond the pH range of interest. Assuming a simple dissociation equilibrium for the telluric acid, the pH is very close to \(\mathrm{p} K_{a_{1}}\) due to the relative weakness of the acid. For this problem, we consider the solution to have a pH of 7.68, which corresponds to \(\mathrm{p} K_{a_{1}}\): $$\text{pH} = \mathrm{p} K_{a_{1}}= 7.68$$

Key concepts

Oxidation State

The oxidation state is a concept used to indicate the degree of oxidation of an atom within a compound. In the compound \(\mathrm{Te(OH)}_6\), we need to determine the oxidation state of tellurium (Te). This molecule comprises tellurium and six hydroxyl (\(\mathrm{OH}\)) groups.

Each hydroxyl group has an oxidation state of +1 for hydrogen and -2 for oxygen. Since there are six \(\mathrm{OH}\) groups:

• The total oxidation state for these groups is \(6 \times (-2 + 1) = -6\).
• Given the molecule is neutral, the sum of all oxidation states must be zero. This means the tellurium must have an oxidation state of +6 to balance the -6 from the hydroxyl groups.

Understanding oxidation states helps predict reactivity and the nature of substances in chemical reactions. In this case, knowing the oxidation state of +6 for tellurium suggests it is in its high oxidation state, contributing to the molecule's properties.

Hydrolysis Reaction

A hydrolysis reaction is a chemical process where a compound reacts with water, resulting in the breakdown of that compound. In this exercise, we see hydrolysis in action when tellurium hexafluoride (\(\mathrm{TeF}_6\)) reacts with water:

\[ \mathrm{TeF}_{6}(g) + 6 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{Te(OH)}_6(aq) + 6 \mathrm{HF}(aq) \]

This reaction decomposes \(\mathrm{TeF}_6\) into telluric acid (\(\mathrm{Te(OH)}_6\)) and hydrofluoric acid (\(\mathrm{HF}\)).

• Water acts as a reactant, breaking the bonds in \(\mathrm{TeF}_6\).
• The result is the formation of an aqueous solution of telluric acid and hydrofluoric acid.

Through hydrolysis, telluric acid is prepared, illustrating how complex molecules can be broken down into simpler, solvable compounds using water. This makes hydrolysis reactions crucial in chemical synthesis and digestion processes.

Diprotic Acid

Diprotic acids are acids that can donate two protons (hydrogen ions) in solution. Telluric acid, \(\mathrm{H}_6\mathrm{TeO}_6\), is an example of a diprotic acid, despite its complex structure compared to simpler diprotic acids like sulfuric acid. When dissolved in water, telluric acid can dissociate in two stages:

• First, it loses one proton, noted by the \(\text{p}K_{a1} = 7.68\).
• Then, it can lose a second proton, with \(\text{p}K_{a2} = 11.29\).

This staged dissociation is due to each proton having a different level of reactivity or strength in the acidic environment.
Understanding diprotic acids is essential when predicting the behavior of acids in various conditions, especially when calculating the pH of solutions. For telluric acid, the first dissociation is relatively weak, contributing to its moderate acidity and pH behavior.

pH Calculation

pH calculation is an important process that quantifies the acidity or basicity of a solution. To calculate the pH of a solution of \(\mathrm{Te(OH)}_6\), we use the known values of its acidic dissociation constant, \(\mathrm{p}K_{a}\).

Given that telluric acid is a weak acid, its dissociation can be simplified as:

• To find the pH, we use the first dissociation constant \(\mathrm{p}K_{a1} = 7.68\).

Since \(\mathrm{pK}_{a1}\) is used over \(\mathrm{pK}_{a2}\), due to its more significant impact within the pH range of interest, the pH of the solution will be around 7.68.

This calculation assumes that the primary contribution to the acidity of the solution is from this first stage, as the second dissociation contributes minimally due to its much higher \(\mathrm{p}K_a\) value. This method provides a reliable approximation for the pH of weak, diprotic acid solutions.