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 \(\operatorname{Te}(\mathrm{OH})_{6} ?\) b. Despite its structural differences with sulfuric and selenic acid, telluric acid is a diprotic acid with \(\mathrm{p} K_{\mathrm{a}_{1}}=7.68\) and \(\mathrm{p} K_{\mathrm{a}_{2}}=11.29 .\) Telluric acid can be prepared by hydrolysis of tellurium hexafluoride according to the equation $$\operatorname{TeF}_{6}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\operatorname{Te}(\mathrm{OH})_{6}(a q)+6 \mathrm{HF}(a q)$$ Tellurium hexafluoride can be prepared by the reaction of elemental tellurium with fluorine gas:$$\operatorname{Te}(s)+3 \mathrm{F}_{2}(g) \longrightarrow \operatorname{Te} \mathrm{F}_{6}(g)$$.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 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 \(\operatorname{Te} \mathrm{F}_{6}(g)\) in \(115 \mathrm{mL}\) solution? Assume \(100 \%\) yield in all reactions.

Short answer

The oxidation state of tellurium in \(Te(OH)_{6}\) is +6. The pH of the solution formed by dissolving the isolated \(TeF_{6}(g)\) in 115 mL of solution is 0.634.

Step-by-step solution

Assign Oxidation States

To determine the oxidation state of Tellurium in \(Te(OH)_{6}\), assign the oxidation state for each atom. You know that Oxygen usually has an oxidation state of -2 and Hydrogen has an oxidation state of +1.

Set up the Equation

Now, set up an equation using the general rule that the sum of oxidation states in a neutral molecule should be equal to zero. So if you let the oxidation state of Tellurium be x, you have the equation \(x + 6*(1+ (-2)) = 0\).

Solve the Equation

Solving the equation obtained in the previous step gives the oxidation state of Tellurium. The equation simplifies to \(x - 6 = 0\). Solving, you get \(x = 6\). So, the oxidation state of Tellurium in \(Te(OH)_{6}\) is +6. #b. Calculation of the pH of the Solution formed by Hydrolysis of Tellurium Hexafluoride#

Calculate the Moles of Tellurium

The given density of tellurium is 6.240 g/cm³. The volume of the Te block is 0.545 cm³. Using the formula for density (density = mass/volume), you can calculate the mass of tellurium: Mass = Density * Volume = 6.240 g/cm³ * 0.545 cm³ = 3.4008 g. The molar mass of tellurium (Te) is 127.60 g/mol. So, the number of moles of Te is given by mass/molar mass = 3.4008 g / 127.60 g/mol = 0.02664 mol.

Apply Stoichiometry and Ideal Gas Law

Given the balanced equation for the formation of TeF6, you know that 1 mol of Te reacts with 3 mol of F2 to produce 1 mol of TeF6. Therefore, 0.02664 mol of Te will react completely with 0.02664 * 3 = 0.07992 mol of F2 to form 0.02664 mol of TeF6. Now, to know if there is enough F2 to react, calculate the number of moles of F2 using the ideal gas law, where PV = nRT. Rearranging it, you get n = PV/RT = (1.06 atm * 2.34 L) / (0.0821 L.atm/(K.mol) * (273 + 25)K) = 0.094 mol F2. Since the moles of F2 available (0.094 mol) are more than what's required by stoichiometry (0.07992 mol), F2 is not the limiting reactant; Te is. Hence, exactly 0.02664 moles of TeF6 may form.

Apply Stoichiometry to find Moles of Telluric Acid

From the balanced equation of hydrolysis, you know that 1 mol of TeF6 forms 1 mol of \(Te(OH)_{6}\). So, 0.02664 moles of TeF6 will form 0.02664 moles of \(Te(OH)_{6}\).

Compute the molar concentration of Telluric Acid and calculate pH

The molar concentration or molarity of \(Te(OH)_{6}\) is calculated by dividing the number of moles (0.02664 mol) by the volume in liters (115 mL is equivalent to 0.115 L). This gives 0.232 mol/L for the molarity of \(Te(OH)_{6}\). Given that \(pK_{a1}=7.68\), you know that the first proton separates out from the acid before reaching the equivalence point. So, the concentration of \(Te(OH)_{5}^{-}\) (which is equivalent to the \(H_{3}O^{+}\) ion concentration) can be calculated from the molarity of \(Te(OH)_{6}\). Remembering that pH = -log[H3O+], you can calculate the pH: pH = -log[H3O+] = - log (0.232) = 0.634.

Key concepts

Telluric Acid

Telluric acid, with the chemical formula \( \text{H}_6\text{TeO}_6 \), is an interesting compound due to its structural formation and chemical properties. Unlike sulfuric acid (\( \text{H}_2\text{SO}_4 \)) or selenic acid (\( \text{H}_2\text{SeO}_4 \)), telluric acid is characterized by the presence of six hydroxyl groups. As a result, it's often represented as Te(OH)\(_6\), highlighting its unique coordination.
This structural difference positions telluric acid as a more complex molecule, yet it remains important in various chemical reactions and has implications in the preparation of other tellurium compounds. It is unique because it behaves as a diprotic acid, which means it can donate two protons despite having more hydrogen atoms in its structure. Understanding telluric acid is crucial for recognizing its role in reactions, especially involving hydrolysis, where it can be synthesized from tellurium hexafluoride.

Diprotic Acid

Diprotic acids, like telluric acid, have the ability to donate two protons (H\(^+\)) per molecule during reactions. This property influences how they dissociate in water, typically in two distinct steps, each characterized by its own dissociation constant, \( pK_a \).
For telluric acid, these dissociation constants are \( pK_{a1} = 7.68 \) and \( pK_{a2} = 11.29 \). This suggests that the first proton releases more readily compared to the second. The knowledge of \( pK_a \) values is important for calculating the acidity of solutions involving diprotic acids, as these values help predict the solution's behavior in acid-base equilibrium.
The stepwise release of protons affects the pH of the solution during titrations or when dissolved in different concentrations, which is pivotal when determining the acidity or basicity during experiments.

Oxidation State Calculation

Calculating the oxidation state is crucial for understanding the chemical behavior of elements in compounds. In \( \text{Te}(\text{OH})_6 \), to determine the oxidation state of tellurium, you rely on known oxidation states of other atoms within the molecule: oxygen typically has a state of \(-2\) and hydrogen \(+1\).
With this information, an equation based on the sum of oxidation states equating to zero in a neutral molecule can be set up: \( x + 6(1 + (-2)) = 0 \), where \( x \) is the oxidation state of tellurium.
Solving the equation demonstrates that tellurium exhibits an oxidation state of \(+6\). Understanding oxidation states helps in predicting the electron transfer in different chemical reactions and is a fundamental concept in redox chemistry.

Ideal Gas Law

The ideal gas law is a valuable tool in calculating the amount of gas involved in chemical reactions. It relates the physical properties of gases, establishing that \( PV = nRT \), where \( P \) represents pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is temperature in Kelvin.
In this context, the ideal gas law is used to determine the number of moles of fluorine gas available to react with tellurium to form tellurium hexafluoride. By rearranging the formula to \( n = \frac{PV}{RT} \), you can calculate the moles of a gas under specific conditions of pressure and temperature.

• Pressure (1.06 ext{ atm})
• Volume (2.34 ext{L})
• Temperature (25\(^\circ\text{C}\) = 298 ext{K})

This calculation informs whether sufficient reactants are available for the reaction to proceed to completion, which is crucial for quantifying the stoichiometry of the product formed.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. In the hydrolysis of tellurium hexafluoride to form telluric acid, stoichiometry helps calculate how much product can be formed based on the amounts of starting materials.
Given that 1 mole of tellurium reacts with 3 moles of fluorine to create 1 mole of \( \text{TeF}_6 \), and 1 mole of \( \text{TeF}_6 \) produces 1 mole of \( \text{Te}(\text{OH})_6 \), stoichiometry guides us in determining the exact quantities of each compound that can react under given conditions.
Using the ideal gas law and the known molar masses of the substances, you can determine the limiting reactant and calculate how much of the product \( \text{Te}(\text{OH})_6 \) will result. This is key to understanding reaction yields and the pH of the solution when dissolved, providing insight into the practical aspects of chemical synthesis.