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 Te(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 $$ \mathrm{TeF}_{6}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Te}(\mathrm{OH})_{6}(a q)+6 \mathrm{HF}(a q) $$ 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 \(\mathrm{atm}\) and \(25^{\circ} \mathrm{C},\) what is the pH of a solution of \(\mathrm{Te}(\mathrm{OH})_{6}\) formed by dissolving the isolated TeF \(_{6}(g)\) in 115 \(\mathrm{mL}\) solution? Assume 100\(\%\) yield in all reactions.

Short answer

First, we calculate the moles of Te and F2 from the given information. For Te, we calculate its mass from the given volume and density and then convert it to moles using molar mass. For F2, we use the ideal gas equation PV = nRT to find the number of moles. Next, we compare the amounts of reactants and determine the limiting reactant. We then calculate the moles of TeF6 produced, which is equal to the moles of \(\mathrm{Te}(\mathrm{OH})_{6}\) formed. By dividing the moles of \(\mathrm{Te}(\mathrm{OH})_{6}\) by the volume of the solution, we find the concentration of \(\mathrm{Te}(\mathrm{OH})_{6}\) in the solution. Finally, we use the first pKa value and the equilibrium constant expression to calculate the concentration of H+ and the pH of the solution.

Step-by-step solution

Calculate moles of Te from the provided dimensions and density

To calculate the moles of tellurium, we need to convert the given information about the cubic block (0.545 cm on edge and density of 6.240 g/cm³) to mass, and then moles. Volume of the cubic block = side^3 Volume = 0.545^3 cm³ mass of tellurium = volume × density = 0.545^3 × 6.240 g Now, we have to convert the mass of Te to moles, using the molar mass of Te (127.6 g/mol): moles of Te = mass of Te / molar mass of Te

Calculate moles of F2 from given volume and pressure

First, convert the given volume of F2 (2.34 L) to moles, using the ideal gas equation: PV = nRT where P is the pressure in atm, V is the volume in L, n is the number of moles of F2, R is the ideal gas constant (0.0821 L⋅atm/mol⋅K), and T is the temperature in K. Solve for n.

Determine the limiting reactant and calculate moles of TeF6 produced

Compare the amounts of Te and F2 to determine the limiting reactant, using the balanced chemical equation for the reaction of Te with F2 to form TeF6. Calculate the moles of TeF6 produced by the reaction.

Calculate the moles of \(\mathrm{Te}(\mathrm{OH})_{6}\) formed in the solution

We know the moles of TeF6 produced and, assuming 100% yield, the number of moles of Te(OH)_6 formed is equal to the number of moles of TeF6 produced.

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

Divide the moles of Te(OH)_6 by the volume of the solution (115 mL) in liters to find the molar concentration of Te(OH)_6.

Calculate the pH of the solution using the two pKa values

Since the pK_a1 value is considerably lower than the pK_a2 value (7.68 vs 11.29), we can consider only the first ionization of telluric acid (H_2TeO_6 -> H^+ + HTeO_6^-) to calculate the pH. Assuming the dissociation of Te(OH)_6 is relatively small compared to the number of moles of Te(OH)_6 in the solution, we can write the equilibrium constant expression, and then solve for pH. K_a1 = [H^+][HTeO_6^-]/[H_2TeO_6] pH = -log([H^+]) Once we find the [H^+], we can calculate the pH of the final solution.

Key concepts

Oxidation State

In chemistry, the oxidation state of an element reflects the degree of oxidation of an atom in a compound. It indicates the number of electrons lost or gained by an atom.
For tellurium in the compound \( \mathrm{Te(OH)}_{6} \), the oxidation state can be calculated by recognizing the structures involved.
Each hydroxide group \( (\mathrm{OH}^{-}) \) contributes a -1 charge.

• There are six \( OH^{-} \) groups, leading to a total negative charge of -6.
• To balance this, tellurium must have an oxidation state of +6, since the overall molecule is neutral.

Thus, in \( \mathrm{Te(OH)}_{6} \), tellurium exhibits an oxidation state of +6. This helps us understand the electron distribution and potentially the reactivity of tellurium in this compound.

Diprotic Acid

A diprotic acid is capable of donating two protons (hydrogen ions) when dissolved in water.
Telluric acid is known to be a diprotic acid. That means it can release two hydrogen ions sequentially, which is indicative of multiple \( pK_a \) values.

• The first ionization involves the release of the first proton: \( \mathrm{H}_2\mathrm{TeO}_6 \rightarrow \mathrm{H}^+ + \mathrm{HTeO}_6^- \).
• The second ionization involves losing another proton: \( \mathrm{HTeO}_6^- \rightarrow \mathrm{H}^+ + \mathrm{TeO}_6^{2-} \).

These sequential steps give rise to the \( pK_{a_{1}} = 7.68 \) and \( pK_{a_{2}} = 11.29 \).The differing \( pK_a \) values indicate the strengths of these ionizations.
The first ionization is stronger and contributes more significantly to the acidity of the solution.

pH Calculation

Calculating the pH of a solution involves determining the concentration of hydrogen ions \( [\mathrm{H}^+] \). For a solution of a diprotic acid like telluric acid, focus on the first ionization step due to its lower \( pK_a \).
Using the equilibrium constant expression for the first ionization:\[K_{a_1} = \frac{[H^+][HTeO_6^-]}{[H_2TeO_6]}\]To find \( [H^+] \), approximate that initially only the first protons are released in significant amounts.
This is solved by rearranging the equation to find \( [H^+] \) from the known \( K_{a_1} \) and \( H_2TeO_6 \) concentration.
After solving for \( [H^+] \), the pH is calculated as:\[pH = -\log([H^+])\]This approach simplifies the pH calculation by addressing the most substantial contribution to acidity.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and number of moles of a gas.
This equation is expressed as:\[PV = nRT\]Where:

• \( P \) is the pressure in atmospheres.
• \( V \) is the volume in liters.
• \( n \) is the number of moles.
• \( R \) is the ideal gas constant (0.0821 L⋅atm/mol⋅K).
• \( T \) is the temperature in Kelvin.

The law is extremely helpful in determining how many moles of gas are present under certain conditions of temperature and pressure.
For instance, using the provided conditions (2.34 L of \( \mathrm{F}_2 \) at 1.06 atm and 25°C), you can determine \( n \) in this reaction.First, convert Celsius to Kelvin by adding 273.15 to the temperature in Celsius.
Then use the rearranged Ideal Gas Law, \( n = \frac{PV}{RT} \), to calculate the number of moles of fluorine gas, which is essential for further reaction stoichiometry.