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 \(\mathrm{p} K_{\mathrm{a}_{1}}=7.68\) and \(\mathrm{P} K_{a_{2}}=11.29 .\) 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 \(\left.=6.240 \mathrm{~g} / \mathrm{cm}^{3}\right)\) 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 \(\mathrm{pH}\) of a solution of \(\mathrm{Te}(\mathrm{OH})_{6}\) formed by dissolving the isolated \(\mathrm{TeF}_{6}(g)\) in \(115 \mathrm{~mL}\) solution? Assume \(100 \%\) yield in all reactions.

The oxidation state of tellurium can be determined by considering the oxidation states of the other atoms in the compound. In \(\mathrm{Te}(\mathrm{OH})_{6}\), each oxygen has an oxidation state of -2 and each hydrogen has an oxidation state of +1. Since there are 6 oxygen and 6 hydrogen atoms, we can write the equation: $$ x + 6(-2) + 6(1) = 0 $$ Where x is the oxidation state of tellurium. Now, simply solve for x to find the oxidation state: $$ x - 12 + 6 = 0 \\ x - 6 = 0 \\ \boxed{x = 6} $$ The oxidation state of tellurium in \(\mathrm{Te}(\mathrm{OH})_{6}\) is +6.

The mass of tellurium can be calculated by multiplying the density by the volume: $$ \text{mass}_\text{Te} = \mathrm{density}_\text{Te} \times \text{volume}_\text{Te} $$ To calculate the volume of the cubic block of tellurium, simply cube the edge length \(0.545 \text{ cm}\): $$ \mathrm{volume}_\text{Te} = (0.545 \mathrm{~cm})^3 = 0.162 \mathrm{~cm}^3 $$ Now calculate the mass: $$ \text{mass}_\text{Te} = 6.240 \frac{\mathrm{g}}{\mathrm{cm}^3} \times 0.162 \mathrm{~cm}^3 = 1.01 \mathrm{~g} $$ Convert the mass to moles using the molar mass of tellurium (approximately 127.6 g/mol): $$ \mathrm{moles}_\text{Te} = \frac{1.01 \mathrm{~g}}{127.6 \frac{\mathrm{g}}{\mathrm{mol}}} = 0.00792 \mathrm{~mol} $$

We can use the ideal gas law to solve for the moles of fluorine gas: $$ \mathrm{PV} = \mathrm{nRT} $$ where P is the pressure (1.06 atm), V is the volume (2.34 L), n is the number of moles of \(\mathrm{F}_2\), R is the ideal gas constant (0.0821 \(\mathrm{L \cdot atm \cdot K^{-1} \cdot mol^{-1}}\)), and T is the temperature in Kelvin (298 K). Solving for n: $$ \mathrm{n} = \frac{\mathrm{PV}}{\mathrm{RT}} = \frac{(1.06 \mathrm{~atm})(2.34 \mathrm{~L})}{(0.0821 \mathrm{~L \cdot atm \cdot K^{-1} \cdot mol^{-1}})(298 \mathrm{~K})} = 0.0941 \mathrm{~mol} $$

Compare the mole ratio of tellurium to fluorine given in the chemical reaction \(\mathrm{Te}(\mathrm{s})+3 \mathrm{F}_{2}(\mathrm{g}) \longrightarrow \mathrm{TeV}_{6}(\mathrm{g})\): $$ \frac{0.00792 \mathrm{~mol \: Te}}{0.0941 \mathrm{~mol \: F2}} = 0.0842 $$ The ideal stoichiometric ratio would be \(\frac{1}{3} = 0.333\ldots\). As the calculated value (0.0842) is less than this ratio, tellurium is the limiting reactant.

All the moles of tellurium react to form \(\mathrm{TeF}_6\), so the number of moles of \(\mathrm{TeF}_6\) is equal to the number of moles of tellurium: $$ \text{moles}_\text{TeF6} = 0.00792 \mathrm{~mol} $$

Dissolve the \(\mathrm{TeF}_6\) in 115 mL of water. Calculate the concentration by dividing the moles of \(\mathrm{Te}(\mathrm{OH})_{6}\) by the volume in liters: $$ \mathrm{concentration}_{\mathrm{Te}(\mathrm{OH})_{6}} = \frac{0.00792 \mathrm{~mol}}{0.115 \mathrm{~L}} = 0.0689 \mathrm{~M} $$ Since telluric acid has two acidic protons and we are given the two pKa values, we can find the total hydrogen ion concentration. First, use an ICE table to represent the acid dissociation reaction: $$ \begin{array}{c c c c} & \mathrm{H_6TeO_6}& \longrightarrow & \mathrm{H^+} + \mathrm{H_5TeO_6^-}\\ \text{Initial} & 0.0689 & 0 & 0 \\ \text{Change} & -x & +x & +x \\ \text{Equilibrium} & 0.0689 - x & x & x \end{array} $$ We can use the pKa₁ value (\(\mathrm{p} K_{\mathrm{a}_{1}}=7.68\)) to find the Ka₁: $$ \mathrm{Ka}_{1} = 10^{-\mathrm{p} K_{\mathrm{a}_{1}}} = 10^{-7.68} = 2.1 \times 10^{-8} $$ Now, write out the acid dissociation equation for the first dissociation using the Ka₁: $$ \mathrm{Ka}_{1} = \frac{\left[\mathrm{H^+}\right]\left[\mathrm{H_5TeO_6^-}\right]}{\left[\mathrm{H_6TeO_6}\right]} = \frac{x^2}{(0.0689 - x)} $$ However, since the \(\mathrm{p} K_{\mathrm{a}_{1}}\) is greater than 3, we can approximate the equilibrium concentrations as \(x \approx \left[\text{H}^+\right] << 0.0689\). Therefore, \(x^2 \approx \mathrm{Ka}_{1}(0.0689)\). Solve for x (which is the concentration of \(\mathrm{H^+}\) ions from the first dissociation step): $$ x \approx \sqrt{2.1 \times 10^{-8} (0.0689)} = 5.54 \times 10^{-5} $$ The concentration of \(\mathrm{H^+}\) ions from the second dissociation step is negligibly small due to the large pKa₂ value. Therefore, the total \(\mathrm{H^+}\) concentration in the solution is approximately \(5.54 \times 10^{-5}\) M.

Use the formula for the pH: $$ \mathrm{pH} = -\log{\left[\mathrm{H^+}\right]} = -\log{(5.54 \times 10^{-5})} $$ Calculate the pH: $$ \boxed{\mathrm{pH} \approx 4.26} $$ The pH of the solution of \(\mathrm{Te}(\mathrm{OH})_{6}\) formed by dissolving the isolated \(\mathrm{TeV}_{6}(\text{g})\) in the 115 mL solution is approximately 4.26.