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?

Step by step solution

01

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: \(\mathrm{Te}(\mathrm{OH})_{6}: \; (+6) + 6 \times (-2 + 1) = 0\)

02

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) $$\text{Moles of Te} = \frac{\text{Mass}}{\text{Molar Mass}}$$ $$\text{Moles of Te} =\frac{0.906\, g}{127.6\, g/mol} = 0.0071\, mol$$

03

Determine the moles of fluorine gas

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

04

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: $$0.0071\, \text{mol}$$

05

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 \(\mathrm{TeF}_{6}(g)\): $$0.0071\, \text{mol}$$

06

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: $$\text{Concentration} = \frac{\text{Moles}}{\text{Volume}}$$ $$\text{Concentration} = \frac{0.0071 \,\text{mol}}{0.115\, L}= 0.062\, M$$

07

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 \(\mathrm{p} K_{a_{1}}\): $$\text{pH} = \mathrm{p} K_{a_{1}}= 7.68$$

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!