Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Chapter 20, Problem 119IP

While selenic acid has the formula H2SeO4 and thus is directly related to sulfuric acid, telluric acid is best visualized as H6TeO6 or Te(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 p K a 1 = 7.68 and p K a 2 = 11.29. Telluric add can be prepared by hydrolysis of tellurium hexafluoride according to the equation

TeF 6 ( g ) + 6H 2 O ( l ) Te ( OH ) 6 ( a q ) + 6HF ( a q )

Tellurium hexafluoride can be prepared by the reaction of elemental tellurium with fluorine gas:

Te ( s ) + 3F 2 ( g ) TeF 6 ( g )

If a cubic block of tellurium (density = 6.240 g/cm3) measuring 0.545 cm on edge is allowed to react with 2.34 L fluorine gas at 1.06 atm and 25°C, what is the pH of a solution of Te(OH)6 formed by dissolving the isolated TeF6(g) in 115 mL solution? Assume 100% yield in all reactions.

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The answers are to be stated for the given options.

Concept introduction: Oxidation state is a number that is assigned to each element in a compound. This number describes the ability to lose or gain electrons.

The pH is calculated using the formula,

pH=log10[H3O+]

To determine: The oxidation state of tellurium in Te(OH)6 .

Explanation of Solution

The oxidation state of tellurium in Te(OH)6 is +6_ .

Oxidation state of oxygen is 2 .

Oxidation state of hydrogen is +1 .

It is assumed that oxidation state of tellurium in Te(OH)6 is x .

The compound Te(OH)6 contains six oxygen and hydrogen atoms and single tellurium atom. Since, stable molecule has a zero overall charge. Therefore,

6(+1)+x+6(2)=0x=+6

Therefore, the oxidation state of tellurium in Te(OH)6 is +6_ .

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The answers are to be stated for the given options.

Concept introduction: Oxidation state is a number that is assigned to each element in a compound. This number describes the ability to lose or gain electrons.

The pH is calculated using the formula,

pH=log10[H3O+]

To determine: The pH value of a solution of Te(OH)6 formed by dissolving the isolated TeF6(g) in 115 mL solution.

Explanation of Solution

Given

Density of tellurium (Te)  is 6.24g/cm3 .

Length of tellurium edge is 0.545cm .

Since, the length of tellurium edge is 0.545cm . The volume of tellurium is,

(0.545cm)3=0.1619cm3

Formula

Mass of Te is calculated using the formula,

MassofTe=Density×Volume

Substitute the values of density and volume of tellurium in the above equation.

MassofTe=Density×Volume=6.24g/cm3×0.1619cm3=1.01g_

Moles of tellurium is 0.00791mol_ .

Mass of tellurium is 1.01g .

Atomic mass of tellurium is 127.6g/mol .

Formula

The number of moles of Te is calculated using the formula,

MolesofTe=MassofTeAtomicmassofTe

Substitute the values of mass and atomic mass of tellurium in the above equation.

MolesofTe=MassofTeAtomicmassofTe=1.01g127.6g/mol=0.00791mol_

Moles of fluorine is 0.101mol_ .

Given

Volume of fluorine gas is 2.34L .

Pressure is 1.06atm .

Temperature is 25°C .

The conversion of degree Celsius (°C) into Kelvin (K) is done as,

T(K)=T(°C)+273

Hence,

The conversion of 25°C into Kelvin is,

T(K)=T(°C)+273T(K)=(25+273)K=298K

Formula

Number of moles is calculated as,

PV=nRTn=PVRT

Where,

  • P is the total pressure.
  • V is the volume.
  • n is the total moles.
  • R is the universal gas constant (0.08206Latm/Kmol) .
  • T is the absolute temperature.

Substitute the values of P,V,R and T in the above equation.

n=PVRT=1.06atm×2.34L(0.08206Latm/Kmol)(298K)=0.101mol_

Molarity of Te(OH)6 is 0.06878M_ .

The reaction of formation of TeF6 is,

Te(s)+3F2(g)TeF6(g)

It is clear from the above equation that, one mole of tellurium reacts with three moles of fluorine to yield one mole of TeF6 . In this reaction, tellurium is limited, whereas fluorine gas is in excess. Hence, the chemical equivalence is,

Number of moles of tellurium is equal to the number of moles of TeF6 formed. The number of moles of TeF6 is 0.00791mol .

Since one mole of TeF6 leads to the formation of one mole of Te(OH)6 . Therefore, 0.00791mol mole of TeF6 leads to the formation of,

0.00791mol×1=0.00791molof Te(OH)6

Formula

The molarity of compound in a solution is calculated using the formula,

Molarityofcompound=MolesofcompoundVolumeofsolution

Substitute the values of number of moles of Te(OH)6 and volume of solution in the above equation.

Molarityofcompound=MolesofcompoundVolumeofsolution=0.00791mol0.115L=0.06878M_

Concentration of [H3O+] is 6.74×103M_ .

Given

The pKa1 value of Te(OH)6 is 7.68 .

The pKa2 value of Te(OH)6 is 11.29 .

Molarity of Te(OH)6 0.06878M .

The Ka value of HF is 6.6×104 . Since, HF is a stronger acid than Te(OH)6 , only HF will be dissociated into protons, whereas dissociation of Te(OH)6 will be suppressed.

The formula to calculate the concentration of proton [H3O+] of HF is,

[H3O+]=Ka(HF)×Molarityof HF

Substitute the values of Ka and molarity of HF in the above equation.

[H3O+]=Ka(HF)×Molarityof HF=6.6×104×0.06878M=6.74×103M_

Required pH value is 2.17_ .

Concentration of [H3O+] is 6.74×103M .

Formula

The pH is calculated using the formula,

pH=log10[H3O+]

Substitute the value of concentration of [H3O+] in the above equation.

pH=log10[H+]=log10[6.74×103]=2.17_

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