Question

While selenic acid has the formula ${\mathbf{\text{H}}}_{\mathbf{\text{2}}}{\mathbf{\text{SeO}}}_{\mathbf{\text{4}}}$and thus is directly related to sulfuric acid, telluric acid is best visualized as ${\mathbf{\text{H}}}_{\mathbf{\text{6}}}{\mathbf{\text{TeO}}}_{\mathbf{\text{6}}}\mathbf{\text{orTe}}{\mathbf{\left(}\mathbf{\text{OH}}\mathbf{\right)}}_{\mathbf{\text{6}}}$.

(a) What is the oxidation state of tellurium in $\mathbf{\text{Te}}{\mathbf{\left(}\mathbf{\text{OH}}\mathbf{\right)}}_{\mathbf{\text{6}}}$?

(b) Despite its structural differences with sulfuric and selenic acid, telluric acid is a diprotic acid with ${\mathbf{\text{pK}}}_{{\mathbf{\text{a}}}_{\mathbf{\text{1}}}}\mathbf{=}{\mathbf{\text{7.68andpK}}}_{{\mathbf{\text{a}}}_{\mathbf{\text{2}}}}\mathbf{=}\mathbf{\text{11.29}}$. Telluric acid can be prepared by hydrolysis of tellurium hexafluoride according to the equation

${\mathbf{\text{TeF}}}_{\mathbf{\text{6}}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}{\mathbf{\text{6H}}}_{\mathbf{\text{2}}}\mathbf{\text{O}}\mathbf{\left(}\mathit{l}\mathbf{\right)}\mathbf{\to }{\mathbf{\text{Te(OH)}}}_{\mathbf{\text{6}}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathbf{\text{6HF}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}$

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

$\mathbf{\text{Te}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathbf{\text{F}}}_{\mathbf{\text{2}}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\to }{\mathbf{\text{TeF}}}_{\mathbf{\text{6}}}\mathbf{\left(}\mathit{g}\mathbf{\right)}$

If a cubic block of tellurium (density localid="1663782855001" $\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{240}\raisebox{1ex}{$\mathbf{\text{g}}$}\!\left/ \!\raisebox{-1ex}{${\mathbf{\text{cm}}}^{\mathbf{\text{3}}}$}\right.$) measuring $\mathbf{\text{0.545cm}}$on edge is allowed to react with $\mathbf{\text{2.34L}}$fluorine gas at $\mathbf{\text{1.06atm}}$and $\mathbf{\text{25°C}}$, what is the $\mathbf{\text{pH}}$of a solution of $\mathbf{\text{Te}}{\mathbf{\left(}\mathbf{\text{OH}}\mathbf{\right)}}_{\mathbf{\text{6}}}$formed by dissolving the isolated ${\mathbf{\text{TeF}}}_{\mathbf{\text{6}}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$in $\mathbf{\text{115mL}}$solution? Assume $\mathbf{\text{100\%}}$ yield in all reactions.

Short answer

(a) As a result, the oxidation state of $\text{Te}$is $+6$.

(b) Therefore the role="math" localid="1663783001741" $\mathrm{pH}$value is $1.76$.

Step-by-step solution

Definition of oxidation state:

An oxidation state, also known as the oxidation number, is the charge that an atom would have if all of its links to other atoms were entirely ionic.

(a) Finding the oxidation state of Te:

The molecular formula for telluric acid is ${\text{H}}_{\text{6}}{\text{TeO}}_{\text{6}}$.

Oxygen has an oxidation state of $-2$, while hydrogen has an oxidation state of $+1$. ${\text{H}}_{\text{6}}{\text{TeO}}_{\text{6}}$is the chemical formula for telluric acid.

You can write an equation in which the $x$represents the oxidation state of $\text{Te}$in role="math" localid="1663825033237" ${\text{H}}_{\text{6}}{\text{TeO}}_{\text{6}}$.

$$\begin{gathered} 6(+1)+x+6(-2)=0 \\ 6+x-12=0 \\ x=+6 \end{gathered}$$

Therefore, the oxidation state of tellurium in ${\text{H}}_{\text{6}}{\text{TeO}}_{\text{6}}$is $+6$.

Hence, the oxidation state of $\text{Te}$is $+6$.

(b) Finding the pH Value:

The hydrolysis of tellurium hexafluoride produces telluric acid, as shown in the balanced reaction below.

${\text{TeF}}_{\text{6}}\left(g\right)+{\text{6H}}_{\text{2}}\text{O}\left(l\right)\to \text{Te}{\left(\text{OH}\right)}_{\text{6}}\left(aq\right)+\text{6HF}\left(aq\right)$

The reaction of elemental tellurium with fluorine gas produces tellurium hexafluoride.

$\text{Te}\left(s\right)+{\text{3F}}_{\text{2}}\left(g\right)\to {\text{TeF}}_{\text{6}}\left(g\right)$

Now, calculating the mass of tellurium

Tellurium's mass may be estimated using its density and volume.

$\begin{array}{rcl}\text{mass}\left(\text{Te}\right)& =& \text{d}\left(\text{Te}\right)\times \text{V}\left(\text{Te}\right)\\ & =& \text{0.24}\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{${\text{cm}}^{\text{3}}$}\right.\times {\text{0.1019cm}}^{\text{3}}\\ & =& \text{1.01g}\end{array}$

The moles of tellurium can then be calculated by dividing the mass of tellurium by the molar mass of tellurium.

$\begin{array}{rcl}\text{n}\left(\text{Te}\right)& =& \frac{\text{mass}}{\text{molarmass}}\\ & =& \frac{\text{1.01g}}{\text{127.6}{\displaystyle \raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.}}\\ & =& \text{0.00791mol}\end{array}$

You can calculate the number of moles of ${\text{F}}_{\text{2}}$from the given data with the ideal gas equation.

$\begin{array}{rcl}\text{n}\left({\text{F}}_{\text{2}}\right)& =& \frac{PV}{RT}\\ & =& \frac{\text{1.06atm}\times \text{2.34L}}{\text{0.08206}{\displaystyle \raisebox{1ex}{$\text{L}\cdot \text{atm}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}\cdot \text{K}$}\right.}\times \text{298K}}\\ & =& 0.101\text{mol}\end{array}$

According to the stoichiometric equation, one mole of$\text{Te}$reacts with of${\text{F}}_2$.

Therefore,$0.00791\text{mol}$of$\text{Te}$reacts with$0.02373\text{mol}$of${\text{F}}_2$.

Hence,$\text{Te}$is the limiting reactant.

The number of moles of${\text{TeF}}_6$formed is equal to the number of moles of$\text{Te}$.

Also, the number of moles of$\text{Te}{\left(\text{OH}\right)}_{\text{6}}$is equal to the number of moles of ${\text{TeF}}_6$.

Hence, the molarity of $\text{Te}{\left(\text{OH}\right)}_{\text{6}}$is,

$\begin{array}{rcl}\text{m}\left(\text{Te}{\left(\text{OH}\right)}_{\text{6}}\right)& =& \frac{\text{n}\left(\text{Te}{\left(\text{OH}\right)}_{\text{6}}\right)}{\text{V}}\\ & =& \frac{\text{0.00791mol}}{\text{0.115L}}\\ & =& \text{0.06878M}\end{array}$

Thus,$\text{Te}{\left(\text{OH}\right)}_{\text{6}}$is a weak acid and it has${\text{pK}}_{\text{a1}}=7.68{\text{andpK}}_{\text{a2}}=11.29$.

Here, $\text{HF}$has${\text{pK}}_a=3.14$hence it is a stronger acid than$\text{Te}{\left(\text{OH}\right)}_{\text{6}}$. Therefore, the dissociation of$\text{Te}{\left(\text{OH}\right)}_{\text{6}}$is suppressed by $\text{HF}$.

The number of moles of$\text{HF}$is$6$times greater than the number of moles of$\text{Te}{\left(\text{OH}\right)}_{\text{6}}$.

Hence, the molarity of $\text{HF}$is $\text{0.41208M}$.

$\begin{array}{rcl}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]& =& \sqrt{{\text{K}}_{\text{a}}\left(\text{HF}\right)\times \text{molarityofHF}}\\ & =& \sqrt{\text{7.2}\times {\text{10}}^{\text{-4}}\times \text{0.41268}}\\ & =& \text{0.01724M}\end{array}$

Therefore, the value of $\text{pH}$is,

$\begin{array}{rcl}\text{pH}& =& -\text{log}\left(\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\right)\\ & =& -\text{log}\left(\text{0.01724}\right)\\ & =& \text{1.76}\end{array}$

Hence, the $\text{pH}$of the solution of $\text{Te}{\left(\text{OH}\right)}_{\text{6}}$formed by dissolving the isolated role="math" localid="1663828363795" ${\text{TeF}}_{\text{6}}$ in$\text{115mL}$ solution is $\text{1.7}6$.