While selenic acid has the formula H2 SeO4 and thus is directly related to sulfuric acid, tellur

While selenic acid has the formula H2 SeO4 and thus is...

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?

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?

Key Concepts

Concentration Determinations and pH Calculations

Calculating concentration, such as molarity, involves determining the number of moles of a substance dissolved in a given volume of solution. This is a precursor to pH calculations, where the degree of ionization (influenced by the acid dissociation constants) determines the hydrogen ion concentration in solution. Understanding these relationships is critical for predicting the acidity or basicity of a solution in acid–base chemistry.

Ideal Gas Law and Gas Phase Reactions

The ideal gas law (PV = nRT) relates the pressure, volume, temperature, and number of moles of a gas, enabling the calculation of gas amounts under known conditions. It is especially useful in reactions involving gaseous reagents, allowing chemists to determine how many moles are involved and to link gas phase reactions with subsequent chemical processes. This concept bridges gas behavior with stoichiometric and equilibrium calculations.

Stoichiometric Calculations and Limiting Reactants

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It involves calculating the amounts of substances consumed and formed using balanced chemical equations and identifying the limiting reactant — the reactant that is completely consumed first, thereby determining the maximum amount of product that can be produced. This concept is fundamental for yielding accurate predictions in synthetic and analytical chemistry.

Polyprotic Acid Behavior and Acid Dissociation Constants

Polyprotic acids can donate more than one proton (H+), and each dissociation step has its own acid dissociation constant (pKa value). In the context of diprotic acids like telluric acid, understanding how each deprotonation step contributes to the observed acid strength and pH of the solution is essential. This concept is central to acid–base equilibrium analysis, titration calculations, and predicting the behavior of acids in aqueous solution.

Oxidation State Determination

This concept involves assigning oxidation numbers to atoms within a molecule based on electronegativity and bonding rules. In compounds like Te(OH)6, one calculates the oxidation state of tellurium by considering the typical oxidation numbers of oxygen and hydrogen and ensuring that the overall charge of the molecule is balanced. This method is critical to understanding redox behavior, reactivity, and overall molecular structure in inorganic chemistry.

Transcript

00:03 For fun problem, we're considering tellurium, and we're told, we're given some information, we're told it's a diprotic acid.

00:16 And our first question, we're asked, what is the oxidation state of the tellurium? and we're considering it, i wrote this in the wrong spot.

00:25 This should have been right here.

00:28 So we've got tellurium, oh, h -6.

00:32 So tellurium has to have an oxidation state of plus six.

00:49 I'm going to write it like this.

00:53 Okay.

00:54 And then despite structural differences, we are told that tellurium is a diprodic acid and we're given the two pkk.

01:02 We're given two equations for how it's performed or how it's made right here.

01:08 And it's a two -step process where we use tellurium and fluorine.

01:14 We're told that we have a block of tellurium.

01:16 We're given the dimensions and it reacts with a known volume of fluorine at conditions of given pressure and temp as to find the ph.

01:28 Okay, so this is sort of a fun problem, like i said.

01:31 So let's figure out first our moles of tullerium.

01:36 Order to do that, i'm going to take my 0 .545 grams, cube that.

01:45 I'm going to multiply that by 6 .240 grams per cubic centimeter.

01:51 That was my given density.

01:54 Then i'm going to take that times the reciprocal of the molar mass.

02:01 And this will give me 7 .92 times 10 to the minus third moles.

02:10 And then let's figure out our moles of f2.

02:15 And we're going to use the n equals pv over rt.

02:20 So our n will equal pv was 2 .34 liters divided by r, be 0 .08206.

02:42 And this will be at 25 degrees t, that'll be 298 kelvin, and that'll equal 0 .101 moles of f2.

02:54 So let me underline these so i know what i'm doing.

03:02 Okay, so my balanced chemical equation requires 1 to 3 mole ratio.

03:14 And let's see what my mole ratio actually is.

03:17 My moles of f2 divided by my moles of te we need 3 to 1 and it's actually going to be 0 .101 moles divided by 7 .92 times 10 to the minus third moles and that'll equal 12 .8.

03:41 So i have plenty of my f2 which whops a tellurium is limiting...

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