Lecture 22 CH131 Fall 2020 12/10/2020 9:42...

Lecture 22 CH131 Fall 2020 12/10/2020 9:42 AM

Copyright © 2020 Dan Dill [email protected] 1

TP The reaction N2 𝑔 2 H2 𝑔 ⇌ 2 N2H4 𝑙 is endothermic. What temperature range will result in the greatest amount of products? Hint: Sketch ln 𝐾 vs 1/𝑇.

1. Very low 𝑇2. Very high T3. The amount will be the same at all 𝑇4. More information needed

Copyright © 2020 Dan Dill [email protected] 22 CH131 Fall 2020

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Lecture 22 CH131 Fall 2020Thursday, December 20, 2020

• Complete: Effect of temperature on equilibrium• Disturbing equilibriumLecture and discussion evaluationCh15: Acid-base equilibria• The pH of water• Weak acids and strong acids• Titration: Reaction of weak acid with strong base

TP The plot shows how ln 𝐾 vs 1/𝑇 for A ⇌ B. The standard entropy change of reaction, ∆ 𝑆°, is …

1. 02. 03. 04. More information is required


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TP The reaction N2 𝑔 2 H2 𝑔 ⇌ 2 N2H4 𝑙 is endothermic. What temperature range will result in the greatest amount of products? Hint: Sketch ln 𝐾 vs 1/𝑇.

1. Very low 𝑇2. Very high T3. The amount will be the same at all 𝑇4. More information needed


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Practice: Problem 14.61Lecture 22 CH131 Fall 2020

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Copyright © 2020 Dan Dill [email protected]

Disturbing equilibriumEssential idea: A system at equilibrium responds to a disturbance by partially offsetting the disturbance.


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Practice: Problem 14.52Lecture 22 CH131 Fall 2020

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Online course evaluation (lecture and discussion)• bu.campuslabs.com/courseeval of lecture and discussion

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Lecture 22 CH131 Fall 2020

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Chapter 15: Acid-base equilibria in aqueous solutions


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The pH of waterpH is defined as log H O .

The chemical equilibrium that accounts for the presence of H O in water is

H2O 𝑙 H2O 𝑙 ⇌ H3O 𝑎𝑞 OH 𝑎𝑞


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The pH of waterPure water at 50℃ is measured to have a pH of 6.63.

This means that the value of the reaction quotient of the water autoionization at equilibrium at 50℃ is

𝑄e 𝐾w H3O OH 10 6.63 2 5.48 10 14


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The pH of waterPure water at 25℃ is measured to have a pH of 7.00.

This means that the value of the reaction quotient of the water autoionization at equilibrium at 25℃ is

𝑄e 𝐾w H3O OH 10 7.00 2 1.00 10 14


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What is an acid?An acid makes H O OH .

An acid HA reacts with water as

HA 𝑎𝑞 H O 𝑙 ⇌ H O 𝑎𝑞 A 𝑎𝑞

Since 𝐾 is fixed at a given temperature , the increase in H O means there is a decrease in pH and a corresponding decrease in OH .


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What is a base?A base makes H O OH .

A base B reacts with water as

B 𝑎𝑞 H O 𝑙 ⇌ BH 𝑎𝑞 OH 𝑎𝑞

Since 𝐾 is fixed at a given temperature , the increase in OH means there is a corresponding decrease in H O and so increase in pH.


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What is a base?Hydroxides, when dissolved in water, also make H O OH .

For example,

NaOH 𝑠 → Na 𝑎𝑞 OH 𝑎𝑞

Since 𝐾 is fixed at a given temperature , the increase in OH means there is a corresponding decrease in H O and so increase in pH.


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TP The pH of pure water is different at different temperatures. This means that as temperature changes …

1. the relative amounts of H O 𝑎𝑞 and OH 𝑎𝑞 in pure water change

2. the acidity of pure water changes3. the value of the equilibrium constant changes4. All the above5. None of the above


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Strong acidsA strong acid reacts with water,

HA 𝑎𝑞 H O 𝑙 ⇌ H O 𝑎𝑞 A 𝑎𝑞 ,

with essentially 100% yield.


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Strong acidsThis means its equilibrium constant,

𝐾 ≫ 1

and that H O is the same as the molarity of the strong acid solution.

For example, ca 0.1 M, 𝐾a 1 10 , H3O 0.1 M.


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Weak acidsA weak acid reacts with water,

HA 𝑎𝑞 H O 𝑙 ⇌ H O 𝑎𝑞 A 𝑎𝑞 ,

with yield much less than 100%.


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Weak acidsThis means its equilibrium constant,

𝐾 ≪ 1

and therefore that H O must be determined by solving the ICE table.


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Weak acidsFor example, at 25℃, ca 0.1 M, 𝐾a 1 10 , H3O 0.001 M.


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“Titrating” a weak acidAn important problem is determining H3O as a result of adding strong base. to a weak acid.

There are two steps.

First, let the added base react with the acid present 100% as a limiting reagent problem.

Then, use the ICE table to solve the weak acid equilibrium


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“Titrating” a weak acid: Incomplete neutralization𝑉 100. mL of 𝑐 0.20 M of OH is combined with 𝑉 100. mL of 𝑐 0.40 M of HA, 𝐾 1.0 10 5 at 25℃.


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“Titrating” a weak acid: Incomplete neutralization𝑉 100. mL of 𝑐 0.20 M of OH is combined with 𝑉 100. mL of 𝑐 0.40 M of HA, 𝐾 1.0 10 5 at 25℃.


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HA 𝑎𝑞 H3O 𝑎𝑞 A 𝑎𝑞 𝑄Initial 0.10 10 7 0.10 10 7 𝐾

ChangeEquilibrium

Approximate

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“Titrating” a weak acid: Incomplete neutralizationVb 100. mL of cb 0.20 M of OH is combined with Va 100. mL of ca 0.40 M of HA, Ka 1.0 10 5 at 25℃.

H3O 𝑥 . .

.1.0 10


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HA 𝑎𝑞 H3O 𝑎𝑞 A 𝑎𝑞 𝑄Initial 0.10 10 7 0.10 10 7 𝐾

Change 𝑥 𝑥 𝑥Equilibrium 0.10 𝑥 10 𝑥 10 𝑥 1.0 10Approximate 0.10 𝑥 0.10 1.0 10

“Titrating” a weak acid: NeutralizationVb 100. mL of cb 0.40 M of OH is combined with Va 100. mL of ca 0.40 M of HA, 𝐾a 1.0 10 6 and 𝐾b 𝐾w / 𝐾a 1.0 10 8 at 25℃.


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A 𝑎𝑞 HA 𝑎𝑞 OH 𝑎𝑞 𝑄Initial

ChangeEquilibriumApproximate


OH 𝑥 𝐾b A 1/2 1.0 10 8 0.20 1/2 4.5 10 5

H3O 𝐾w / OH 1.0 10 14 / 4.5 10 5 2.2 10 10


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A 𝑎𝑞 HA 𝑎𝑞 OH 𝑎𝑞 𝑄Initial 0.20 0 10 7 0

Change 𝑥 𝑥 𝑥Equilibrium 0.20 𝑥 𝑥 10 7 𝑥 1.0 10 8

Approximate 0.20 𝑥 𝑥 1.0 10 8

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“Titrating” a weak acid: Excess base𝑉b 200. mL of 𝑐b 0.30 M of OH is combined with 𝑉a 200. mL of 𝑐a 0.20 M of HA, 𝐾a 1.0 10 6 and 𝐾b 1.0 10 8 at 25℃.


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A 𝑎𝑞 HA 𝑎𝑞 OH 𝑎𝑞 𝑄


OH 0.050 easy!

H3O Kw / OH 1.0 10 14 / 0.050 2.0 10 13

HA x Kb A- / OH- 1.0 10 8 0.10 / 0.050 2.0 10 8 tiny!


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A 𝑎𝑞 HA 𝑎𝑞 OH 𝑎𝑞 𝑄Initial 0.10 0 0.050 0

Change 𝑥 𝑥 𝑥Equilibrium 0.10 𝑥 𝑥 0.050 𝑥 1.0 10 8

Approximate 0.10 x 0.050 1.0 10 8

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Lecture 22 CH131 Fall 2020 12/10/2020 9:42...

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Transcript of Lecture 22 CH131 Fall 2020 12/10/2020 9:42...