Lecture 22 CH131 Fall 2020 12/10/2020 9:42 AM
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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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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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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
• Use BU login and Kerberos password
• Anonymous and seen by instructors only after grades submitted
• Comments in text fields especially valued and encouraged.
• Please try to answer all questions
• When done, please close your browser
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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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“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
“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℃.
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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“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
“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℃.
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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