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Chapter 17: Additional Aspects of Aqueous Equilibria
Section 2: Buffered Solutions
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Introduction
Solutions prepared with common ions have a tendency to resist drastic pH changes even when subjected to the addition of strong acids or bases.
Such solutions are called buffered solutions.
Blood is buffered to a pH of about 7.4.
Surface seawater is buffered to a pH between 8.1 and 8.3.
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Composition and Action
A buffer resists pH change because it contains … an acid to neutralize added OH− ions a base to neutralize added H+ ions
But we cannot have the acid and base consume one another through neutralization.
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Composition and Action
Therefore, we use weak conjugate acids and bases to prepare our buffers.
For example: HCH3COO and CH3COO−
HF and F−
NH4+ and NH3
HClO and ClO−
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Composition and Action
If we have a solution of HF and NaF, we have HF(aq) and F−(aq) in solution.
If we add a small amount of a strong acid, we use the F− to neutralize it:
F−(aq) + H+(aq) → HF(aq) If we add a small amount of a strong
base, we use the HF to neutralize it:
HF (aq) + OH−(aq) → F−(aq) + H2O(l)
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Composition and Action
Let’s consider another buffer composed of the weak acid HX and the strong electrolyte MX (M is a metal from Group 1, for example).
The acid dissociation involves both the weak acid and its conjugate base.
HX(aq) ⇄ H+(aq) + X−(aq)
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Composition and Action
Let’s consider another buffer composed of the weak acid HX and the strong electrolyte MX (M is a metal from Group 1, for example).
The acid dissociation involves both the weak acid and its conjugate base.
HX(aq) ⇄ H+(aq) + X−(aq) The acid dissociation constant is
Ka =
[H+][X−][HX]
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Composition and Action
Let’s consider another buffer composed of the weak acid HX and the strong electrolyte MX (M is a metal from Group 1, for example).
The acid dissociation involves both the weak acid and its conjugate base.
HX(aq) ⇄ H+(aq) + X−(aq) Solving for [H+]
[H+] = Ka
[HX][X−]
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Composition and Action
[H+] = Ka[HX][X−]
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Composition and Action
[H+] = Ka
The [H+] and pH depend on two things:
[HX][X−]
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Composition and Action
[H+] = Ka
The [H+] and pH depend on two things: Ka
the ratio of the concentrations of conjugate acid-base pair.
[HX][X−]
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Composition and Action
[H+] = Ka
If OH− ions are added, they will react with the acid component to produce water and X−.
[HX][X−]
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Composition and Action
[H+] = Ka
If OH− ions are added, they will react with the acid component to produce water and X−.
OH−(aq) + HX(aq) → H2O(l) + X−(aq)
[HX][X−]
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Composition and Action
[H+] = Ka
If OH− ions are added, they will react with the acid component to produce water and X−.
OH−(aq) + HX(aq) → H2O(l) + X−(aq) This causes a decrease in [HX] and
an increase in [X−].
[HX][X−]
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Composition and Action
[H+] = Ka
If OH− ions are added, they will react with the acid component to produce water and X−.
OH−(aq) + HX(aq) → H2O(l) + X−(aq) This causes a decrease in [HX] and an
increase in [X−]. So long as the amounts of HX and X− are
large, the ratio of [HX]/[X−] does not change.
[HX][X−]
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Composition and Action
[H+] = Ka
If H+ ions are added, they will react with the base component to produce the acid.
[HX][X−]
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Composition and Action
[H+] = Ka
If H+ ions are added, they will react with the base component to produce the acid.
H+(aq) + X−(aq) → HX (aq)
[HX][X−]
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Composition and Action
[H+] = Ka
If H+ ions are added, they will react with the base component to produce the acid.
H+(aq) + X−(aq) → HX (aq) This causes a decrease in [X−] and
an increase in [HX].
[HX][X−]
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Composition and Action
[H+] = Ka
If H+ ions are added, they will react with the base component to produce the acid.
H+(aq) + X−(aq) → HX (aq) This causes a decrease in [X−] and an
increase in [HX]. So long as the amounts of HX and X− are
large, the ratio of [HX]/[X−] does not change.
[HX][X−]
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Calculating the pH of a Buffer
We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer.
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Calculating the pH of a Buffer
We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer.
But, there is an alternative method.
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Calculating the pH of a Buffer
We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer.
But, there is an alternative method. We start with the expression for
finding [H+] of a buffer.
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Calculating the pH of a Buffer
We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer.
But, there is an alternative method. We start with the expression for
finding [H+] of a buffer.[HX][X−][H+] =
Ka
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Calculating the pH of a Buffer
We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer.
But, there is an alternative method. We then find the negative log of both
sides.[HX][X−][H+] =
Ka
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We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer.
But, there is an alternative method. We then find the negative log of both
sides.−log[H+] = −log(Ka
)
Calculating the pH of a Buffer
[HX][X−]
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We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer.
But, there is an alternative method. We then find the negative log of both
sides.−log[H+] = −logKa
− log
Calculating the pH of a Buffer
[HX][X−]
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We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer.
But, there is an alternative method. We note that −log[H+] = pH and
−logKa = pKa
−log[H+] = −logKa − log
Calculating the pH of a Buffer
[HX][X−]
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We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer.
But, there is an alternative method. We note that −log[H+] = pH and
−logKa = pKa
pH = pKa − log
Calculating the pH of a Buffer
[HX][X−]
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We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer.
But, there is an alternative method. We note that − log[HX]/[X−] = +
log[X−]/[HX]pH= pKa − log
Calculating the pH of a Buffer
[HX][X−]
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We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer.
But, there is an alternative method. We note that − log[HX]/[X−] = +
log[X−]/[HX]pH= pKa + log
Calculating the pH of a Buffer
[X−][HX]
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We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer.
But, there is an alternative method. We can generalize HX as an acid and
X− as a base.pH= pKa + log
Calculating the pH of a Buffer
[X−][HX]
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We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer.
But, there is an alternative method. We can generalize HX as an acid and
X− as a base.pH= pKa + log
Calculating the pH of a Buffer
[base][acid]
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We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer.
But, there is an alternative method. This is called the Henderson-
Hasselbach equation.pH= pKa + log
Calculating the pH of a Buffer
[base][acid]
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What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate, NaCH3CH(OH)COO? For lactic acid, Ka = 1.4 × 10−4.
Sample Exercise 17.3 (pg. 725)
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What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate, NaCH3CH(OH)COO? For lactic acid, Ka = 1.4 × 10−4.
Known:
Sample Exercise 17.3 (pg. 725)
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What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate, NaCH3CH(OH)COO? For lactic acid, Ka = 1.4 × 10−4.
Known: Ka = 1.4 × 10−4
Sample Exercise 17.3 (pg. 725)
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What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate, NaCH3CH(OH)COO? For lactic acid, Ka = 1.4 × 10−4.
Known: Ka = 1.4 × 10−4 ⇒ pKa = −logKa
Sample Exercise 17.3 (pg. 725)
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What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate, NaCH3CH(OH)COO? For lactic acid, Ka = 1.4 × 10−4.
Known: Ka = 1.4 × 10−4 ⇒ pKa = −logKa = 3.85
Sample Exercise 17.3 (pg. 725)
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What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate, NaCH3CH(OH)COO? For lactic acid, Ka = 1.4 × 10−4.
Known: Ka = 1.4 × 10−4 ⇒ pKa = −logKa = 3.85
[base] = 0.10 M
Sample Exercise 17.3 (pg. 725)
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What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate, NaCH3CH(OH)COO? For lactic acid, Ka = 1.4 × 10−4.
Known: Ka = 1.4 × 10−4 ⇒ pKa = −logKa = 3.85
[base] = 0.10 M[acid] = 0.12 M
Sample Exercise 17.3 (pg. 725)
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What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate, NaCH3CH(OH)COO? For lactic acid, Ka = 1.4 × 10−4.
Known: Ka = 1.4 × 10−4 ⇒ pKa = −logKa = 3.85
[base] = 0.10 M[acid] = 0.12 M
pH = pKa + log
Sample Exercise 17.3 (pg. 725)
[base][acid]
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What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate, NaCH3CH(OH)COO? For lactic acid, Ka = 1.4 × 10−4.
Known: Ka = 1.4 × 10−4 ⇒ pKa = −logKa = 3.85
[base] = 0.10 M[acid] = 0.12 M
pH = 3.85+ log
Sample Exercise 17.3 (pg. 725)
0.100.12
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What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate, NaCH3CH(OH)COO? For lactic acid, Ka = 1.4 × 10−4.
Known: Ka = 1.4 × 10−4 ⇒ pKa = −logKa = 3.85
[base] = 0.10 M[acid] = 0.12 M
pH = 3.85+ log = 3.85 + log(0.83)
Sample Exercise 17.3 (pg. 725)
0.100.12
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What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate, NaCH3CH(OH)COO? For lactic acid, Ka = 1.4 × 10−4.
Known: Ka = 1.4 × 10−4 ⇒ pKa = −logKa = 3.85
[base] = 0.10 M[acid] = 0.12 M
pH = 3.85+ log = 3.85 − 0.08
Sample Exercise 17.3 (pg. 725)
0.100.12
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What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate, NaCH3CH(OH)COO? For lactic acid, Ka = 1.4 × 10−4.
Known: Ka = 1.4 × 10−4 ⇒ pKa = −logKa = 3.85
[base] = 0.10 M[acid] = 0.12 M
pH = 3.85+ log = 3.85 − 0.08 = 3.77
Sample Exercise 17.3 (pg. 725)
0.100.12
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Use Sample Exercise 17.3 to help solve homework exercises 17.21 through 17.24 (pages 759 − 760).
Sample Exercise 17.3 (pg. 725)
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How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH4Cl.)
Sample Exercise 17.4 (pg. 726)
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How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH4Cl.)
Known:
Sample Exercise 17.4 (pg. 726)
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How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH4Cl.)
Known: Kb = 1.8 × 10−5
Sample Exercise 17.4 (pg. 726)
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How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH4Cl.)
Known: Kb = 1.8 × 10−5 ⇒
Sample Exercise 17.4 (pg. 726)
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How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH4Cl.)
Known: Kb = 1.8 × 10−5 ⇒ Ka = 5.6 × 10−10
Sample Exercise 17.4 (pg. 726)
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How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH4Cl.)
Known: Kb = 1.8 × 10−5 ⇒ Ka = 5.6 × 10−10
pH = 9.00
Sample Exercise 17.4 (pg. 726)
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How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH4Cl.)
Known: Kb = 1.8 × 10−5 ⇒ Ka = 5.6 × 10−10
pH = 9.00 ⇒
Sample Exercise 17.4 (pg. 726)
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How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH4Cl.)
Known: Kb = 1.8 × 10−5 ⇒ Ka = 5.6 × 10−10
pH = 9.00 ⇒ [H+] = 1.0 × 10−9
Sample Exercise 17.4 (pg. 726)
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How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH4Cl.)
Known: Kb = 1.8 × 10−5 ⇒ Ka = 5.6 × 10−10
pH = 9.00 ⇒ [H+] = 1.0 × 10−9
[base] = 0.10 M
Sample Exercise 17.4 (pg. 726)
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How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH4Cl.)
Known: Kb = 1.8 × 10−5 ⇒ Ka = 5.6 × 10−10
pH = 9.00 ⇒ [H+] = 1.0 × 10−9
[base] = 0.10 M
[H+] = Ka
Sample Exercise 17.4 (pg. 726)
[acid][base]
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How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH4Cl.)
Known: Kb = 1.8 × 10−5 ⇒ Ka = 5.6 × 10−10
pH = 9.00 ⇒ [H+] = 1.0 × 10−9
[base] = 0.10 M
[H+] = Ka ⇒ [acid] =
Sample Exercise 17.4 (pg. 726)
[acid][base]
[H+][base]Ka
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How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH4Cl.)
Known: Kb = 1.8 × 10−5 ⇒ Ka = 5.6 × 10−10
pH = 9.00 ⇒ [H+] = 1.0 × 10−9
[base] = 0.10 M
[acid] =
Sample Exercise 17.4 (pg. 726)
[H+][base]Ka
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How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH4Cl.)
Known: Kb = 1.8 × 10−5 ⇒ Ka = 5.6 × 10−10
pH = 9.00 ⇒ [H+] = 1.0 × 10−9
[base] = 0.10 M
[acid] = =
Sample Exercise 17.4 (pg. 726)
[H+][base]Ka
(1.0 × 10−9)(0.10)5.6 × 10−10
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How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH4Cl.)
Known: Kb = 1.8 × 10−5 ⇒ Ka = 5.6 × 10−10
pH = 9.00 ⇒ [H+] = 1.0 × 10−9
[base] = 0.10 M
[acid] = = 0.18 M
Sample Exercise 17.4 (pg. 726)
[H+][base]Ka
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How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH4Cl.)
Known: Kb = 1.8 × 10−5 ⇒ Ka = 5.6 × 10−10
pH = 9.00 ⇒ [H+] = 1.0 × 10−9
[base] = 0.10 M
[acid] = = 0.18 M = [NH4+]
Sample Exercise 17.4 (pg. 726)
[H+][base]Ka
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How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH4Cl.)
Known: Kb = 1.8 × 10−5 ⇒ Ka = 5.6 × 10−10
pH = 9.00 ⇒ [H+] = 1.0 × 10−9
[base] = 0.10 M
[NH4+] = 0.18 M
Sample Exercise 17.4 (pg. 726)
![Page 63: Chapter 17 : Additional Aspects of Aqueous Equilibria](https://reader037.vdocuments.us/reader037/viewer/2022102609/568139ce550346895da17f65/html5/thumbnails/63.jpg)
How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH4Cl.)
Known: Kb = 1.8 × 10−5 ⇒ Ka = 5.6 × 10−10
pH = 9.00 ⇒ [H+] = 1.0 × 10−9
[base] = 0.10 M
[NH4+] = 0.18 M ⇒
Sample Exercise 17.4 (pg. 726)
![Page 64: Chapter 17 : Additional Aspects of Aqueous Equilibria](https://reader037.vdocuments.us/reader037/viewer/2022102609/568139ce550346895da17f65/html5/thumbnails/64.jpg)
How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH4Cl.)
Known: Kb = 1.8 × 10−5 ⇒ Ka = 5.6 × 10−10
pH = 9.00 ⇒ [H+] = 1.0 × 10−9
[base] = 0.10 M
[NH4+] = 0.18 M ⇒ n = M × V
Sample Exercise 17.4 (pg. 726)
![Page 65: Chapter 17 : Additional Aspects of Aqueous Equilibria](https://reader037.vdocuments.us/reader037/viewer/2022102609/568139ce550346895da17f65/html5/thumbnails/65.jpg)
How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH4Cl.)
Known: Kb = 1.8 × 10−5 ⇒ Ka = 5.6 × 10−10
pH = 9.00 ⇒ [H+] = 1.0 × 10−9
[base] = 0.10 M
[NH4+] = 0.18 M ⇒ n = (0.18 M)(2.0 L)
Sample Exercise 17.4 (pg. 726)
![Page 66: Chapter 17 : Additional Aspects of Aqueous Equilibria](https://reader037.vdocuments.us/reader037/viewer/2022102609/568139ce550346895da17f65/html5/thumbnails/66.jpg)
How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH4Cl.)
Known: Kb = 1.8 × 10−5 ⇒ Ka = 5.6 × 10−10
pH = 9.00 ⇒ [H+] = 1.0 × 10−9
[base] = 0.10 M
[NH4+] = 0.18 M ⇒ n = 0.36 moles
Sample Exercise 17.4 (pg. 726)
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Use Sample Exercise 17.4 to help solve homework exercises 17.25 through 17.26 (page 760).
Sample Exercise 17.4 (pg. 726)
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There are two very important characteristics of buffers. the buffer capacity the effective pH range
Buffer Capacity and pH Range
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The buffer capacity is the amount of acid or base a buffer can neutralize before the pH begins to change to an appreciable degree.
Buffer capacity depends on … the amount of buffer materials the identity of the buffer materials
Buffer Capacity and pH Range
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The pH range of any buffer is the range of pH over which a buffer works to effectively neutralize an acid or a base.
The optimal situation is where the concentration of the buffer acid is equal to the concentration of the buffer base.
In that case, pH = pKa. Buffers usually have a pH range of about
±1 pH unit of pKa.
Buffer Capacity and pH Range
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Next, we’ll look at what happens when we add a strong acid or strong base to a buffer.
This is going to be a quantitative approach (that means numbers!). Remember …▪ Strong acids react with weak bases to completion.▪ H+ + ClO− → HClO
▪ Strong bases react with weak acids to completion.▪ OH− + CH3COOH → H2O + CH3COO−
Addition of Strong Acid or Base
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To calculate how the pH of a buffer changes when adding a strong acid or base, we follow the following strategy. First, consider the acid-base
neutralization reaction and determine its effect on [HX] and [X−].
Use the Ka and new values for [HX] and [X−] to calculate [H+].▪ Use i-c-e table or Henderson-Hasselbach
(easiest).
Addition of Strong Acid or Base
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A buffer is made by adding 0.300 mol CH3COOH and 0.300 mol CH3COONa to enough water to make 1.00 L of solution.
Sample Exercise 17.5 (page 728)
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A buffer is made by adding 0.300 mol CH3COOH and 0.300 mol CH3COONa to enough water to make 1.00 L of solution. The pH of the buffer is 4.74 (Sample Exercise 17.1).
Sample Exercise 17.5 (page 728)
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A buffer is made by adding 0.300 mol CH3COOH and 0.300 mol CH3COONa to enough water to make 1.00 L of solution. The pH of the buffer is 4.74 (Sample Exercise 17.1).
a) Calculate the pH of this solution after 0.020 mol NaOH is added.
Sample Exercise 17.5 (page 728)
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A buffer is made by adding 0.300 mol CH3COOH and 0.300 mol CH3COONa to enough water to make 1.00 L of solution. The pH of the buffer is 4.74 (Sample Exercise 17.1).
a) Calculate the pH of this solution after 0.020 mol NaOH is added.
b) For comparison, calculate the pH that would result if 0.020 mol NaOH were added to pure water.
Sample Exercise 17.5 (page 728)
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a) Calculate the pH of this solution after 0.020 mol NaOH is added.
First we need to do the neutralization.
Sample Exercise 17.5 (page 728)
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a) Calculate the pH of this solution after 0.020 mol NaOH is added.
First we need to do the neutralization.
OH− + CH3COOH → CH3COO− + H2O
Sample Exercise 17.5 (page 728)
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a) Calculate the pH of this solution after 0.020 mol NaOH is added.
First we need to do the neutralization.
OH− + CH3COOH → CH3COO− + H2O
Sample Exercise 17.5 (page 728)
ninitial 0.020 0.300 0.300
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a) Calculate the pH of this solution after 0.020 mol NaOH is added.
First we need to do the neutralization.
OH− + CH3COOH → CH3COO− + H2O
Sample Exercise 17.5 (page 728)
ninitial 0.020 0.300 0.300
nchg −0.020 −0.020 +0.020
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a) Calculate the pH of this solution after 0.020 mol NaOH is added.
First we need to do the neutralization.
OH− + CH3COOH → CH3COO− + H2O
Sample Exercise 17.5 (page 728)
ninitial 0.020 0.300 0.300
nchg −0.020 −0.020 +0.020
nfinal 0.000 0.280 0.320
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a) Calculate the pH of this solution after 0.020 mol NaOH is added.
First we need to do the neutralization.OH− + CH3COOH → CH3COO− +
H2O
This give us: [CH3COOH] = 0.280 M
Sample Exercise 17.5 (page 728)
ninitial 0.020 0.300 0.300
nchg −0.020 −0.020 +0.020
nfinal 0.000 0.280 0.320
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a) Calculate the pH of this solution after 0.020 mol NaOH is added.
First we need to do the neutralization.OH− + CH3COOH → CH3COO− +
H2O
This give us: [CH3COOH] = 0.280 M[CH3COO−] = 0.320 M
Sample Exercise 17.5 (page 728)
ninitial 0.020 0.300 0.300
nchg −0.020 −0.020 +0.020
nfinal 0.000 0.280 0.320
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a) Calculate the pH of this solution after 0.020 mol NaOH is added.
First we need to do the neutralization.OH− + CH3COOH → CH3COO− +
H2O
This give us: [CH3COOH] = 0.280 M[CH3COO−] = 0.320 M
Sample Exercise 17.5 (page 728)
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a) Calculate the pH of this solution after 0.020 mol NaOH is added.
First we need to do the neutralization.
This give us: [CH3COOH] = 0.280 M[CH3COO−] = 0.320 M
Sample Exercise 17.5 (page 728)
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a) Calculate the pH of this solution after 0.020 mol NaOH is added.
First we need to do the neutralization.
This give us: [CH3COOH] = 0.280 M[CH3COO−] = 0.320 M
Sample Exercise 17.5 (page 728)
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a) Calculate the pH of this solution after 0.020 mol NaOH is added.
First we need to do the neutralization.
This give us: [CH3COOH] = 0.280 M[CH3COO−] = 0.320 M
Using Henderson-Hasselbach:
Sample Exercise 17.5 (page 728)
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a) Calculate the pH of this solution after 0.020 mol NaOH is added.
First we need to do the neutralization.
This give us: [CH3COOH] = 0.280 M[CH3COO−] = 0.320 M
Using Henderson-Hasselbach:
pH = pKa + log
Sample Exercise 17.5 (page 728)
[base][base]
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a) Calculate the pH of this solution after 0.020 mol NaOH is added.
First we need to do the neutralization.
This give us: [CH3COOH] = 0.280 M[CH3COO−] = 0.320 M
Using Henderson-Hasselbach:
pH = 4.74+ log
Sample Exercise 17.5 (page 728)
[0.280]
[0.320]
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a) Calculate the pH of this solution after 0.020 mol NaOH is added.
First we need to do the neutralization.
This give us: [CH3COOH] = 0.280 M[CH3COO−] = 0.320 M
Using Henderson-Hasselbach:
pH = 4.74+ log(1.14)
Sample Exercise 17.5 (page 728)
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a) Calculate the pH of this solution after 0.020 mol NaOH is added.
First we need to do the neutralization.
This give us: [CH3COOH] = 0.280 M[CH3COO−] = 0.320 M
Using Henderson-Hasselbach:
pH = 4.74+ log(1.14) = 4.74 + 0.06
Sample Exercise 17.5 (page 728)
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a) Calculate the pH of this solution after 0.020 mol NaOH is added.
First we need to do the neutralization.This give us: [CH3COOH] = 0.280 M
[CH3COO−] = 0.320 M Using Henderson-Hasselbach:
pH = 4.74+ log(1.14) = 4.74 + 0.06 = 4.80
Sample Exercise 17.5 (page 728)
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b) For comparison, calculate the pH that would result if 0.020 mol NaOH were added to pure water.
Sample Exercise 17.5 (page 728)
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b) For comparison, calculate the pH that would result if 0.020 mol NaOH were added to pure water.
In pure water, NaOH is fully dissociated.
Sample Exercise 17.5 (page 728)
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b) For comparison, calculate the pH that would result if 0.020 mol NaOH were added to pure water.
In pure water, NaOH is fully dissociated. Therefore, [OH−] = 0.020 mol/1.00 L
Sample Exercise 17.5 (page 728)
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b) For comparison, calculate the pH that would result if 0.020 mol NaOH were added to pure water.
In pure water, NaOH is fully dissociated. Therefore, [OH−] = 0.020 mol/1.00 L = 0.020 M.
Sample Exercise 17.5 (page 728)
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b) For comparison, calculate the pH that would result if 0.020 mol NaOH were added to pure water.
In pure water, NaOH is fully dissociated. Therefore, [OH−] = 0.020 mol/1.00 L = 0.020 M.
pOH = −log[OH−]
Sample Exercise 17.5 (page 728)
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b) For comparison, calculate the pH that would result if 0.020 mol NaOH were added to pure water.
In pure water, NaOH is fully dissociated. Therefore, [OH−] = 0.020 mol/1.00 L = 0.020 M.
pOH = −log[OH−] = −log(0.020)
Sample Exercise 17.5 (page 728)
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b) For comparison, calculate the pH that would result if 0.020 mol NaOH were added to pure water.
In pure water, NaOH is fully dissociated. Therefore, [OH−] = 0.020 mol/1.00 L = 0.020 M.
pOH = −log[OH−] = −log(0.020) = 1.70
Sample Exercise 17.5 (page 728)
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b) For comparison, calculate the pH that would result if 0.020 mol NaOH were added to pure water.
In pure water, NaOH is fully dissociated. Therefore, [OH−] = 0.020 mol/1.00 L = 0.020 M.
pOH = −log[OH−] = −log(0.020) = 1.70pH = 14.00 − pOH
Sample Exercise 17.5 (page 728)
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b) For comparison, calculate the pH that would result if 0.020 mol NaOH were added to pure water.
In pure water, NaOH is fully dissociated. Therefore, [OH−] = 0.020 mol/1.00 L = 0.020 M.
pOH = −log[OH−] = −log(0.020) = 1.70pH = 14.00 − pOH = 14.00 − 1.70
Sample Exercise 17.5 (page 728)
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b) For comparison, calculate the pH that would result if 0.020 mol NaOH were added to pure water.
In pure water, NaOH is fully dissociated. Therefore, [OH−] = 0.020 mol/1.00 L = 0.020 M.
pOH = −log[OH−] = −log(0.020) = 1.70pH = 14.00 − pOH = 14.00 − 1.70 = 12.30
Sample Exercise 17.5 (page 728)