![Chapter 19 Ionic Equilibria in Aqueous Systems Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.](http://fsd.intolimp.org/html/2023/10/30/i_653f673333a31/img_phpx7sK07_L-19---bis_0.jpg)
Chapter 19
Ionic Equilibria in Aqueous Systems
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
![The effect of addition of acid or base to an unbuffered solution base added acid added The effect of addition of acid or base to a buffered solution acid added base added 2](http://fsd.intolimp.org/html/2023/10/30/i_653f673333a31/img_phpx7sK07_L-19---bis_1.jpg)
The effect of addition of acid or base to an unbuffered solution
base added
acid added
The effect of addition of acid or base to a buffered solution
acid added
base added
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![An acid-base buffer is a solution that lessens the impact on pH from the addition of acid or base. Most often, the components of a buffer are the conjugate acid-base pair of a weak acid (or base). Buffers work through a phenomenon known as the common-ion effect . CH 3 COOH( aq ) + H 2 O( l ) H 3 O + ( aq ) + CH 3 COO - ( aq ) If some CH 3 COO - ion is added, the equilibrium position shifts to the left; thus, [H 3 O + ] decreases, lowering the extent of acid dissociation. Similarly, if acetic acid is dissolved in a sodium acetate solution, acetate ion and H 3 O + ion from the acid enter the solution. The acetate ion already present in the solution prevents the acid from dissociating as much as it would in pure water, thus lowering [H 3 O + ]. In this case, acetate ion is called the common ion . The common-ion effect occurs when a given ion is added to an equilibrium mixture that already contains that ion, and the position of the equilibrium shifts away from forming more of it.](http://fsd.intolimp.org/html/2023/10/30/i_653f673333a31/img_phpx7sK07_L-19---bis_2.jpg)
An acid-base buffer is a solution that lessens the impact on pH from the addition of acid or base.
Most often, the components of a buffer are the conjugate acid-base pair of a weak acid (or base).
Buffers work through a phenomenon known as the common-ion effect .
CH 3 COOH( aq ) + H 2 O( l ) H 3 O + ( aq ) + CH 3 COO - ( aq )
If some CH 3 COO - ion is added, the equilibrium position shifts to the left; thus, [H 3 O + ] decreases, lowering the extent of acid dissociation.
Similarly, if acetic acid is dissolved in a sodium acetate solution, acetate ion and H 3 O + ion from the acid enter the solution. The acetate ion already present in the solution prevents the acid from dissociating as much as it would in pure water, thus lowering [H 3 O + ].
In this case, acetate ion is called the common ion .
The common-ion effect occurs when a given ion is added to an equilibrium mixture that already contains that ion, and the position of the equilibrium shifts away from forming more of it.
![2](http://fsd.intolimp.org/html/2023/10/30/i_653f673333a31/img_phpx7sK07_L-19---bis_3.jpg)
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![How a buffer works 2](http://fsd.intolimp.org/html/2023/10/30/i_653f673333a31/img_phpx7sK07_L-19---bis_4.jpg)
How a buffer works
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![The Henderson-Hasselbalch Equation 2 3 + - 3 + - K a 3 + K a - - 3 + K a [base] pH = p K a + log [acid] 6](http://fsd.intolimp.org/html/2023/10/30/i_653f673333a31/img_phpx7sK07_L-19---bis_5.jpg)
The Henderson-Hasselbalch Equation
2 3 + -
3 + -
K a
3 + K a
-
-
3 + K a
[base]
pH = p K a + log
[acid]
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![Buffer Capacity and Buffer Range Buffer capacity is the ability to resist pH change A buffer has the highest capacity when the component concentrations are equal (in other terms, its pH ≈ p K a of its acid component) Buffer range is the pH range over which the buffer acts effectively Buffers have a usable range within ± 1 pH unit of the p K a of its acid component. 6](http://fsd.intolimp.org/html/2023/10/30/i_653f673333a31/img_phpx7sK07_L-19---bis_6.jpg)
Buffer Capacity and Buffer Range
Buffer capacity is the ability to resist pH change
A buffer has the highest capacity when the component concentrations are equal (in other terms, its pH ≈ p K a of its acid component)
Buffer range is the pH range over which the buffer acts effectively
Buffers have a usable range within ± 1 pH unit of the p K a of its acid component.
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![The relation between buffer capacity and pH change 6](http://fsd.intolimp.org/html/2023/10/30/i_653f673333a31/img_phpx7sK07_L-19---bis_7.jpg)
The relation between buffer capacity and pH change
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![Colors and approximate pH range of some common acid-base indicators An acid-base indicator is a weak acid, with a different color than its conjugate base and with the color change occuring over a specific and narrow pH range. 6](http://fsd.intolimp.org/html/2023/10/30/i_653f673333a31/img_phpx7sK07_L-19---bis_8.jpg)
Colors and approximate pH range of some common acid-base indicators
An acid-base indicator is a weak acid, with a different color than its conjugate base and with the color change occuring over a specific and narrow pH range.
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![The color change of the indicator bromthymol blue Basic Acidic 6](http://fsd.intolimp.org/html/2023/10/30/i_653f673333a31/img_phpx7sK07_L-19---bis_9.jpg)
The color change of the indicator bromthymol blue
Basic
Acidic
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![6](http://fsd.intolimp.org/html/2023/10/30/i_653f673333a31/img_phpx7sK07_L-19---bis_10.jpg)
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![Curve for a weak acid-strong base titration. 6](http://fsd.intolimp.org/html/2023/10/30/i_653f673333a31/img_phpx7sK07_L-19---bis_11.jpg)
Curve for a weak acid-strong base titration.
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![Sample Problem 19.4 Calculate the pH during the titration of 40.00 mL of 0.1000 M propanoic acid (HPr; K a = 1.3 x 10 -5 ) after adding the following volumes of 0.1000 M NaOH: (a) 0.00 mL (b) 30.00 mL (c) 40.00 mL (d) 50.00 mL The amounts of HPr and Pr - will be changing during the titration. Remember to adjust the total volume of solution after each addition. Find the starting pH using the methods of Sample Problem 18.8. [Pr - ][H 3 O + ] [Pr - ] = x = [H 3 O + ] K a = [HPr] x = 1.1 x 10 -3 ; pH = 2.96 HPr( aq ) + OH - ( aq ) Pr - ( aq ) + H 2 O ( l ) - 0.004000 0 - Initial 0.003000 - Change - - Final 0 - 0.001000 0.003000 13](http://fsd.intolimp.org/html/2023/10/30/i_653f673333a31/img_phpx7sK07_L-19---bis_12.jpg)
Sample Problem 19.4
Calculate the pH during the titration of 40.00 mL of 0.1000 M propanoic acid (HPr; K a = 1.3 x 10 -5 ) after adding the following volumes of 0.1000 M NaOH:
(a) 0.00 mL
(b) 30.00 mL
(c) 40.00 mL
(d) 50.00 mL
The amounts of HPr and Pr - will be changing during the titration. Remember to adjust the total volume of solution after each addition.
Find the starting pH using the methods of Sample Problem 18.8.
[Pr - ][H 3 O + ]
[Pr - ] = x = [H 3 O + ]
K a =
[HPr]
x = 1.1 x 10 -3 ; pH = 2.96
HPr( aq ) + OH - ( aq ) Pr - ( aq ) + H 2 O ( l )
-
0.004000
0
-
Initial
0.003000
-
Change
-
-
Final
0
-
0.001000
0.003000
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![Sample Problem 19.4 continued (2 of 3) 0.001000 mol [H 3 O + ] = 1.3 x 10 -5 x = 4.3 x 10 -6 M pH = 5.37 0.003000 mol (c) When 40.00 mL of NaOH are added, all of the HPr will be reacted and the [Pr - ] will be (0.004000 mol) = 0.05000 M (0.004000 L) + (0.004000 L) 1.0 x 10 -14 K w K a x K b = K w K b = = = 7.7 x 10 -10 1.3 x 10 -5 K a K w [H 3 O + ] = = 1.6 x 10 -9 M pH = 8.80 14](http://fsd.intolimp.org/html/2023/10/30/i_653f673333a31/img_phpx7sK07_L-19---bis_13.jpg)
Sample Problem 19.4
continued (2 of 3)
0.001000 mol
[H 3 O + ] = 1.3 x 10 -5 x
= 4.3 x 10 -6 M
pH = 5.37
0.003000 mol
(c) When 40.00 mL of NaOH are added, all of the HPr will be reacted and the [Pr - ] will be
(0.004000 mol)
= 0.05000 M
(0.004000 L) + (0.004000 L)
1.0 x 10 -14
K w
K a x K b = K w
K b = = = 7.7 x 10 -10
1.3 x 10 -5
K a
K w
[H 3 O + ] = = 1.6 x 10 -9 M
pH = 8.80
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![Sample Problem 19.4 continued (3 of 3) (d) 50.00 mL of NaOH will produce an excess of OH - . mol excess base = (0.1000 M )(0.05000 L - 0.04000 L) = 0.00100 mol 1.0 x 10 -14 M = (0.00100 mol) M = 0.01111 [H 3 O + ] = = 9.0 x 10 -13 M (0.0900 L) 0.01111 pH = 12.05 15](http://fsd.intolimp.org/html/2023/10/30/i_653f673333a31/img_phpx7sK07_L-19---bis_14.jpg)
Sample Problem 19.4
continued (3 of 3)
(d) 50.00 mL of NaOH will produce an excess of OH - .
mol excess base = (0.1000 M )(0.05000 L - 0.04000 L) = 0.00100 mol
1.0 x 10 -14
M = (0.00100 mol)
M = 0.01111
[H 3 O + ] = = 9.0 x 10 -13 M
(0.0900 L)
0.01111
pH = 12.05
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![Curve for a weak base-strong acid titration 15](http://fsd.intolimp.org/html/2023/10/30/i_653f673333a31/img_phpx7sK07_L-19---bis_15.jpg)
Curve for a weak base-strong acid titration
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![Formation of acidic precipitation 15](http://fsd.intolimp.org/html/2023/10/30/i_653f673333a31/img_phpx7sK07_L-19---bis_16.jpg)
Formation of acidic precipitation
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