Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question
![point is a buffer (since [HA] = [A]). From the graph above it can be seen that adding OH- does not significantly affect the
pH in the region from about 8 mL to 18 mL. The pH in this region is intermediate between that of the weak acid at the start
of the titration and the pH at the equivalence point.
In any solution containing a conjugate acid/base pair (HA and A-), this equilibrium occurs:
НА(аq) +
H2O(1)
А (аq) +
H3O*(aq)
[H3O*][A¯]
[HA]
for which the equilibrium constant is Ka
(where stronger acids have larger Ka values)
[A]
which can be rewritten as: pH = pka + log10HAT) where pKa =-log10(Ka).
[equation 1]
An effective buffer solution needs to be able to consume added acid and base so the concentrations of the acid and conjugate
base in the buffer need to be approximately equal: [A¯V[HA] - 1 which means the pH of the buffer will be close to the pKa
of the acid used to prepare it (log10(1) = 0 in equation 1).
When a small quantity of a strong acid is added to a buffer, it reacts with the A. For example, say you have a buffer made
from a weak acid with pKa = 5.20 and the concentrations of the weak acid and base are: [HA] = 0.130 M and [A¯] = 0.130
M. The pH of this buffer will be 5.20 (from equation 1). If you add 1.0 mL of 0.10 M HCI (which contains 0.00010 mol H*)
to 100 mL of the buffer the following reaction happens:
A(aq) +
H30*(aq, added) →
HA(aq) + H2O(1)
Initial quantities
0.0130
0.00010
0.0130
mol
Final quantities
0.0129
0.0131
mol
Hence, the amount of HA increases slightly while that of A is decreased by the same amount. The H* from the strong acid
is gone. The total volume of the solution is 101.0 mL and the concentrations are now: [HA] = 0.0131 mol/0.101 L= 0.1297
M and [A-]=0.0129 mol/0.101 L = 0.1277 M and the pH = 5.20 + log10(0.1277/0.1297) = 5.19. For comparison, if you add
1.0 mL of 0.10 M HCl to 100 mL of water, the pH of the solution changes from 7.0 to 3.0 (the actual concentration of H*
increases by a factor of about 10,000 times).
Buffers can be prepared for almost any pH by appropriate choice of a weak acid and conjugate base pair. Since a buffer is
most efficient at a pH close to the pKa of the acid (when the weak acid and conjugate base concentrations are approximately
equal), this provides a way to select the appropriate pair for the desired pH of the buffer. The following table gives some
common weak acid/base conjugate pairs and the pH range over which they can be used to prepare buffers.
Acid/Base Pair
pka of Acid
pH Range of Effective Buffers
H3PO4/H2PO4-
2.15
1.1 to 3.1
HF/F-
3.17
2.2 to 4.2
НСООН/НСО
3.74
2.7 to 4.7
4.76
CH3COOH/CH3CO2-
H2PO4-/HPO42-
NH4*/NH3
HCO3-/CO32-
HPO42-/PO43
3.8 to 5.8
7.20
5.9 to 7.9
9.24
8.2 to 10.2
10.33
9.3 to 11.3
12.38
11.4 to 13.4
Buffers are most effective at pH values in the range pka -1 to pKa + 1, e.g., a buffer based on NH4*/NH3 will only maintain
a stable pH in the range 8.24 to 10.24. The further the pH of the buffer is away from the pKa value, the less effective it is
because the concentrations of weak acid and conjugate basc must be uncqual to produce such pH's.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa99082a6-8fdc-4e0d-a6b2-be446400ef13%2F4ca4fa85-4f2d-48c0-93ff-903518038f5d%2Fgroq7nq_processed.png&w=3840&q=75)
Transcribed Image Text:point is a buffer (since [HA] = [A]). From the graph above it can be seen that adding OH- does not significantly affect the
pH in the region from about 8 mL to 18 mL. The pH in this region is intermediate between that of the weak acid at the start
of the titration and the pH at the equivalence point.
In any solution containing a conjugate acid/base pair (HA and A-), this equilibrium occurs:
НА(аq) +
H2O(1)
А (аq) +
H3O*(aq)
[H3O*][A¯]
[HA]
for which the equilibrium constant is Ka
(where stronger acids have larger Ka values)
[A]
which can be rewritten as: pH = pka + log10HAT) where pKa =-log10(Ka).
[equation 1]
An effective buffer solution needs to be able to consume added acid and base so the concentrations of the acid and conjugate
base in the buffer need to be approximately equal: [A¯V[HA] - 1 which means the pH of the buffer will be close to the pKa
of the acid used to prepare it (log10(1) = 0 in equation 1).
When a small quantity of a strong acid is added to a buffer, it reacts with the A. For example, say you have a buffer made
from a weak acid with pKa = 5.20 and the concentrations of the weak acid and base are: [HA] = 0.130 M and [A¯] = 0.130
M. The pH of this buffer will be 5.20 (from equation 1). If you add 1.0 mL of 0.10 M HCI (which contains 0.00010 mol H*)
to 100 mL of the buffer the following reaction happens:
A(aq) +
H30*(aq, added) →
HA(aq) + H2O(1)
Initial quantities
0.0130
0.00010
0.0130
mol
Final quantities
0.0129
0.0131
mol
Hence, the amount of HA increases slightly while that of A is decreased by the same amount. The H* from the strong acid
is gone. The total volume of the solution is 101.0 mL and the concentrations are now: [HA] = 0.0131 mol/0.101 L= 0.1297
M and [A-]=0.0129 mol/0.101 L = 0.1277 M and the pH = 5.20 + log10(0.1277/0.1297) = 5.19. For comparison, if you add
1.0 mL of 0.10 M HCl to 100 mL of water, the pH of the solution changes from 7.0 to 3.0 (the actual concentration of H*
increases by a factor of about 10,000 times).
Buffers can be prepared for almost any pH by appropriate choice of a weak acid and conjugate base pair. Since a buffer is
most efficient at a pH close to the pKa of the acid (when the weak acid and conjugate base concentrations are approximately
equal), this provides a way to select the appropriate pair for the desired pH of the buffer. The following table gives some
common weak acid/base conjugate pairs and the pH range over which they can be used to prepare buffers.
Acid/Base Pair
pka of Acid
pH Range of Effective Buffers
H3PO4/H2PO4-
2.15
1.1 to 3.1
HF/F-
3.17
2.2 to 4.2
НСООН/НСО
3.74
2.7 to 4.7
4.76
CH3COOH/CH3CO2-
H2PO4-/HPO42-
NH4*/NH3
HCO3-/CO32-
HPO42-/PO43
3.8 to 5.8
7.20
5.9 to 7.9
9.24
8.2 to 10.2
10.33
9.3 to 11.3
12.38
11.4 to 13.4
Buffers are most effective at pH values in the range pka -1 to pKa + 1, e.g., a buffer based on NH4*/NH3 will only maintain
a stable pH in the range 8.24 to 10.24. The further the pH of the buffer is away from the pKa value, the less effective it is
because the concentrations of weak acid and conjugate basc must be uncqual to produce such pH's.
![PRE-LAB CALCULATION
(Compulsory, must be completed to be allowed to start experiment. To be done before your lab class.)
The tables below show the solutions you will have in the laboratory and the pH of the buffer you will need to prepare (based
on your course and workspace). Select the reagents to prepare your buffer (refer to the table of pka values on the previous
page) and in the space below calculate the volumes of cach solution required to prepare 20.0 ml. of the buffer. Circle the
buffer pH and reagents you will use.
Solutions available to prepare buffers
0.10 M CH3COOH
0.10 M NaH2PO4
0.10 M NH4CI
0.10 M NaHCO3
Workspace
Buffer pH
Buffer pH
0.10 M CH3COON
0.10 M NazHPO4
0.10 M NH3
0.10 M NaɔCO3
(CHEM1011)
(CHEM1031/
CHEM1051)
My selected Buffer
1 and 7
4.76
4,50
pH
2 and 8
7.20
7.30
3 and 6
9.24
9.40
4 and 9
10.33
10.40
5 and 10
4.76
4.60
1. Choose the acid/base pair you will be using to make your buffer in the laboratory. Refer to the table of pKa's in the
introduction and the table of solutions available in the laboratory.
0.10 M NHẠCI
0.10 M NH3
2. Calculate the volumes of cach solution you will need to mix to make your buffer.
Use the pH of your buffer and the pk, of the acid component of your buffer to calculate [A¯[HA].
pH = pka + logio(
HA]
) which gives: log1o(HA = pH - pKa =
which means that
(HA]
Since the concentrations of the acids and conjugate bases are all the same in the solutions provided, the ratio
(A-V[HA] equals the ratio of the volume of base (V½) to the volume of acid (V,) (remember, A¯ is the base).
[A/HA] = V½/V½ = ,
and to simplify the algebra we will call this ratio 'C.
Volume of buffer = 20.0 mL., so V, + V½ = 20.0 or /, = 20.0 – Vz-
VWV2 =C=(20.0– VY) so that Cx V =(20.0 – V) which gives Va= 20.0 /(1+ C). Use this to calculate the
volumes of acid and base solutions required to make 20.0 mL of your assigned buffer.
Volume of acid (V)=
mL.
Volume of base (V) =
mL](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa99082a6-8fdc-4e0d-a6b2-be446400ef13%2F4ca4fa85-4f2d-48c0-93ff-903518038f5d%2Frl87nmk_processed.png&w=3840&q=75)
Transcribed Image Text:PRE-LAB CALCULATION
(Compulsory, must be completed to be allowed to start experiment. To be done before your lab class.)
The tables below show the solutions you will have in the laboratory and the pH of the buffer you will need to prepare (based
on your course and workspace). Select the reagents to prepare your buffer (refer to the table of pka values on the previous
page) and in the space below calculate the volumes of cach solution required to prepare 20.0 ml. of the buffer. Circle the
buffer pH and reagents you will use.
Solutions available to prepare buffers
0.10 M CH3COOH
0.10 M NaH2PO4
0.10 M NH4CI
0.10 M NaHCO3
Workspace
Buffer pH
Buffer pH
0.10 M CH3COON
0.10 M NazHPO4
0.10 M NH3
0.10 M NaɔCO3
(CHEM1011)
(CHEM1031/
CHEM1051)
My selected Buffer
1 and 7
4.76
4,50
pH
2 and 8
7.20
7.30
3 and 6
9.24
9.40
4 and 9
10.33
10.40
5 and 10
4.76
4.60
1. Choose the acid/base pair you will be using to make your buffer in the laboratory. Refer to the table of pKa's in the
introduction and the table of solutions available in the laboratory.
0.10 M NHẠCI
0.10 M NH3
2. Calculate the volumes of cach solution you will need to mix to make your buffer.
Use the pH of your buffer and the pk, of the acid component of your buffer to calculate [A¯[HA].
pH = pka + logio(
HA]
) which gives: log1o(HA = pH - pKa =
which means that
(HA]
Since the concentrations of the acids and conjugate bases are all the same in the solutions provided, the ratio
(A-V[HA] equals the ratio of the volume of base (V½) to the volume of acid (V,) (remember, A¯ is the base).
[A/HA] = V½/V½ = ,
and to simplify the algebra we will call this ratio 'C.
Volume of buffer = 20.0 mL., so V, + V½ = 20.0 or /, = 20.0 – Vz-
VWV2 =C=(20.0– VY) so that Cx V =(20.0 – V) which gives Va= 20.0 /(1+ C). Use this to calculate the
volumes of acid and base solutions required to make 20.0 mL of your assigned buffer.
Volume of acid (V)=
mL.
Volume of base (V) =
mL
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