ACCESS Test Bank for Biochemistry Concepts and Connections 2nd Edition
Appling
Chapter 02
1. a. Equation 2.3 predicts that the interaction energy between the ions will be greater in the lower
dielectric medium; thus, the attraction between the Na+ and Cl− will be greater in pentane.
b. Since the length of the ionic bonds is 2 × 1.16 Å in each case (i.e., the value of r is the same for
Ca-F and Na-F), Equation 2.3 predicts that the interaction energy between the ion will be greater as
the values of q increase; thus, the attraction between the Ca2+ (q = 2) and F− (q = −1) will be greater
than the attraction between the Na+ (q = 1) and F− (q = −1).
c. Ca2+ will be bound more tightly by a –COOH group that is fully deprotonated. At pH = 3
the –COOH form will predominate. At pH = 4.2 the –COOH and –COO− forms will be in equal
concentration. At pH 8 the –COO− will predominate; thus, expect the greatest Ca2+ binding at pH = 8.
2. In the red/black pair, NH is the donor and O is the acceptor. This is the more likely interaction. In
the blue/black pair, the black NH is the donor and the blue N is the acceptor.
H
O
N
H
O
O
N H
N H
H
O
H
The resonance structure for formamide
makes H-bonding to oxygen more
likely than H-bonding to nitrogen
because the nitrogen lone pair is
N H
partially tied up in a double bond as
shown.
H
3. The curve shown in the question describes the situation where the water dipoles are parallel and
side by side. Assuming the antiparallel dipoles are also side by side (rather than head to tail), the
curve would be the mirror image reflected across the Energy = 0 line on the y-axis.
4. a. HCl is a strong acid; thus, [HCl] ≈ [H+] and pH = −log[H+] = −log[0.35] = 0.456.
b. Acetic acid is a weak acid with Ka = 1.74 × 10−5; thus, use ICE table to solve this problem:
1
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Appling Appling
Chapter 2 Chapter 2
H2O + CH3COOH H3O+ + CH3COO− x2
Ka = = 5.62 × 10−10
NH4+ H+ NH3 1− x
Initial 1M 0M 0M Solve for x = 2.371 × 10−5 = [H+]; thus, pH = −log(2.371 × 10−5) = 4.63
Change −x +x +x
Equilibrium 1−x M xM xM b. NH4+ is a weak acid with pKa = 9.25. Here, NaOH is consuming H+ from the NH4+ thus, use this
alternate version of the ICE table to solve for [HA] and [A−] after addition of NaOH
[CH 3COO − ][H3O + ] x2 (a source of –OH). Note: the activity of H2O is assumed to be unity (see Equation 2.7), so it does not
Ka = =
[CH 3COOH] [ 3
CH COOH ]− x appear in these calculations.
HA + −OH A− + H2O
Assume [CH3COO−] = [H3O+] and [CH3COOH] >> x; therefore: NH4+ −
OH NH3
Initial 0.040 mol 0.010 mol ~0
x2 Change −0.010 mol −0.010 mol +0.010 mol
1.74 × 10−5 =
0.35 − x Equilibrium 0.030 mol 0 0.010 mol
x = 2.47 × 10−3 = [ H 3O + ]
Solve for [H+] using the Henderson–Hasselbalch equation:
−3
pH = −log(2.47 × 10 ) = 2.61
This answer verifies the initial assumption that [CH3COOH] >> x. 0.010 mol
[ A− ] 0.050 L
c. Here [CH3COOH] >> x cannot be assumed, so use the ICE table approach with the quadratic pH = pK a + log = 9.25 + log = 9.25 + −0.602 = 8.85
[ HA] 0.040 mol
equation to solve this problem: 0.050 L
x2 c. Solve as in part (b)
Ka =
0.035 − x HA + −OH A− + H2O
Rearrange to 0 = x 2 + K a x − 0.035 K a = x 2 + (1.74 × 10 −5 ) x − (6.09 × 10 −7 ) NH4+ −
OH NH3
Solve using the quadratic equation: Initial 0.040 mol 0.030 mol ~0
−(1.74 × 10−5 ) ± (1.74 × 10−5 )2 − (4 × − 6.09 × 10−7 ) Change −0.030 mol −0.030 mol +0.030 mol
x= Equilibrium 0.010 mol 0 0.030 mol
2
x = 7.717 × 10−4 = [ H + ] ; thus, pH = −log[7.717 × 10−4] = 3.11 Solve for [H+] using the Henderson–Hasselbalch equation:
d. pH is lowest for the strongest acid (HCl) because it completely dissociates into H+ and conjugate 0.030 mol
base (i.e., Cl−). For acetic acid more H+ dissociates as the concentration of HA increases. This follows
[ A− ] 0.070 L
from Le Chatelier’s principle for the equilibrium: HA H + + A − . pH = pK a + log = 9.25 + log = 9.25 + 0.477 = 9.73
[ HA] 0.010 mol
0.070 L
5. a. See Table 2.6, which indicates that NH4+ is a weak acid with Ka = 5.62 × 10−10; thus, use the ICE
table to solve this problem:
HA H+ + A−
6. a. HA H+ + A−
Initial [] 0.2M 0 0
NH4+ H+ NH3
Change [] −x +x +x
Initial 1M 0M 0M
Equilibrium [] 0.2M−x +x +x
Change −x +x +x
x = 0.2M × 2% = 0.004M
Equilibrium 1−x M xM xM
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Appling Appling
Chapter 2 Chapter 2
And 9. a.
H 2 PO 4 − + H 2 O HPO 4 −2 + H 3O +
2 2
x 0.004 pK a = 6.86
Ka = = = 8.2 × 10−5
0.2 − x 0.2 − 0.004 pH = 7.00
[H 2 PO4 − ] = 0.1M
6. b. pH = − log[H + ] = − log[0.004] = 2.40
A−
7. The titration curve can be generated using a spreadsheet to perform multiple calculations of pH as a pH = pK a + log = 7.0
function of added titrant (as was done in problems 5b and 5c). The calculated pH values can then be HA
plotted as a function of added titrant. For Example: A−
7.0 = 6.86 + log
0.1
Total volume of pH Titration of weak acid with base A −
KOH added 0.14 = log
0mL 3.38 12.5 0.1
50mL 3.80 [ A− ] = 0.138M
10.0
150mL 4.38
250mL 4.76 7.5 9. b.
pH
350mL 5.14 [ H 2 PO4 − ] + [ HPO4 −2 ] = 0.3M
400mL 5.36 5.0
[ H 2 PO4 − ] = 0.3M − [ HPO4 −2 ]
450mL 5.71
475mL 6.04
2.5
HPO4 −2 0.3 − H 2 PO4 −
7.00 = pK a + log −
= 6.86 + log −
500mL 8.23 0.0 H 2 PO4 H 2 PO4
550mL 10.63 0 250 500 750 1000 1250
600mL 10.96 0.01M KOH added/mL
0.3 − H 2 PO4 −
750mL 11.30 log − = 0.14
800mL 11.41 H 2 PO4
1000mL 11.52 0.3 − H PO −
100.14 = 2
−
4
= 1.38
H 2 PO 4
8. a. Moles of acetic acid = (1 M) × (0.100 L) = 0.10 mol 0.3 − x
Moles of acetate = (0.5 M) × (0.100 L) = 0.050 mol 1.38 =
x
Total volume = 0.200 L; thus, [acetic acid] = 0.50 M, and [acetate] = 0.25 M
solve : x
[ A− ] 0.25 x = 0.126M = [ KH 2 PO4 ]
pH = pK a + log = 4.76 + log = 4.46
[ HA] 0.50 [ Na2 HPO4 ] = 0.174M
8. b. Moles of H3PO4 = (0.3 M) × (0.250 L) = 0.075 mol 10. Formic acid is a weak acid with pKa = 3.75. Since the formate buffer is at pH = pKa, [HA]=[A−]
Moles of H2PO4− = (0.8 M) × (0.250 L) = 0.20 mol initially. Here, KOH is consuming H+ from the HCOOH; thus, use this alternate version of the ICE
Total volume = 0.500 L; thus, [H2PO4−] = 0.150 M, and [HPO42−] = 0.40 M table to solve for [HA] and [A−] after addition of KOH (a source of –OH). Note: the activity of H2O is
assumed to be unity (see Equation 2.7), so it does not appear in these calculations.
[ A− ] 0.40
pH = pK a + log = 2.14 + log = 2.57
[ HA] 0.150
4 5
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Appling Appling
Chapter 2 Chapter 2
HA + −
OH A− + H2O 13. The relevant describes the dissociation of carbonic acid to bicarbonate ion and H+: pKa = 6.3.
−
HCOOH OH HCOO−
Initial 0.025 mol 0.005 mol 0.025 mol CO2 + H2O H2CO3 H+ + HCO3− H+ + CO32−
Change −0.005 mol −0.005 mol +0.005 mol
Equilibrium 0.020 mol 0 0.030 mol From the Henderson–Hasselbalch equation:
Solve for [H+] using the Henderson–Hasselbalch equation: [HCO3− ] [HCO3− ]
7.4 = 6.3 + log → = 12.56
[H 2 CO3 ] [H 2CO3 ]
0.030 mol mol H 2 CO3 1
0.505 L mole fraction H 2 CO3 = = = 0.0737
[ A− ] mol HCO3− + mol H 2 CO3 12.56 + 1
pH = pK a + log = 3.75 + log = 3.75 + 0.176 = 3.93
[ HA] 0.020 mol
0.505 L
Thus ~ 7.4% is H2CO3 and ~ 92.6% is HCO3− (as expected when pH > pKa). Since the pKa for
dissociation of HCO3− is 10.25, we expect the [CO32−] to be negligible at pH = 7.4.
11. The best choice would be a buffer of H2PO4− and HPO42− because the pKa for this conjugate
14. Protein molecules in aqueous solution become increasingly protonated as the pH decreases. Thus,
acid/base pair is 6.86, which is close to the target pH of 7.0.
proteins become more positively charged because carboxylic acids become less negatively charged as
pH drops, whereas amines become more positively charged. Proteins become more negatively
12. Glycine is a zwitterion at pH 7, so upon dissolving neutral solid glycine in water we have almost
entirely NH3+CH2COO−. This species is “HA” in the Henderson–Hasselbalch equation used below. charged as pH increases, because acidic groups become more negatively charged while the basic
Moles of glycine = (0.100 M) × (2.00 L) = 0.200 mol → 15.02 g of glycine required for 2 L of 0.100 groups become less positively charged.
M buffer.
15. a. Species III is the isoelectric species (not net charge); thus, the pKas that describe ionization
[ A− ] [ A− ] events that include this species will be used to calculate the pI. Since Species I is not one of these (and
9.0 = 9.6 + log → = 0.251
[ HA] [ HA] Species I is present at an insignificant concentration), we can ignore it for this simple case (i.e., a
molecule with only three ionizable groups).
Thus, at pH = 9.0, for every 1 mol of HA, there are 0.251 mol of A−. With this information the mole
fractions of A− and HA can be calculated: pK a1 + pK a 2 8.99 + 12.5
pI = = = 10.75
2 2
mol A− 0.251
mole fraction A− = = = 0.201
mol A− + mol HA 0.251 + 1 15. b. At pH = 9.20 the deprotonation of the α–carboxylic acid will be essentially 100% (pKa = 1.82 is
mol HA 1 >7 pH units below the pH 9.20; thus, the deprotonated form will predominate by 7 orders of
mole fraction HA = = = 0.799 magnitude). Thus, it is safe to assume a charge of −1 on the α–carboxylate at pH 9.20.
mol A− + mol HA 0.251 + 1
Thus, of the 0.200 mol of glycine in the buffer, 20.1% will be in the A− form and 79.9% will be At pH = 9.20 the deprotonation of the α–amino group will be closer to 50% (pKa = 8.99 is close to the
in the HA form. At pH 9.0, there will be (0.200 mol glycine) × (0.201) = 0.040 mol A− and pH 9.20; thus, the [HA] and [A−] will be within a factor of ten). The fractional charge on the α–amino
(0.200 mol glycine) × (0.799) = 0.160 mol HA. The source of A− is the titration of HA with NaOH; group can be calculated using the Henderson–Hasselbalch equation:
thus, to generate 0.040 mol of A−, 0.040 mol of NaOH are need. Since the molarity of NaOH is 1 M,
40 mL of NaOH are needed. [ A− ] [ A− ]
9.20 = 8.99 + log → = 1.62
[ HA] [ HA]
In summary, 15.02 g of glycine would be dissolved in water, 40 mL of 1 M NaOH would be added,
then more water would be added to bring the final volume to 2.00 L. In practice, the glycine would
be added to ~ 1.8 L water, and the 1 M NaOH would be added while monitoring the pH with a pH Thus, at pH = 9.2, for every 1 mol of HA, there are 1.62 mol of A−. With this information the mole
meter. Once pH 9.0 was achieved, water would be added to bring the final volume to 2.00 L. fractions of A− (i.e., –NH2) and HA (i.e., –NH3+) can be calculated:
6 7
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Appling
Chapter 02
1. a. Equation 2.3 predicts that the interaction energy between the ions will be greater in the lower
dielectric medium; thus, the attraction between the Na+ and Cl− will be greater in pentane.
b. Since the length of the ionic bonds is 2 × 1.16 Å in each case (i.e., the value of r is the same for
Ca-F and Na-F), Equation 2.3 predicts that the interaction energy between the ion will be greater as
the values of q increase; thus, the attraction between the Ca2+ (q = 2) and F− (q = −1) will be greater
than the attraction between the Na+ (q = 1) and F− (q = −1).
c. Ca2+ will be bound more tightly by a –COOH group that is fully deprotonated. At pH = 3
the –COOH form will predominate. At pH = 4.2 the –COOH and –COO− forms will be in equal
concentration. At pH 8 the –COO− will predominate; thus, expect the greatest Ca2+ binding at pH = 8.
2. In the red/black pair, NH is the donor and O is the acceptor. This is the more likely interaction. In
the blue/black pair, the black NH is the donor and the blue N is the acceptor.
H
O
N
H
O
O
N H
N H
H
O
H
The resonance structure for formamide
makes H-bonding to oxygen more
likely than H-bonding to nitrogen
because the nitrogen lone pair is
N H
partially tied up in a double bond as
shown.
H
3. The curve shown in the question describes the situation where the water dipoles are parallel and
side by side. Assuming the antiparallel dipoles are also side by side (rather than head to tail), the
curve would be the mirror image reflected across the Energy = 0 line on the y-axis.
4. a. HCl is a strong acid; thus, [HCl] ≈ [H+] and pH = −log[H+] = −log[0.35] = 0.456.
b. Acetic acid is a weak acid with Ka = 1.74 × 10−5; thus, use ICE table to solve this problem:
1
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Appling Appling
Chapter 2 Chapter 2
H2O + CH3COOH H3O+ + CH3COO− x2
Ka = = 5.62 × 10−10
NH4+ H+ NH3 1− x
Initial 1M 0M 0M Solve for x = 2.371 × 10−5 = [H+]; thus, pH = −log(2.371 × 10−5) = 4.63
Change −x +x +x
Equilibrium 1−x M xM xM b. NH4+ is a weak acid with pKa = 9.25. Here, NaOH is consuming H+ from the NH4+ thus, use this
alternate version of the ICE table to solve for [HA] and [A−] after addition of NaOH
[CH 3COO − ][H3O + ] x2 (a source of –OH). Note: the activity of H2O is assumed to be unity (see Equation 2.7), so it does not
Ka = =
[CH 3COOH] [ 3
CH COOH ]− x appear in these calculations.
HA + −OH A− + H2O
Assume [CH3COO−] = [H3O+] and [CH3COOH] >> x; therefore: NH4+ −
OH NH3
Initial 0.040 mol 0.010 mol ~0
x2 Change −0.010 mol −0.010 mol +0.010 mol
1.74 × 10−5 =
0.35 − x Equilibrium 0.030 mol 0 0.010 mol
x = 2.47 × 10−3 = [ H 3O + ]
Solve for [H+] using the Henderson–Hasselbalch equation:
−3
pH = −log(2.47 × 10 ) = 2.61
This answer verifies the initial assumption that [CH3COOH] >> x. 0.010 mol
[ A− ] 0.050 L
c. Here [CH3COOH] >> x cannot be assumed, so use the ICE table approach with the quadratic pH = pK a + log = 9.25 + log = 9.25 + −0.602 = 8.85
[ HA] 0.040 mol
equation to solve this problem: 0.050 L
x2 c. Solve as in part (b)
Ka =
0.035 − x HA + −OH A− + H2O
Rearrange to 0 = x 2 + K a x − 0.035 K a = x 2 + (1.74 × 10 −5 ) x − (6.09 × 10 −7 ) NH4+ −
OH NH3
Solve using the quadratic equation: Initial 0.040 mol 0.030 mol ~0
−(1.74 × 10−5 ) ± (1.74 × 10−5 )2 − (4 × − 6.09 × 10−7 ) Change −0.030 mol −0.030 mol +0.030 mol
x= Equilibrium 0.010 mol 0 0.030 mol
2
x = 7.717 × 10−4 = [ H + ] ; thus, pH = −log[7.717 × 10−4] = 3.11 Solve for [H+] using the Henderson–Hasselbalch equation:
d. pH is lowest for the strongest acid (HCl) because it completely dissociates into H+ and conjugate 0.030 mol
base (i.e., Cl−). For acetic acid more H+ dissociates as the concentration of HA increases. This follows
[ A− ] 0.070 L
from Le Chatelier’s principle for the equilibrium: HA H + + A − . pH = pK a + log = 9.25 + log = 9.25 + 0.477 = 9.73
[ HA] 0.010 mol
0.070 L
5. a. See Table 2.6, which indicates that NH4+ is a weak acid with Ka = 5.62 × 10−10; thus, use the ICE
table to solve this problem:
HA H+ + A−
6. a. HA H+ + A−
Initial [] 0.2M 0 0
NH4+ H+ NH3
Change [] −x +x +x
Initial 1M 0M 0M
Equilibrium [] 0.2M−x +x +x
Change −x +x +x
x = 0.2M × 2% = 0.004M
Equilibrium 1−x M xM xM
2 3
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Appling Appling
Chapter 2 Chapter 2
And 9. a.
H 2 PO 4 − + H 2 O HPO 4 −2 + H 3O +
2 2
x 0.004 pK a = 6.86
Ka = = = 8.2 × 10−5
0.2 − x 0.2 − 0.004 pH = 7.00
[H 2 PO4 − ] = 0.1M
6. b. pH = − log[H + ] = − log[0.004] = 2.40
A−
7. The titration curve can be generated using a spreadsheet to perform multiple calculations of pH as a pH = pK a + log = 7.0
function of added titrant (as was done in problems 5b and 5c). The calculated pH values can then be HA
plotted as a function of added titrant. For Example: A−
7.0 = 6.86 + log
0.1
Total volume of pH Titration of weak acid with base A −
KOH added 0.14 = log
0mL 3.38 12.5 0.1
50mL 3.80 [ A− ] = 0.138M
10.0
150mL 4.38
250mL 4.76 7.5 9. b.
pH
350mL 5.14 [ H 2 PO4 − ] + [ HPO4 −2 ] = 0.3M
400mL 5.36 5.0
[ H 2 PO4 − ] = 0.3M − [ HPO4 −2 ]
450mL 5.71
475mL 6.04
2.5
HPO4 −2 0.3 − H 2 PO4 −
7.00 = pK a + log −
= 6.86 + log −
500mL 8.23 0.0 H 2 PO4 H 2 PO4
550mL 10.63 0 250 500 750 1000 1250
600mL 10.96 0.01M KOH added/mL
0.3 − H 2 PO4 −
750mL 11.30 log − = 0.14
800mL 11.41 H 2 PO4
1000mL 11.52 0.3 − H PO −
100.14 = 2
−
4
= 1.38
H 2 PO 4
8. a. Moles of acetic acid = (1 M) × (0.100 L) = 0.10 mol 0.3 − x
Moles of acetate = (0.5 M) × (0.100 L) = 0.050 mol 1.38 =
x
Total volume = 0.200 L; thus, [acetic acid] = 0.50 M, and [acetate] = 0.25 M
solve : x
[ A− ] 0.25 x = 0.126M = [ KH 2 PO4 ]
pH = pK a + log = 4.76 + log = 4.46
[ HA] 0.50 [ Na2 HPO4 ] = 0.174M
8. b. Moles of H3PO4 = (0.3 M) × (0.250 L) = 0.075 mol 10. Formic acid is a weak acid with pKa = 3.75. Since the formate buffer is at pH = pKa, [HA]=[A−]
Moles of H2PO4− = (0.8 M) × (0.250 L) = 0.20 mol initially. Here, KOH is consuming H+ from the HCOOH; thus, use this alternate version of the ICE
Total volume = 0.500 L; thus, [H2PO4−] = 0.150 M, and [HPO42−] = 0.40 M table to solve for [HA] and [A−] after addition of KOH (a source of –OH). Note: the activity of H2O is
assumed to be unity (see Equation 2.7), so it does not appear in these calculations.
[ A− ] 0.40
pH = pK a + log = 2.14 + log = 2.57
[ HA] 0.150
4 5
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Appling Appling
Chapter 2 Chapter 2
HA + −
OH A− + H2O 13. The relevant describes the dissociation of carbonic acid to bicarbonate ion and H+: pKa = 6.3.
−
HCOOH OH HCOO−
Initial 0.025 mol 0.005 mol 0.025 mol CO2 + H2O H2CO3 H+ + HCO3− H+ + CO32−
Change −0.005 mol −0.005 mol +0.005 mol
Equilibrium 0.020 mol 0 0.030 mol From the Henderson–Hasselbalch equation:
Solve for [H+] using the Henderson–Hasselbalch equation: [HCO3− ] [HCO3− ]
7.4 = 6.3 + log → = 12.56
[H 2 CO3 ] [H 2CO3 ]
0.030 mol mol H 2 CO3 1
0.505 L mole fraction H 2 CO3 = = = 0.0737
[ A− ] mol HCO3− + mol H 2 CO3 12.56 + 1
pH = pK a + log = 3.75 + log = 3.75 + 0.176 = 3.93
[ HA] 0.020 mol
0.505 L
Thus ~ 7.4% is H2CO3 and ~ 92.6% is HCO3− (as expected when pH > pKa). Since the pKa for
dissociation of HCO3− is 10.25, we expect the [CO32−] to be negligible at pH = 7.4.
11. The best choice would be a buffer of H2PO4− and HPO42− because the pKa for this conjugate
14. Protein molecules in aqueous solution become increasingly protonated as the pH decreases. Thus,
acid/base pair is 6.86, which is close to the target pH of 7.0.
proteins become more positively charged because carboxylic acids become less negatively charged as
pH drops, whereas amines become more positively charged. Proteins become more negatively
12. Glycine is a zwitterion at pH 7, so upon dissolving neutral solid glycine in water we have almost
entirely NH3+CH2COO−. This species is “HA” in the Henderson–Hasselbalch equation used below. charged as pH increases, because acidic groups become more negatively charged while the basic
Moles of glycine = (0.100 M) × (2.00 L) = 0.200 mol → 15.02 g of glycine required for 2 L of 0.100 groups become less positively charged.
M buffer.
15. a. Species III is the isoelectric species (not net charge); thus, the pKas that describe ionization
[ A− ] [ A− ] events that include this species will be used to calculate the pI. Since Species I is not one of these (and
9.0 = 9.6 + log → = 0.251
[ HA] [ HA] Species I is present at an insignificant concentration), we can ignore it for this simple case (i.e., a
molecule with only three ionizable groups).
Thus, at pH = 9.0, for every 1 mol of HA, there are 0.251 mol of A−. With this information the mole
fractions of A− and HA can be calculated: pK a1 + pK a 2 8.99 + 12.5
pI = = = 10.75
2 2
mol A− 0.251
mole fraction A− = = = 0.201
mol A− + mol HA 0.251 + 1 15. b. At pH = 9.20 the deprotonation of the α–carboxylic acid will be essentially 100% (pKa = 1.82 is
mol HA 1 >7 pH units below the pH 9.20; thus, the deprotonated form will predominate by 7 orders of
mole fraction HA = = = 0.799 magnitude). Thus, it is safe to assume a charge of −1 on the α–carboxylate at pH 9.20.
mol A− + mol HA 0.251 + 1
Thus, of the 0.200 mol of glycine in the buffer, 20.1% will be in the A− form and 79.9% will be At pH = 9.20 the deprotonation of the α–amino group will be closer to 50% (pKa = 8.99 is close to the
in the HA form. At pH 9.0, there will be (0.200 mol glycine) × (0.201) = 0.040 mol A− and pH 9.20; thus, the [HA] and [A−] will be within a factor of ten). The fractional charge on the α–amino
(0.200 mol glycine) × (0.799) = 0.160 mol HA. The source of A− is the titration of HA with NaOH; group can be calculated using the Henderson–Hasselbalch equation:
thus, to generate 0.040 mol of A−, 0.040 mol of NaOH are need. Since the molarity of NaOH is 1 M,
40 mL of NaOH are needed. [ A− ] [ A− ]
9.20 = 8.99 + log → = 1.62
[ HA] [ HA]
In summary, 15.02 g of glycine would be dissolved in water, 40 mL of 1 M NaOH would be added,
then more water would be added to bring the final volume to 2.00 L. In practice, the glycine would
be added to ~ 1.8 L water, and the 1 M NaOH would be added while monitoring the pH with a pH Thus, at pH = 9.2, for every 1 mol of HA, there are 1.62 mol of A−. With this information the mole
meter. Once pH 9.0 was achieved, water would be added to bring the final volume to 2.00 L. fractions of A− (i.e., –NH2) and HA (i.e., –NH3+) can be calculated:
6 7
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