3.6: ACIDS AND BASES Workbook pgs 145- 148 Buffered Solutions
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Transcript of 3.6: ACIDS AND BASES Workbook pgs 145- 148 Buffered Solutions
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3.6: ACIDS AND BASES… Workbook pgs 145- 148… Buffered Solutions…
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BUFFER SOLUTIONS Buffer
Causes solutions to be resistant to a change in pH when a strong acid or base is added
Mixture of a weak acid and salts of conjugate bases If you add 0.010 mol of NaOH to 1.0 L of pure water the
pH drops from 7 to 2 If you add 0.010 mol of NaOH to 1.0 L of blood (pH of 7.4)
the pH will drop to 7.3 Two requirements for a buffer:
An acid capable of reacting with added OH- ions and a base that can consume added H3O+ ions.
The acid and base must not react with each other (acetic acid and acetate ion or ammonia and
ammunium ions)
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BUFFERS If a small amount of hydroxide is added to an
equimolar solution of HF in NaF, for example, the HF reacts with the OH− to make F− and water.
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BUFFERS: Similarly, if acid is added, the F− reacts with
it to form HF and water.
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MATH TIME: What is the pH of a 0.700M solution of acetic acid?
Ka = 1.8 x 10-5
pH = 2.45 What is the pH of a 0.600M solution of sodium
acetate? Kb = 5.6 x 10-10
pH = 9.26 What is the pH of an acetic acid/sodium acetate
buffer with [CH3CO2H] = 0.700M and [CH3CO2-] =
0.600M? Ka = 1.8 x 10-5
pH = 4.68
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BUFFER CALCULATIONS: Consider the equilibrium constant expression
for the dissociation of a generic acid, HA:
HA + H2O H3O+ + A−
[H3O+] [A−][HA]
Ka =
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© 2009, Prentice-Hall, Inc.
BUFFER CALCULATIONS
Rearranging slightly, this becomes
[A−][HA]Ka = [H3O+]
Taking the negative log of both side, we get
[A−][HA]−log Ka = −log [H3O+] + −log
pKa
pHacid
base
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© 2009, Prentice-Hall, Inc.
BUFFER CALCULATIONS So
pKa = pH − log [base][acid]
• Rearranging, this becomes
pH = pKa + log [base][acid]
• This is the Henderson–Hasselbalch equation.
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HENDERSON-HASSELBALCH EQUATION: For when the concentrations of the acid and
conjugate base are quite large… … or when you want a short cut to a buffer
solution
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PRACTICE: 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? Ka for lactic acid is 1.4 10−4.
pH = pKa + log [base][acid]
pH = −log (1.4 10−4) + log(0.10)(0.12)
pH = 3.85 + (−0.08) = 3.77
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PRACTICE: Benzoic acid (C6H5CO2H, 2.00g) and sodium
benzoate (NaC6H5CO2, 2.00g) are dissolved in enough water to make 1.00L of solution. Calculate the pH of the solution using the Henderson-Hasselbalch equation. Ka = 6.3 x 10-5
2.00g benzoic acid = 0.0164 mol 2.00g sodium benzoate = 0.0139 mol pH = 4.20 + log(0.0139/0.0164) = 4.13
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PREPARING BUFFER SOLUTIONS Buffers work on a pH range:
The pH range is the range of pH values over which a buffer system works effectively.
It is best to choose an acid with a pKa close to the desired pH
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PREPARING BUFFER SOLUTIONS: To be useful, a buffer solution must have 2
characteristics: pH control:
pH = pKa + log[conjugatebbase]/[acid] Acid is chosen whose pKa is near the intended value of
the pH Exact value of pH is then achieved by adjusting the
acid-to-conjugate base ratio Buffer Capacity:
Buffer should have the ability to keep the pH approximately constant after the addition of reasonable amounts of acid and base.
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PREPARING BUFFER SOLUTIONS: You wish to prepare 1.0L of a buffer solution
with a pH of 4.30. A list of possible acids (and their conjugate bases) follows:
Which combination should be selected, and what should the ratio of acid to conjugate base be?
Acid Conjugate Base
Ka pKa
HSO4- SO4
- 1.2 x 10-2 1.92CH3CO2H CH3CO2
- 1.8 x 10-5 4.75HCO3 CO3
- 4.8 x 10-11 10.32
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PREPARING BUFFER SOLUTIONS:
[CH3CO2−]
[CH3CO2H]Ka = [H3O+]
[CH3CO2−]
[CH3CO2H]Ka =
[H3O+]
[CH3CO2−]
[CH3CO2H]5.0 x 10-5 =
1.8 x 10-5
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PREPARING BUFFER SOLUTIONS:
Ratio of conjugate base to acid = 0.36
pH = pKa + log [base][acid]
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PREPARING BUFFER SOLUTIONS: The relative number of moles of acid and
conjugate base is important in determining the pH of a buffer solution, not the concentration. Why?
Volume cancels out! That means that diluting a buffer solution will not
change its pH