1. Buffers Buffers are solutions of a weak conjugate acid-base pair.
They are particularly resistant to pH changes, even when strong acid or base is added.
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2. 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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3. Buffers
Similarly, if acid is added, the F− reacts with it to form HF and water.
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4. 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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5. Buffer Calculations Rearranging slightly, this becomes
[HA]
Ka =
[H3O+]
Taking the negative log of both side, we get
base
[A−]
[HA]
−log Ka =
−log [H3O+] + (− log)
pKa pH
acid
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6. 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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7. Henderson–Hasselbalch Equation
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.
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8. Henderson–Hasselbalch Equation
pH = pKa + log
[base]
[acid]
pH = −log (1.4  10−4) + log
(0.10)
(0.12)
pH = (−0.08) pH pH
pH = 3.77
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9. 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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10. When Strong Acids or Bases Are Added to a Buffer…
…it is safe to assume that all of the strong acid or base is consumed in the reaction.
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11. Addition of Strong Acid or Base to a Buffer
Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution.
Use the Henderson–Hasselbalch equation to determine the new pH of the solution.
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12. Calculating pH Changes in Buffers
A buffer is made by adding mol HC2H3O2 and mol NaC2H3O2 to enough water to make 1.00 L of solution. The pH of the buffer is Calculate the pH of this solution after mol of NaOH is added.
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13. Calculating pH Changes in Buffers
Before the reaction, since
mol HC2H3O2 = mol C2H3O2−
pH = pKa = −log (1.8  10−5) = 4.74
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14. Calculating pH Changes in Buffers
The mol NaOH will react with mol of the acetic acid:
HC2H3O2(aq) + OH−(aq)  C2H3O2−(aq) + H2O(l)
HC2H3O2
C2H3O2−
OH−
Before reaction
0.300 mol
0.020 mol
After reaction
0.280 mol
0.320 mol
0.000 mol
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15. Calculating pH Changes in Buffers
Now use the Henderson–Hasselbalch equation to calculate the new pH:
pH = log
(0.320)
(0.200) pH
pH = 4.80
pH =
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16. Sample Exercise 17.3 Calculating the pH of a Buffer
What is the pH of a buffer that is 0.12 M in lactic acid [CH3CH(OH)COOH, or HC3H5O3] and 0.10 M in sodium lactate [CH3CH(OH)COONa or NaC3H5O3]? For lactic acid, Ka = 1.4 × 10-4.
Solution

17. Henderson–Hasselbalch equation can be used to calculate pH directly:
Sample Exercise 17.3 Calculating the pH of a Buffer
Solution (Continued)
NOTE:
Henderson–Hasselbalch equation can be used to calculate pH directly:

18. Sample Exercise 17.5 Calculating pH Changes in Buffers
A buffer is made by adding mol CH3COOH and 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 mol of NaOH is added. (b) For comparison, calculate the pH that would result if mol of NaOH were added to 1.00 L of pure water (neglect any volume changes).
Solution

19. Equilibrium Calculation:
Sample Exercise Calculating pH Changes in Buffers
Solution (Continued)
Equilibrium Calculation:
(b) To determine the pH of a solution made by adding mol of NaOH to 1.00 L of pure water, we can first determine pOH using Equation and subtracting from 14.
NOTE: Although the small amount of NaOH changes the pH of water significantly, the pH of the buffer changes very little.