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Buffer SolutionsDr. Sobers’ Lecture Notes
Buffer Solutions
Buffer solutions resist changes to pH
A buffer solution contains a weak acid and a weak base (usually the conjugate base of the acid).
Example: an acetic acid and acetate ion buffer
Weak Acid
CH3CO2H
Weak Base
CH3CO2 -
Addition of strong base should raise the solution’s pH
+
WeakAcid
CH3CO2H +CH3CO2 - H2OStrongBase
HO-
The pH does not change significantly
Example: an acetic acid and acetate ion buffer
Weak AcidCH3CO2H
Weak BaseCH3CO2 -
The strong base is neutralized by the weak acid:
Addition of strong acid should lower the solution’s pH
+CH3CO2H H2O
The pH does not change significantly
Example: an acetic acid and acetate ion buffer
Weak Acid Weak BaseCH3CO2H CH3CO2 -
The strong acid is neutralized by the weak base:
StrongAcid
H3O+
WeakBase
CH3CO2 -+
Which combinations of acid and base may not be used to make a buffer solution?
HF and NaF HCN and KCN
HNO2 and NaNO2 HCl and NaCl
HCl is a strong acid. Its conjugate base is not actually basic.
Calculate the pH of a buffer solution prepared by dissolving 0.20 mole of cyanic acid (HCNO) and 0.80 mole of sodium cyanate (NaCNO) in enough water to make 1.0 liter of solution.
Ka(HCNO) = 2.0 × 10–4
Weak acid = HCNO Weak base = CNO-
Acid ionization for HCNO:
HCNO(aq) + H2O(l) ⇄ CNO- (aq) + H3O+ (aq)
Acid ionization for HCNO:
Ka = [H3O+][OCN-1]
[HCNO]
HCNO(aq) + H2O(l) ⇄ CNO- (aq) + H3O+ (aq)
Calculate the pH of a buffer solution prepared by dissolving 0.20 mole of cyanic acid (HCNO) and 0.80 mole of sodium cyanate (NaCNO) in enough water to make 1.0 liter of solution.
Ka(HCNO) = 2.0 × 10–4
Ka(HCNO) = 2.0 × 10–4
[ ]initial
[ ]final
Δ
0.20 0.80 0
-x +x +x
0.20 - x 0.80 + x x
HCNO(aq) + H2O(l) ⇄ CNO- (aq) + H3O+ (aq)
Ka = [H3O+][OCN-1]
[HCNO]
2.0 x 10-4 = (x)[0.80+x][0.20-x]
[ ]final xHCNO(aq) + H2O(l) ⇄ CNO- (aq) + H3O+ (aq)
0.20 - x 0.80 + x
2.0 x 10-4 ≈ (x)[0.8][0.20] [H3O+] = x = 5.0x10-5 M
pH = -log[H3O+] = 4.30
Ka = [H3O+][OCN-1]
[HCNO]
Henderson–Hasselbalch Equation
Lawrence Joseph Henderson
Henderson, L. J. (1908). Concerning the relationship between the strength of acids and their capacity to preserve neutrality. American Journal of Physiology, 21, 173–179.
Karl Albert Hasselbalch
Hasselbalch, K. A. (1917). Die Berechnung der Wasserstoffzahl des Blutes aus der freien und gebundenen Kohlensäure desselben, und die Sauerstoffbindung des Blutes als Funktion der Wasserstoffzahl. Biochemische Zeitschrift, 78, 112–144.
Henderson–Hasselbalch Equation
pH = pKa + log [base][acid]
[base] = concentration of the weak base[acid] = concentration of the weak acid
pH = -log [H3O+] pKa = -log Ka
The Meaning of pKa
Recall the pH scale:
The pKa indicates the strength of the acid
pH = -log [H3O+]
pKa = -log Ka
The lower the pH, the more acidic the solution
The lower the pKa, the stronger the acid
Acid and Base StrengthpKa = - log Ka
Acid Ka
CH3CO2H 1.8x10-5
NH4+ 5.6x10-10
HF 7.4x10-4
HCl 1.3x106
acid
stre
ngth
pKa
4.749.25
3.13-6.20
Stronger acids have smaller pKa values. The same for bases.
Derivation of Henderson–Hasselbalch Equation
Ka = [H3O+][An-1]
[HAn ]
HAn
WeakAcid
H2O ⇄ H2O+An-1 ++WeakBase
Ka = [H3O+][base]
[acid]
Equilibrium Expression:
Ka = [H3O+][base]
[acid]Take the cologarithm of each side:
-logKa = -log [H3O+][base]
[acid]⎛⎝⎜
⎞⎠⎟
-logKa = -log[H3O+] + -log [base]
[acid]⎛⎝⎜
⎞⎠⎟
log (AB) = logA + logB:
-logKa = -log[H3O+] + -log [base]
[acid]⎛⎝⎜
⎞⎠⎟
pKa = pH - log [base][acid]
⎛⎝⎜
⎞⎠⎟
pH = pKa + log [base][acid]
⎛⎝⎜
⎞⎠⎟
Calculate the pH of a buffer solution prepared by dissolving 0.20 mole of cyanic acid (HCNO) and 0.80 mole of sodium cyanate (NaCNO) in enough water to make 1.0 liter of solution.
pH = pKa + log [base][acid]
pH = 3.70 + log (0.80)(0.20)
Ka = 2.0 × 10–4
pKa = 3.70
pH = 4.30
Moles and MolarityWhen using the Henderson–Hasselbalch Equation, both the base and the acid are in the same solution.
pH = pKa + log [base][acid]
The ratio of the molarities is also the ratio of the moles (n) present. (Same solution volume)
pH = pKa + log nbase
nacid
⎛⎝⎜
⎞⎠⎟
Calculate the pH of a buffer solution prepared by dissolving 0.20 mole of cyanic acid (HCNO) and 0.80 mole of sodium cyanate (NaCNO) in enough water to make 1.2 liter of solution.
pH = 3.70 + log (0.80)(0.20)
pH = pKa + log nbase
nacid
⎛⎝⎜
⎞⎠⎟
pH = 4.30
What if the concentration of the weak acid and conjugate base are equal?
pH = pKa + log nbase
nacid
⎛⎝⎜
⎞⎠⎟
Example: An 800ml acetic acid/ sodium acetate buffer that contains 0.50 moles of each.
CH3CO2H Ka = 1.8x10-5 pKa = 4.75
pH = 4.75+ log 0.500.50
⎛⎝⎜
⎞⎠⎟ = 4.75
The pH of the buffer will be close to the pKa. It will be slightly higher or slightly lower depending on whether there is more acid or base present.
CH3CO2H Ka = 1.8x10-5 pKa = 4.75
Ratiobase / acid
1.0
pH of Buffer4.754.821.24.650.80
Equal amounts:Slightly more base:Slightly more acid:
Some more ProblemsThe pKa of acetic acid is 4.75. What is the ratio of acetate ion to acetic acid in a solution with a pH of 3.50?The pKa of acetic acid is 4.75. What ratio of acetate ion to acetic acid is required to make a solution with a pH of 3.50?
3.50 = 4.75+ log baseacid
⎛⎝⎜
⎞⎠⎟
Some more ProblemsThe pKa of acetic acid is 4.75. What is the ratio of acetate ion to acetic acid in a solution with a pH of 3.50?The pKa of acetic acid is 4.75. What ratio of acetate ion to acetic acid is required to make a solution with a pH of 3.50?
baseacid
⎛⎝⎜
⎞⎠⎟ = 0.056
Compare the change in pH on addition of 10ml of 1M HCl to 1.0 L of water and to 1.0 L of of a buffer containing 0.50 moles of both acetate and acetic acid (pKa = 4.75).
Assume volume change is small.
Assume volume change is small.
Moles of HCl (H3O+) added:
0.010L 1.0mol1L
⎛⎝⎜
⎞⎠⎟ = 0.010 moles
Compare the change in pH on addition of 10ml of 1M HCl to 1.0 L of water and to 1.0 L of of a buffer containing 0.50 moles of both acetate and acetic acid (pKa = 4.75).
Assume volume change is small.
Neutral water pH: 7.00
[H3O+] = 0.010/1.0L = 0.010 M
pH = -log [H3O+] = 2.00
ΔpH = 2.00 - 7.00 = -5.00
Compare the change in pH on addition of 10ml of 1M HCl to 1.0 L of water and to 1.0 L of of a buffer containing 0.50 moles of both acetate and acetic acid (pKa = 4.75).
+
WeakAcid
CH3CO2H +CH3CO2 - H2OStrongBase
HO-
+CH3CO2H H2OStrongAcid
H3O+
WeakBase
CH3CO2 -+
Which reaction occurs?
Compare the change in pH on addition of 10ml of 1M HCl to 1.0 L of water and to 1.0 L of of a buffer containing 0.50 moles of both acetate and acetic acid (pKa = 4.75).
Reaction Table:
+CH3CO2H H2OH3O+ CH3CO2 -+
Initial 0.010 0.50 0.50
react - 0.010 - 0.010 + 0.010
Final 0.49 0.51
pH = pKa + log nbase
nacid
⎛⎝⎜
⎞⎠⎟
+CH3CO2H H2OH3O+ CH3CO2 -+
Final 0.490 0.510
pH = 4.75+ log 0.490.51
⎛⎝⎜
⎞⎠⎟ = 4.73
ΔpH = 4.73 - 4.75 = -0.02