Buffer Solutions and HH Equation - Rick Sobers Solutions and HH Equation.pdfBuffer Solutions Buffer...

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


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


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]



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


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 = 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.


The ratio of the molarities is also the ratio of the moles (n) present. (Same solution volume)

pH = pKa + log nbase

nacid

⎛⎝⎜

⎞⎠⎟


pH = 3.70 + log (0.80)(0.20)


nacid

⎛⎝⎜

⎞⎠⎟

pH = 4.30

What if the concentration of the weak acid and conjugate base are equal?


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.


Moles of HCl (H3O+) added:

0.010L 1.0mol1L

⎛⎝⎜

⎞⎠⎟ = 0.010 moles



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


+

WeakAcid

CH3CO2H +CH3CO2 - H2OStrongBase

HO-

+CH3CO2H H2OStrongAcid

H3O+

WeakBase

CH3CO2 -+

Which reaction occurs?


Reaction Table:

+CH3CO2H H2OH3O+ CH3CO2 -+

Initial 0.010 0.50 0.50

react - 0.010 - 0.010 + 0.010

Final 0.49 0.51


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

Buffer Solutions and HH Equation - Rick Sobers Solutions and HH Equation.pdfBuffer Solutions Buffer...

Documents

Transcript of Buffer Solutions and HH Equation - Rick Sobers Solutions and HH Equation.pdfBuffer Solutions Buffer...