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The Henderson-Hasselbalch Equation. The Henderson-Hasselbalch Equationthe significance of pHthe predominant solution species The significance of pH in solutions which contain many different acid/base pairs Diprotic Acids - PowerPoint PPT Presentation
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The Henderson-Hasselbalch Equation
1. The Henderson-Hasselbalch Equationthe significance of pH
the predominant solution species
2. The significance of pH in solutions which contain many different acid/base pairs
3. Diprotic Acids
4. Triprotic AcidsExample - calculating
concentrations of all phosphate containing species in a solution of KH2PO4 at known pH
5. pH buffer solutions- choosing the conjugate
acid/base pair - calculating the pH
Henderson-Hasselbalch Equation
HA = H+ + A- Ka = [H+] [A-][HA]
An especially convenient form of the equilibrium equation is obtained by re-writing the equilibrium expression using logs -
log10 Ka = log10 [H+] + log10
[A-
][HA]
- pKa = - pH + log10
[A-
][HA]
pH = pKa + log10
[A-
][HA]
Henderson-Hasselbalch
Equation
Use of the Henderson-Hasselbalch Equation
pH = pKa + log10
[A-
][HA]
Henderson-Hasselbalch
Equation
In most practical cases, the pH of the solution is known either from
(1) direct measurement with a pH meter
(2) use of a pH buffer in the solution
When the pH is known, the H.-H. Equation is much more convenient to use than the equilibrium constant expression. It immediately gives the ratio of concentrations of all conjugate acid/base pairs in the solution.
measured or set by buffer
known (from tabulations)
calculate ratio of (base/acid) concentrations
pKa and the Dissociation of Weak Acids
For any conjugate acid/base pair,
HA = H+ + A-
pH = pKa + log10
[A-]
[HA]
Note that pH = pKa when
[base]
[acid]
[A-]
[HA]=1=
because log10 (1) = 0
The Use of pH plots
pH
0 7 14
pKa=4.7
5
Mostly CH3COO-
mostly CH3COOH
CH3COOH = H+ + CH3COO- pKa = 4.75When pH < pKa, acetic acid is mostly protonated.
When pH > pKa, acetic acid is mostly deprotonated.When pH = pKa, concentrations of the protonated and deprotonated forms are equal.
pH, pKa and % Dissociation
When the pH = pKa, half the conjugate acid/base pair
is in the protonated form, half is de-
protonated.
At pH 4.7 [CH3COO-] / [CH3COOH] = 1 (equal)
pH 5.7 [CH3COO-] / [CH3COOH] = 10 (mostly de-protonated)
pH 3.7 [CH3COO-] / [CH3COOH] = 0.1 (mostly protonated)
If the pKa = 4.7 (as for acetic acid):
An Exercise in % Dissociation
A 0.050 M acetic acid solution is made pH 7.00 with added NaOH.
Find [CH3COOH], [CH3COO-], and [H+] in this solution.
pH = pKa + log10
[CH3COO-]
[CH3COOH]
4.75
7.00
ratio = 180
Thus, [CH3COO-] ≈ 0.050 M
[CH3COOH] ≈ 2.8 x 10-4 M
most of the acetic acid is dissociated (%
undissociated is 0.56%)
How to Use the [base] / [acid] Ratio
[base]
[acid]pH = pKa + log10
[acetic acid] + [acetate] = 0.050 M
[acetate] / [acetic acid] = r
[base]
[acid]r =
pH = pKa + log10
(r)
where
[acetic acid] = r * [acetate]
[acetic acid] (1 + r) = 0.050 M
[acetic acid] = 0.050 M / (1 + r) = 2.8 x 10-4 M [acetate] = 0.050 M - 2.8 x 10-4 M ≈ 0.050 M
1. The solution may contain many conjugate acid/base pairs (biological solutions usually do). In order to reproduce a sample, you need to reproduce the pH. This guarantees that all conjugate acid/base pairs will have the same ratio of protonated/deprotonated concentrations as in the original sample.
2. When the pH is known, you can readily calculate the ratio of (protonated/deprotonated) forms of any acid for which you know the pKa.
Summary: Significance of the pH
Ratio of protonated and deprotonated species
Knowledge of the pH completely determines the state of protonation / deprotonation of every Bronsted conjugate acid/base pair in solution.
An example:
A solution at pH 6.9 contains lactic acid (pKa = 3.9). Is lactic acid predominantly in the protonated form or the deprotonated form (lactate ion)?
Ans: Predominantly deprotonated. The ratio
[Lac-]/[HLac] = 103
Diprotic Weak Acid - H2CO3
H2CO3 = H+ + HCO3- pKa1
= 6.4
HCO3- = H+ + CO3
2- pKa2 = 10.3
pH
0 7 14
pKa1=6.
4 mostly CO3
2-
mostly H2CO3
pKa2=10.
3mostly HCO3
-
Triprotic Weak Acid - H3PO4
H3PO4 = H+ + H2PO4- pKa1
= 2.12
H2PO4- = H+ + HPO4
2- pKa2 = 7.21
HPO42- = H+ + PO4
3- pKa3 = 12.67
pH
0 7 14
pKa1=2.1
2mostly H3PO4
pKa2=7.2
1mostly H2PO4
-
mostly HPO4
2-mostly PO4
3-
pKa3=12.67
Effect of pH on Solution Composition: H3PO4
pH
0 7 14
pKa1=2.1
2mostly H3PO4
pKa2=7.2
1mostly H2PO4
-
mostly HPO4
2-
mostly PO4
3-
pKa3=12.6
7
Problem:
A solution is prepared by dissolving 0.100 mole of NaH2PO4 in water to produce 1.00 L of solution. The pH
is then adjusted to pH 8.50 with NaOH.
What are the concentrations of H3PO4, H2PO4-, HPO4
2-, PO4
3-, and H+?
Calculating the Concentrations
pH = pKa + log
[base]
[acid]
Apply H-H eqn to [H2PO4-] and [HPO4
2-] using pKa2 :
8.50 = 7.21 + log10 ( [HPO42-] /
[H2PO4-] )
This gives [HPO42-] / [H2PO4
-] = 20Apply H-H eqn to [H3PO4] and [H2PO4
-] using pKa1
:8.50 = 2.12 + log10 ( [H2PO4
-] / [H3PO4] )
This gives [H2PO4-] / [H3PO4] =
2.4 x 106Apply H-H eqn to [PO43-] and [HPO4
2-] using pKa3 :
8.50 = 12.67 + log10 ( [PO42-] /
[HPO42-] )
This gives [PO43-] / [HPO4
2-] = 6.8 x 10-5
(1)
(2)
(3)
Answers to the Exercise
pH
0 7 14
pKa1=2.12mostl
y H3PO4
pKa2=7.21mostl
y H2PO4
-
mostly HPO4
2-
mostly PO4
3-
pKa3=12.67
solution
Ans: [HPO42-] = 0.95 x10-1 M
[H2PO4-] = 0.047 x10-1
M
[PO43-] = 1.6 x10-4 *
[HPO42-]
[H3PO4] = 2.0 x106 * [H2PO4
-][H3O+] = 3.2 x10-8 M
Problem
Calculate the pH of a 500 mL solution prepared from:
0.050 mol of acetic acid and 0.020 mol sodium acetate.
(a)
HAc = H+ + Ac- pKa = 4.75
?
pH = pKa + log10
[Ac-]
[HAc]
4.75
?
?
Problem
Suppose that 0.010 mol NaOH is added to the buffer of
part (a). What is the pH?
(b)
HAc + OH- = H2O + Ac-pKa = 4.75
pH = pKa + log10
[Ac-][HAc]
4.75
(0.020 + 0.010) / 0.50
(0.050 - 0.010) / 0.50
4.63
As long as the buffer capacity is not exceeded, the change of pH is small, in
this case, 4.35 to 4.63
pH Buffers
1. pH buffers resist a change in pH upon addition of small amounts of either acid or base.
2. Buffer solutions should contain roughly equal concentrations of a conjugate acid and its conjugate base.
3. The conjugate acid/base pair of the buffer should have a pKa that approximately equals the pH.
H+ + C2H3O2- = HC2H3O2OH- + HC2H3O2 = H2O + C2H3O2
-
Added H+ is neutralized by the conjugate base
Added OH- is neutralized by the conjugate acid
For example: a buffer will result from mixing 0.1 M acetic acid and 0.1 M sodium acetate.
pH of Buffer Solutions
Problem:
Prepare a pH 5.00 buffer using sodium acetate and acetic acid
Any solution with this composition (i.e., this ratio of base / acid), will form a
buffer, but higher concentrations provide higher buffering capacity. For example, one
could use
0.178 M sodium acetate + 0.100 M acetic acid
pH = pKa + log10
[base]
[acid]
4.75
5.00 ratio
[base]/[acid] = 1.78
Exercises with Buffers
Use the data in Table 10.2 to design buffers at:
pH 6.9
pH 9.3
pH 3.6
Find the weight of solid compounds you would use to produce 100 mL each buffer.
Choosing a Weak Acid/Base Pair for a Buffer
6.9