The Henderson-Hasselbalch Equation. The Henderson-Hasselbalch Equation the significance of pH the predominant solution species The significance of pH in solutions which contain many different acid/base pairs Diprotic Acids
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The Henderson-Hasselbalch Equation the significance of pH the predominant solution species
The significance of pH in solutions which contain many different acid/base pairs
Triprotic Acids Example - calculating concentrations of all phosphate containing species in a solution of KH2PO4 at known pH
pH buffer solutions - choosing the conjugate acid/base pair - calculating the pH
log10 Ka = log10 [H+] + log10
- pKa = - pH + log10
pH = pKa + log10
HA = H+ + A- Ka =
An especially convenient form of the equilibrium equation is obtained by re-writing the equilibrium expression using logs -
pH = pKa + log10
[HA]Use of the 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.
calculate ratio of (base/acid) concentrations
measured or set by buffer
known (from tabulations)
When the pH = pKa, half the conjugate acid/base pair is in the protonated form, half is de-protonated.
If the pKa = 4.7 (as for acetic acid):
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)
[CH3COOH]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.
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%)
pH = pKa + log10
[acid]How to Use the [base] / [acid] Ratio
[acetic acid] + [acetate] = 0.050 M
[acetate] / [acetic acid] = r
[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
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.
Knowledge of the pH completely determines the state of protonation / deprotonation of every Bronsted conjugate acid/base pair in solution.
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
pHEffect of pH on Solution Composition: H3PO4
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-, HPO42-, PO43-, and H+?
[acid]Calculating the Concentrations
pH = pKa + log
Apply H-H eqn to [H2PO4-] and [HPO42-] using pKa2 :
8.50 = 7.21 + log10 ( [HPO42-] / [H2PO4-] )
This gives [HPO42-] / [H2PO4-] = 20
Apply H-H eqn to [H3PO4] and [H2PO4-] using pKa1 :
8.50 = 2.12 + log10 ( [H2PO4-] / [H3PO4] )
This gives [H2PO4-] / [H3PO4] = 2.4 x 106
Apply H-H eqn to [PO43-] and [HPO42-] using pKa3 :
8.50 = 12.67 + log10 ( [PO42-] / [HPO42-] )
This gives [PO43-] / [HPO42-] = 6.8 x 10-5
Suppose that 0.010 mol NaOH is added to the buffer of part (a). What is the pH?
HAc + OH- = H2O + Ac- pKa = 4.75
(0.020 + 0.010) / 0.50
pH = pKa + log10
(0.050 - 0.010) / 0.50
As long as the buffer capacity is not exceeded, the change of pH is small, in this case, 4.35 to 4.63
1. pH buffers resist a change in pH upon addition of small amounts of either acid or base.
Buffer solutions should contain roughly equal concentrations of a conjugate acid and its conjugate base.
The conjugate acid/base pair of the buffer should have a pKa that approximately equals the pH.
For example: a buffer will result from mixing 0.1 M acetic acid and 0.1 M sodium acetate.
Added OH- is neutralized by the conjugate acid
Added H+ is neutralized by the conjugate base
H+ + C2H3O2- = HC2H3O2
OH- + HC2H3O2 = H2O + C2H3O2-
[acid]pH of Buffer Solutions
Prepare a pH 5.00 buffer using sodium acetate and acetic acid
ratio [base]/[acid] = 1.78
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
Use the data in Table 10.2 to design buffers at:
Find the weight of solid compounds you would use to produce 100 mL each buffer.