The henderson hasselbalch equation
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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

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The Henderson-Hasselbalch Equation

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The henderson hasselbalch equation

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

Triprotic AcidsExample - 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


Henderson hasselbalch equation

[A-]

log10 Ka = log10 [H+] + log10

[HA]

[A-]

- pKa = - pH + log10

[HA]

[A-]

pH = pKa + log10

[HA]

Henderson-Hasselbalch Equation

[H+] [A-]

HA = H+ + A-Ka =

[HA]

An especially convenient form of the equilibrium equation is obtained by re-writing the equilibrium expression using logs -

Henderson-Hasselbalch Equation


Use of the henderson hasselbalch equation

[A-]

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.

Henderson-Hasselbalch Equation

calculate ratio of (base/acid) concentrations

measured or set by buffer

known (from tabulations)


Pk a and the dissociation of weak acids

=

=1

[A-]

[base]

[A-]

[HA]

[acid]

[HA]

pKa and the Dissociation of Weak Acids

For any conjugate acid/base pair,

HA = H+ + A-

pH = pKa + log10

Note that pH = pKa when

because log10 (1) = 0


The use of ph plots

mostly CH3COOH

Mostly CH3COO-

pKa=4.75

CH3COOH = H+ + CH3COO-pKa = 4.75

0 7 14

When pH < pKa, acetic acid is mostly protonated.

pH

When pH > pKa, acetic acid is mostly deprotonated.

When pH = pKa, concentrations of the protonated and deprotonated forms are equal.

The Use of pH plots


Ph pk a and dissociation

pH, pKa and % Dissociation

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)


An exercise in dissociation

pH = pKa + log10

[CH3COO-]

[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

7.00

4.75

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

pH = pKa + log10

r =

pH = pKa + log10

(r)

where

[base]

[base]

[acid]

[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


Summary significance of the ph

Summary: Significance of the pH

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.


Ratio of protonated and deprotonated species

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 h 2 co 3

0 7 14

pH

Diprotic Weak Acid - H2CO3

H2CO3 = H+ + HCO3-pKa1 = 6.4

HCO3-= H+ + CO32-pKa2 = 10.3

pKa1=6.4

pKa2=10.3

mostly H2CO3

mostly CO32-

mostly HCO3-


Triprotic weak acid h 3 po 4

0 7 14

pH

Triprotic Weak Acid - H3PO4

H3PO4 = H+ + H2PO4-pKa1 = 2.12

H2PO4- = H+ + HPO42-pKa2 = 7.21

HPO42- = H+ + PO43-pKa3 = 12.67

pKa1=2.12

pKa2=7.21

pKa3=12.67

mostly H3PO4

mostly H2PO4-

mostly HPO42-

mostly PO43-


Effect of ph on solution composition h 3 po 4

0 7 14

pH

Effect of pH on Solution Composition: H3PO4

pKa1=2.12

pKa2=7.21

pKa3=12.67

mostly H3PO4

mostly H2PO4-

mostly HPO42-

mostly PO43-

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-, HPO42-, PO43-, and H+?


Calculating the concentrations

[base]

[acid]

Calculating the Concentrations

pH = pKa + log

(1)

Apply H-H eqn to [H2PO4-] and [HPO42-] using pKa2 :

8.50 = 7.21 + log10 ( [HPO42-] / [H2PO4-] )

This gives [HPO42-] / [H2PO4-] = 20

(2)

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 :

(3)

8.50 = 12.67 + log10 ( [PO42-] / [HPO42-] )

This gives [PO43-] / [HPO42-] = 6.8 x 10-5


Answers to the exercise

0 7 14

pH

Answers to the Exercise

pKa1=2.12

pKa2=7.21

pKa3=12.67

mostly H3PO4

mostly H2PO4-

mostly HPO42-

mostly PO43-

solution

Ans:

[HPO42-] = 0.95 x10-1 M

[PO43-] = 1.6 x10-4 * [HPO42-]

[H3PO4] = 2.0 x106 * [H2PO4-]

[H2PO4-] = 0.047 x10-1 M

[H3O+] = 3.2 x10-8 M


Problem

(a)

Calculate the pH of a 500 mL solution prepared from:

0.050 mol of acetic acid and 0.020 mol sodium acetate.

HAc = H+ + Ac- pKa= 4.75

?

[Ac-]

pH = pKa + log10

[HAc]

4.75

?

?

Problem


Problem1

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

(0.020 + 0.010) / 0.50

[Ac-]

pH = pKa + log10

[HAc]

(0.050 - 0.010) / 0.50

4.75

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

pH Buffers

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-


Ph of buffer solutions

pH = pKa + log10

5.00

4.75

[base]

[acid]

pH of Buffer Solutions

Problem:

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


Exercises with buffers

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

Choosing a Weak Acid/Base Pair for a Buffer

6.9


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