ch 18 buffers

29
The Color Change of the Indicator Bromthymol Blue Fig. 19.5

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The Color Change of the Indicator Bromthymol Blue

Fig. 19.5

Acid-Base IndicatorsUsually dyes that are weak acids and display different

colours in protonated/deprotonated forms.

HIn(aq.) H+ (aq.) +In- (aq.)

In general we seek an indicator whose transition range (±1pH unit from the indicator pKa) overlaps the steepest part of the titration curve as closely as possible

HIn

In HK

-

a

Strong acids give weak bases.

Weak acids give strong bases.

CH3 CO

O H

+ HCH3 CO

O

H Cl + HCl

H C H

H

H

H C

H

H

+ H

-7

pKa

4.74

~50

Strong acids give weak bases.

Weak acids give strong bases.

CH3 CO

O H

+ HCH3 CO

O

H Cl + HCl

H C H

H

H

H C

H

H

+ H

acidity

basicity

Colors and Approximate pH Range of Some Common Acid-Base Indicators

Which molecule is the stronger acid, ethanol or acetic acid?

CH3 C

O

O

acetic acid

ethanol

Ka

10-16

10-4.74

more stable anion because of resonance and inductive effects

the stronger acid

C C

O

OH

H

H

H

H+C C

O

O

H

H

H

Predict whether trifluoroacetic acid will be a stronger or weaker acid than acetic acid.

C C

O

OH

F

F

F

trifluoroacetic acid

acetic acid

H+C C

O

O

F

F

F

Ka

10-0.23

10-4.74

more acidic acidmore acidic acid

Fluorine is more electro-Fluorine is more electro-negative than hydrogen. negative than hydrogen. Anion is more stable.Anion is more stable.

pKpKaa

4848

3838

15.715.7

C H

H

H

H C

H

H

H + H

N H

H

H N

H

H + H

O

H

O

H

+ HH

increasing increasing electronegativitelectronegativityy

It is a general principle that the more stable the anion the more acidic is the acid.

The principle is also successful across a row of the periodic table.

Biological fluids are often buffered (constant pH) an it is useful to know the predominant species present at a given pH.

In chemistry, particularly biology, a large number of compounds are acids and bases.

HO

HO

NH2HO

HO

NH3

H+

CO2HHO

CO2H CO2H

+ H

CO2HO

CO2H CO2H

dopaminedopamine

citric acidcitric acid

at pH = 4.74

Consider acetic acid with a Ka = 10-4.74

at lower pH, more acidic than 4.74, acetic acid is the major species present

[CH3CO2H] [CH3CO2]=

H

at pH = 4.74

Consider acetic acid with a Ka = 10-4.74

at lower pH, more acidic than 4.74, acetic acid is the major species present

[CH3CO2H] [CH3CO2] >

H

[CH3CO2H]

at pH = 4.74

Consider acetic acid with a Ka = 10-4.74

at lower pH, more acidic than 4.74, acetic acid is the major species present

at higher pH, less acidic than 4.74, acetate ion is the major species present

[CH3CO2H] [CH3CO2]<

[CH3CO2H]

[CH3CO2]

OH

The pH of blood is maintained at 7.4If the pH of blood was 4.74 then the acetate ion would be equal to the acetic acid ion concentration.

If acetic acid is introduced into the blood what will be the predominant species present? Will it be acetate ion or acetic acid?

[HOCCH3] =

O

[ OCCH3]

O

at pH = 4.74

If the pH is raised to 7.4 will the concentration of acetate ion increase or decrease?

If acetic acid is introduced into the blood what will be the predominant species present? Will it be acetate ion or acetic acid?

[HOCCH3] =

O

[ OCCH3]

O

pH 4.74 7.4

If the pH is raised to 7.4 (more basic) will the concentration of acetate ion increase or decrease?

O H

If acetic acid is introduced into the blood what will be the predominant species present? Will it be acetate ion or acetic acid?

[HOCCH3] =

O

[ OCCH3]

O

at pH = 4.74

If the pH is raised to 7.0 will the concentration of acetate ion increase or decrease?

Acetate ion is the major species present if acetic acid is introduced into blood.

The Henderson-Hasselbalch Equation

Take the equilibrium ionization of a weak acid:

HA(aq) + H2O(aq) = H3O+(aq) + A-

(aq) Ka =[H3O+] [A-]

[HA]Solving for the hydronium ion concentration gives:

[H3O+] = Ka x[HA][A-]

Taking the negative logarithm of both sides:

-log[H3O +] = -log Ka - log ( )[HA] [A-]

pH = -log Ka - log( )[HA] [A-]

Generalizing for any conjugate acid-base pair :

pH = -log Ka + log ( )[base][acid]

Henderson-Hasselbalch equation

H-A +H A

when [H-A] [ A ]=

a useful concept:

p H = -log [H]

Biological fluids are often buffered (constant pH) an it is useful to know the predominant species present at a given pH.

pH = pKa + log [A-][HA]

HENDERSON-HASSELBALCH EQUATIONHENDERSON-HASSELBALCH EQUATION

[HA]

]][A[HKa

[HA]]][A[H

logKlog- a

[HA]][A

log][Hlog

For acids:

For bases:][BH

[B]logpKpH a

B + H2O BH+ + OH-

acidbase acid base

Ka

Kb

pKa applies to this acid

When [A-] = [HA], pH = pKa

Derivation: HA H+ + A-

[HA]

][AlogpHpKa

[HA]][A

logpKpH a

[HA]][A

logpKpH a

BUFFERS

Mixture of an acid and its conjugate base.

Buffer solution resists change in pH when acids or bases are added or when dilution occurs.

Mix:

A moles of weak acid + B moles of conjugate base

Find:

• moles of acid remains close to A, and

• moles of base remains close to B Very little reaction

HA H+ + A- Le Chatelier’s principle

Why does a buffer resist change in pH when small amounts of strong acid or bases is added?

The acid or base is consumed by A- or HA respectively

A buffer has a maximum capacity to resist change to pH.

Buffer capacity, :

Measure of how well solution resists change in pH when strong acid/base is added.

pH

C

pH

C ab

d

d

d

d

?

Larger more resistance to pH change

How a Buffer Works

Consider adding H3O+ or OH- to water and also to a buffer

For 0.01 mol H3O+ to 1 L water: [H3O+] = 0.01 mol/1.0 L = 0.01 M pH = -log([H3O+]) = 2.0

So, change in pH from pure water: pH = 7.00 – 2.00 = 5.0

For the H2CO3- / HCO3

- system: pH of buffer = 7.38

Addition of 0.01 mol H3O+ changes pH to 7.46

So change in pH from buffer: pH = 7.46 – 7.38 = 0.08 !!!

How a Buffer Works

Consider a buffer made from acetic acid and sodium acetate:

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+

(aq)

Ka = or [CH3COO-] [H3O+]

[CH3COOH]

[CH3COOH]

[CH3COO-][H3O+] = Ka x

How a Buffer WorksLet’s consider a buffer made by placing 0.25 mol of acetic acid and 0.25 mol of sodium acetate per liter of solution.

What is the pH of the buffer?

And what will be the pH of 100.00 mL of the buffer before and after 1.00 mL of concentrated HCl (12.0 M) is added to the buffer? What will be the pH of 300.00 mL of pure water if the same acid is added?

pH = -log[H3O+] = -log(1.8 x 10-5) = pH = 4.74 Before acid added!

[H3O+] = Ka x = 1.8 x 10-5 x[CH3COOH]

[CH3COO-] (0.25)

(0.25)= 1.8 x 10-5

How a Buffer Works

1.00 mL conc. HCl 1.00 mL x 12.0 mol/L = 0.012 mol H3O+

Added to 300.00 mL of water :

0.012 mol H3O+

301.00 mL soln.= 0.0399 M H3O+ pH = -log(0.0399 M)

pH = 1.40 Without buffer!

What is pH if added to pure water?

How a Buffer WorksAfter acid is added to buffer: Conc. (M) CH3COOH(aq) + H2O(aq) CH3COO- + H3O+

Initial 0.250 ---- 0.250 0Change +0.012 ---- -0.012 0.012Equilibrium 0.262 ---- 0.238 0.012

Solving for the quantity ionized:

Initial 0.262 ---- 0.238 0Change -x ---- +x +xEquilibrium 0.262 - x ---- 0.238 + x x

Assuming: 0.262 - x = 0.262 & 0.238 + x = 0.238

pH = -log(1.982 x 10-5) = 5.000 - 0.297 = 4.70 After the acid is added!

Conc. (M) CH3COOH(aq) + H2O(aq) CH3COO- + H3O+

[CH3COOH]

[CH3COO-][H3O+] = Ka x =1.8 x 10-5 x (0.262)

(0.238)= 1.982 x 10-5

How a Buffer Works

Suppose we add 1.0 mL of a concentrated base instead of an acid. Add1.0 mL of 12.0 M NaOH to pure water and our buffer, and let’s see what the impact is: 1.00 mL x 12.0 mol OH-/1000mL = 0.012 mol OH-

This will reduce the quantity of acid present and force the equilibrium to produce more hydronium ion to replace that neutralized by the addition of the base!

Conc. (M) CH3COOH(aq) + H2O(aq) CH3COO- + H3O+

Initial 0.250 ---- 0.250 0Change - 0.012 ---- +0.012 +0.012Equilibrium 0.238 ---- 0.262 +0.012

Assuming: Again, using x as the quantity of acid dissociated we get: our normal assumptions: 0.262 + x = 0.262 & 0.238 - x = 0.238

[H3O+] = 1.8 x 10-5 x = 1.635 x 10-50.2380.262

pH = -log(1.635 x 10-5) = 5.000 - 0.214 = 4.79 After base is added!

How a Buffer Works

By adding the 1.00mL base to 300.00 mL of pure water we would get a hydroxide ion concentration of:

301.00 mL0.012 mol OH-

[OH-] = = 3.99 x 10-5 M OH-

This calculates out to give a pH of:

The hydrogen ion concentration is:

[H3O+] = = = 2.506 x 10-10Kw

[OH-]1 x 10-14

3.99 x 10-5 M

pH = -log(2.5 6 x 10-10) = 10.000 - 0.408 = 9.59 With 1.0 mL of the base in pure water!

In summary: Buffer alone pH = 4.74 Buffer plus 1.0 mL base pH = 4.79 Base alone, pH = 9.59 Buffer plus 1.0 mL acid pH = 4.70 Acid alone, pH = 1.40

Problem:

Calculate the pH of a solution containing 0.200 M NH3 and 0.300 M NH4Cl given that the acid dissociation constant for NH4

+ is 5.7x10-10.

NH3 + H2O NH4+ + OH-

acidbaseKa

][BH

[B]logpKpH a pKa applies

to this acid

pKa = 9.244

pH = 9.244 + log (0.200)

(0.300)

pH = 9.07