Ch 18 buffers
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Transcript of Ch 18 buffers
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
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