pH and Buffers The Whole Story

Acid & Base and pH 0 7 14 1 M 10-7 M 10-14 M Very Neutral Very This should be a review but we will go over it to help refresh your memory 0 7 14 1 M 10-7 M 10-14 M Very Neutral Very Acidic Basic

Trivia time: Why was the concept of pH developed?

Trivia time: Why was the concept of pH developed? By the Danish Biochemist Sorensen to test the acidity of the beer he was making Who said biochemistry isn’t cool?

pH - pouvoir hydrogene (the power of hydrogen) Water undergoes ionization Water ionizes to form the hydronium (hydroxyl) ion and hydroxide ions Water can act as both an acid and base The equilibrium constant for the ionization of water is: The concentration of pure water 1 liter = 1000g MW of water is 10.015 the final concentration of water is 55M and H+ concentration is about 1.8 x 10-9 Very little water actually dissociates So Keq is very small – not easily measured or easy to use

pH - pouvoir hydrogene (the power of hydrogen) Water undergoes ionization Water ionizes to form the hydronium (hydroxyl) ion and hydroxide ions Water can act as both an acid and base The equilibrium constant for the ionization of water is: The concentration of pure water 1 liter = 1000g MW of water is 10.015 the final concentration of water is 55M and H+ concentration is about 1.8 x 10-9 Very little water actually dissociates So Keq is very small – not easily measured or easy to use H 2 O 3 + OH _

pH - pouvoir hydrogene (the power of hydrogen) Water undergoes ionization Water ionizes to form the hydronium (hydroxyl) ion and hydroxide ions Water can act as both an acid and base The equilibrium constant for the ionization of water is: The concentration of pure water 1 liter = 1000g MW of water is 10.015 the final concentration of water is 55M and H+ concentration is about 1.8 x 10-9 Very little water actually dissociates So Keq is very small – not easily measured or easy to use H 2 O 3 + OH _ K eq = [products] [reactants] [H+ ] [OH- ] [H2O]2

pH - pouvoir hydrogene (the power of hydrogen) Water undergoes ionization Water ionizes to form the hydronium (hydroxyl) ion and hydroxide ions Water can act as both an acid and base The equilibrium constant for the ionization of water is: The concentration of pure water 1 liter = 1000g MW of water is 10.015 the final concentration of water is 55M and H+ concentration is about 1.8 x 10-9 Very little water actually dissociates So Keq is very small – not easily measured or easy to use H 2 O 3 + OH _ [H2O] is effectively constant. [products] [H+ ] [OH- ] K = = eq [reactants] [H2O]2

Instead a different constant is used where the denominator is ignored Kw= 1.0 X 10 -14 pH is a measure of the acidity and basicity of a solution when [H+] = [OH-] the solution is neutral and pH is 7 when [H+] > [OH-] the solution is acidic and pH is less than 7 when [H+] < [OH-] the solution is basic and pH is more than 7 a change in 1 pH units is = a ten fold change in hydrogen ion concentration

Instead a different constant is used where the denominator is ignored Kw= 1.0 X 10 -14 pH is a measure of the acidity and basicity of a solution when [H+] = [OH-] the solution is neutral and pH is 7 when [H+] > [OH-] the solution is acidic and pH is less than 7 when [H+] < [OH-] the solution is basic and pH is more than 7 a change in 1 pH units is = a ten fold change in hydrogen ion concentration _ + Therefore: K = [H ] [OH ] w And as measured in pure water [H ] = + [OH ] = 1 X 10 _ - 7

Instead a different constant is used where the denominator is ignored Kw= 1.0 X 10 -14 pH is a measure of the acidity and basicity of a solution pH = -log[H+] when [H+] = [OH-] the solution is neutral and pH is 7 when [H+] > [OH-] the solution is acidic and pH is less than 7 when [H+] < [OH-] the solution is basic and pH is more than 7 a change in 1 pH units is = a ten fold change in hydrogen ion concentration _ + Therefore: K = [H ] [OH ] w And as measured in pure water [H ] = + [OH ] = 1 X 10 _ - 7

The extent of ionization of a weak acid is a function of its acid dissociation constant pKa Bronsted and Lowry acid and bases acid donates protons bases accepts protons Strong acids dissociate nearly fully Weak acids only partially dissociate Acids with Ka < 1 are considered weak acids Ka for acetic acid is 1.76 x 10-5 -> difficult to work with so instead use log scale: pKa = -log Ka So the pKa of acetic acid is = -log 1.76 x 10-5 = 4.75 The pH is a measure of acidity and the pKa is a measure of acid strength

The extent of ionization of a weak acid is a function of its acid dissociation constant pKa Bronsted and Lowry acid and bases acid donates protons bases accepts protons Strong acids dissociate nearly fully Weak acids only partially dissociate Acids with Ka < 1 are considered weak acids Ka for acetic acid is 1.76 x 10-5 -> difficult to work with so instead use log scale: pKa = -log Ka So the pKa of acetic acid is = -log 1.76 x 10-5 = 4.75 The pH is a measure of acidity and the pKa is a measure of acid strength HA + H O H O + A 2 3 + _ acid base conjugate conjugate acid base

[H+] = [ acid] and thus pH = -log [acid] The extent of ionization of a weak acid is a function of its acid dissociation constant pKa Bronsted and Lowry acid and bases acid donates protons bases accepts protons Strong acids dissociate nearly fully [H+] = [ acid] and thus pH = -log [acid] Weak acids only partially dissociate Acids with Ka < 1 are considered weak acids Ka for acetic acid is 1.76 x 10-5 -> difficult to work with so instead use log scale: pKa = -log Ka So the pKa of acetic acid is = -log 1.76 x 10-5 = 4.75 The pH is a measure of acidity and the pKa is a measure of acid strength HA + H O H O + A 2 3 + _ acid base conjugate conjugate acid base

[H+] = [ acid] and thus pH = -log [acid] The extent of ionization of a weak acid is a function of its acid dissociation constant pKa Bronsted and Lowry acid and bases acid donates protons bases accepts protons Strong acids dissociate nearly fully [H+] = [ acid] and thus pH = -log [acid] Weak acids only partially dissociate Acids with Ka < 1 are considered weak acids Ka for acetic acid is 1.76 x 10-5 -> difficult to work with so instead use log scale: pKa = -log Ka So the pKa of acetic acid is = -log 1.76 x 10-5 = 4.75 The pH is a measure of acidity and the pKa is a measure of acid strength + _ HA + H O H O + A 2 3 acid base conjugate conjugate acid base K = [H ] + [A ] _ [HA] a pKa and Ka are used for all acids.

Dissociation constants and pKa values More Acidic C hemi c al K a ( M ) p K a F o r mic aci d -4 3 . 8 1 . 77 x 10 C a rb oni c a ci d -7 6 . 4 4 . 30 x 10 B ic a r b o na te -11 10 . 2 5 . 61 x 10 Am m o n ium -10 9 . 2 5 . 62 x 10 Note relationship between pKA, Ka and acid strength.

Titrations and pKa pKa of a weak acid is determined experimentally by titration. pKa is when the concentration of acid and base is equal in a titration (ask yourself if that then equal to pH 7). There is a point in a titration of a weak acid where the change in pH is very little. This is the buffer action of the acid. Many acids have more than one ionizable group (polyprotic)

The relationship between pH and pKa FOR A WEAK ACID is described by the Henderson-Hasselbalch equation HA H+ + A-

The relationship between pH and pKa is described by the Henderson-Hasselbalch equation HA H+ + A- Start with a weak acid

The relationship between pH and pKa is described by the Henderson-Hasselbalch equation HA H+ + A- Start with a weak acid K = [ H+] [A-] HA ] a

The relationship between pH and pKa is described by the Henderson-Hasselbalch equation HA H+ + A- Start with a weak acid K = [ H+] [A-] HA ] a Arrange to get

The relationship between pH and pKa is described by the Henderson-Hasselbalch equation HA H+ + A- Start with a weak acid K = [ H+] [A-] HA ] a Arrange to get [HA] [H+] K = a [A-]

Multiply each side by log The relationship between pH and pKa is described by the Henderson-Hasselbalch equation HA H+ + A- Start with a weak acid K = [ H+] [A-] HA ] a Arrange to get [HA] [H+] K = a Multiply each side by log [A-]

Multiply each side by log The relationship between pH and pKa is described by the Henderson-Hasselbalch equation HA H+ + A- Start with a weak acid K = [ H+] [A-] HA ] a Arrange to get [HA] [H+] K = a Multiply each side by log [A-] [HA] Log[H+] = Log K + Log a [A-]

Multiply each side by log The relationship between pH and pKa is described by the Henderson-Hasselbalch equation HA H+ + A- Start with a weak acid K = [ H+] [A-] HA ] a Arrange to get [HA] [H+] K = a Multiply each side by log [A-] [HA] Log[H+] = Log K + Log a [A-] Multiply by -1

Multiply each side by log The relationship between pH and pKa is described by the Henderson-Hasselbalch equation HA H+ + A- Start with a weak acid K = [ H+] [A-] HA ] a Arrange to get [HA] [H+] K = a Multiply each side by log [A-] [HA] Log[H+] = Log K + Log a [A-] Multiply by -1 [HA] -Log[H+] = -Log K - Log a [A-]

Multiply each side by log The relationship between pH and pKa is described by the Henderson-Hasselbalch equation HA H+ + A- Start with a weak acid K = [ H+] [A-] HA ] a Arrange to get [HA] [H+] K = a Multiply each side by log [A-] [HA] Log[H+] = Log K + Log a [A-] Multiply by -1 [HA] Substitute pH = -Log[H+] pKa = Log Ka -Log[H+] = -Log K - Log a [A-]

Multiply each side by log The relationship between pH and pKa is described by the Henderson-Hasselbalch equation HA H+ + A- Start with a weak acid K = [ H+] [A-] HA ] a Arrange to get [HA] [H+] K = a Multiply each side by log [A-] [HA] Log[H+] = Log K + Log a [A-] Multiply by -1 [HA] Substitute pH = -Log[H+] pKa = Log Ka -Log[H+] = -Log K - Log a [A-] [HA] pH = pKa - Log [A-]

Multiply each side by log Remove (-) and invert last term The relationship between pH and pKa is described by the Henderson-Hasselbalch equation HA H+ + A- Start with a weak acid K = [ H+] [A-] HA ] a Arrange to get [HA] [H+] K = a Multiply each side by log [A-] [HA] Log[H+] = Log K + Log a [A-] Multiply by -1 [HA] Substitute pH = -Log[H+] pKa = Log Ka -Log[H+] = -Log K - Log a [A-] [HA] pH = pKa - Log [A-] Remove (-) and invert last term

Multiply each side by log Remove (-) and invert last term The relationship between pH and pKa is described by the Henderson-Hasselbalch equation HA H+ + A- Start with a weak acid K = [ H+] [A-] HA ] a Arrange to get [HA] [H+] K = a Multiply each side by log [A-] [HA] Log[H+] = Log K + Log a [A-] Multiply by -1 [HA] Substitute pH = -Log[H+] pKa = Log Ka -Log[H+] = -Log K - Log a [A-] [HA] pH = pKa - Log [A-] Remove (-) and invert last term [A-] pH = pKa + Log [HA]

Multiply each side by log Remove (-) and invert last term The relationship between pH and pKa is described by the Henderson-Hasselbalch equation HA H+ + A- Start with a weak acid K = [ H+] [A-] HA ] a Arrange to get [HA] [H+] K = a Multiply each side by log [A-] [HA] Log[H+] = Log K + Log a [A-] Multiply by -1 [HA] Substitute pH = -Log[H+] pKa = Log Ka -Log[H+] = -Log K - Log a [A-] [HA] pH = pKa - Log [A-] Remove (-) and invert last term [A-] pH = pKa + Log [HA] Henderson -Hasselbalch

What is the H-H Equation Used For? This is used to determine the concentration of acid and base at a given pH. It is Also used to determine the pH of a known solution. These concepts are used to calculate buffer strength and understand the pH of a biological solution. Remember that buffers are mixtures of weak acids and their conjugate bases that resist pH by shifting the equilibrium between the acid and base in response to the pH of a solution.

Case 1) when the concentration of base equals the acid. _ [A ] pH = pKa + Log [HA] [ x ] = pKa + Log [ x ] pH = pKa When pH = pKa 50% of the acid is dissociated

Case 2) when the pH is above or below 1 pH unit of the pKa _ [A ] 5.00 = 4.00 + Log [HA] _ [A ] 1.00 = Log [HA] _ [A ] 10.0 = [HA] Then 90% of the buffer is in the conjugate base for If pH is 2 units different then 99% of buffer is in the conjugate base form What does this mean about the buffering ability if more acid or base is added?

Calculate the pH of a mixture of 250 mM acetic acid and 100 mM Na acetate. The pKa of acetic acid is 4.75.

Calculate the pH of a mixture of 250 mM acetic acid and 100 mM Na acetate. The pKa of acetic acid is 4.75. Start with the HH equation pH = pKa + Log [A-]/[HA]

Calculate the pH of a mixture of 250 mM acetic acid and 100 mM Na acetate. The pKa of acetic acid is 4.75. Start with the HH equation pH = pKa + Log [A-]/[HA] Recognize that a compound that ends with -ate is the base form of a weak acid and plug in the known terms

Calculate the pH of a mixture of 250 mM acetic acid and 100 mM Na acetate. The pKa of acetic acid is 4.75. Start with the HH equation pH = pKa + Log [A-]/[HA] Recognize that a compound that ends with -ate is the base form of a weak acid and plug in the known terms pH = 4.75 + Log (0.1/0.25)

Remember [X] indicates molar concentration Calculate the pH of a mixture of 250 mM acetic acid and 100 mM Na acetate. The pKa of acetic acid is 4.75. Start with the HH equation pH = pKa + Log [A-]/[HA] Recognize that a compound that ends with -ate is the base form of a weak acid and plug in the known terms pH = 4.75 + Log (0.1/0.25) Remember [X] indicates molar concentration

Remember [X] indicates molar concentration Calculate the pH of a mixture of 250 mM acetic acid and 100 mM Na acetate. The pKa of acetic acid is 4.75. Start with the HH equation pH = pKa + Log [A-]/[HA] Recognize that a compound that ends with -ate is the base form of a weak acid and plug in the known terms pH = 4.75 + Log (0.1/0.25) Remember [X] indicates molar concentration pH =4.35

What is the ratio of lactic acid to lactate in a buffer at pH of 5. 00 What is the ratio of lactic acid to lactate in a buffer at pH of 5.00. The pKa of lactic acid is 3.86?

What is the ratio of lactic acid to lactate in a buffer at pH of 5. 00 What is the ratio of lactic acid to lactate in a buffer at pH of 5.00. The pKa of lactic acid is 3.86? Again we recognize this is a weak acid and start with the HH equation

What is the ratio of lactic acid to lactate in a buffer at pH of 5. 00 What is the ratio of lactic acid to lactate in a buffer at pH of 5.00. The pKa of lactic acid is 3.86? Again we recognize this is a weak acid and start with the HH equation pH = pKa + Log [A-]/[HA]

What is the ratio of lactic acid to lactate in a buffer at pH of 5. 00 What is the ratio of lactic acid to lactate in a buffer at pH of 5.00. The pKa of lactic acid is 3.86? Again we recognize this is a weak acid and start with the HH equation pH = pKa + Log [A-]/[HA] Plug in the known terms

What is the ratio of lactic acid to lactate in a buffer at pH of 5. 00 What is the ratio of lactic acid to lactate in a buffer at pH of 5.00. The pKa of lactic acid is 3.86? Again we recognize this is a weak acid and start with the HH equation pH = pKa + Log [A-]/[HA] Plug in the known terms 5.00 = 3.86 + Log [A-]/[HA]

What is the ratio of lactic acid to lactate in a buffer at pH of 5. 00 What is the ratio of lactic acid to lactate in a buffer at pH of 5.00. The pKa of lactic acid is 3.86? Again we recognize this is a weak acid and start with the HH equation pH = pKa + Log [A-]/[HA] Plug in the known terms 5.00 = 3.86 + Log [A-]/[HA] Arrange

What is the ratio of lactic acid to lactate in a buffer at pH of 5. 00 What is the ratio of lactic acid to lactate in a buffer at pH of 5.00. The pKa of lactic acid is 3.86? Again we recognize this is a weak acid and start with the HH equation pH = pKa + Log [A-]/[HA] Plug in the known terms 5.00 = 3.86 + Log [A-]/[HA] Arrange 1.14 = Log [A-]/[HA]

What is the ratio of lactic acid to lactate in a buffer at pH of 5. 00 What is the ratio of lactic acid to lactate in a buffer at pH of 5.00. The pKa of lactic acid is 3.86? Again we recognize this is a weak acid and start with the HH equation pH = pKa + Log [A-]/[HA] Plug in the known terms 5.00 = 3.86 + Log [A-]/[HA] Arrange 1.14 = Log [A-]/[HA] Inv Log on both sides

What is the ratio of lactic acid to lactate in a buffer at pH of 5. 00 What is the ratio of lactic acid to lactate in a buffer at pH of 5.00. The pKa of lactic acid is 3.86? Again we recognize this is a weak acid and start with the HH equation pH = pKa + Log [A-]/[HA] Plug in the known terms 5.00 = 3.86 + Log [A-]/[HA] Arrange 1.14 = Log [A-]/[HA] Inv Log on both sides Ratio = 13.80

That means that for each mole of acid there are 13.8 moles of base What is the ratio of lactic acid to lactate in a buffer at pH of 5.00. The pKa of lactic acid is 3.86? Again we recognize this is a weak acid and start with the HH equation pH = pKa + Log [A-]/[HA] Plug in the known terms 5.00 = 3.86 + Log [A-]/[HA] Arrange 1.14 = Log [A-]/[HA] Inv Log on both sides Ratio = 13.80 That means that for each mole of acid there are 13.8 moles of base

That means that for each mole of acid there are 13.8 moles of base What is the ratio of lactic acid to lactate in a buffer at pH of 5.00. The pKa of lactic acid is 3.86? Again we recognize this is a weak acid and start with the HH equation pH = pKa + Log [A-]/[HA] Plug in the known terms 5.00 = 3.86 + Log [A-]/[HA] Arrange 1.14 = Log [A-]/[HA] Inv Log on both sides Ratio = 13.80 That means that for each mole of acid there are 13.8 moles of base

What is the concentration of base and acid you need to add to make a 50 mM solution of lactate buffer at pH 4.0? The MW of Lactic acid is 91 amu and sodium lactate is 102 amu. This is for you to take home try and we will calculate an answer with Dr Provost - Also start to do the pH Homework!