Chemistry 130

Acid and Base equilibria

Dr. John F. C. Turner

409 Buehler Hall

[email protected]

Chemistry 130 Acids and bases

The Brønsted­Lowry definition of an acid states that any material that produces the hydronium ion is an acid. A Brønsted­Lowry acid is a proton donor:

+ ­ HA  H2 Ol  H3 Oaq  Aaq

+ + The hydronium ion, H 3 O  a q  or H  a q  has the same structure as ammonia. It is pyramidal and has one lone pair on O.

Chemistry 130 Acids and bases

The Brønsted­Lowry definition of a base states that any material that accepts a proton is a base. A Brønsted­Lowry base is a proton acceptor:

­ ­ + Baq  H2 Ol  HOaq  BHaq In aqueous solution, a base forms the hydroxide ion:

Chemistry 130 Conjugate acids and bases

For any acid­base reaction, the original acid and base are complemented by the conjugate acid and conjugate base:

­ + NH3aq  H2Ol  HOaq  NH4aq On the RHS, On the LHS,

Water is the proton donor Hydroxide ion is the proton acceptor

Ammonia is the proton acceptor Ammonium ion is the proton donor

Water and hydroxide ion are conjugate acid and base

Ammonia and ammonium ion are also conjugate acid and base

Chemistry 130 Conjugate acids and bases

The anion of every acid is the conjugate base of that acid

The cation of every base is the conjugate acid of that base

Acid­base conjugates exist due to the dynamic equilibrium that exists in solution. ­ + NH3aq  H2Ol  HOaq  NH4aq

­ + NH3aq  H2Ol  HOaq  NH4aq

Chemistry 130 Acid and base constants

We use equilibrium constants to define the position of equilibrium and to reflect the dynamic nature of the system.

For an acid, we define

+ ­ HClaq  H2 Ol  H3 Oaq  Claq

+ ­ [Products] [H3 Oaq][Claq] K A = = [Reactants] [HClaq] Water is not included as the change in concentration of water is negligible (water is ~55 M when pure) for a dilute solution.

Chemistry 130 Acid and base constants

We use equilibrium constants to define the position of equilibrium and to reflect the dynamic nature of the system.

For a base, we define

+ ­ NaOHaq  H2 Ol  Naaq  OHaq

+ ­ [Products] [Naaq][OHaq] K B = = [Reactants] [NaOHaq] Again, water is not included as the change in concentration of water is negligible (water is ~55 M when pure) for a dilute solution.

Chemistry 130 Acid and base constants

We also use the logarithm of the acid or base constant as an indicator of acid or base strength:

+ ­ + ­ [H3 Oaq][Claq] [Naaq][OHaq] K A = K B = [HClaq] [NaOHaq]

pK A = −lg K A pK B = −lg K B

Note that the logarithm used here is to base 10, not the natural log

lg = log10

ln = loge

Chemistry 130 Conjugate acid and base strengths

The dynamic nature of the equilibrium between the conjugate acid­base pair means that a strong acid will have a weak conjugate base

A strong base will have a weak conjugate acid.

In both these cases, 'strong' and 'weak' are defined qualitatively the position of the acid or base equilibrium.

A strong acid forms almost exclusively the hydronium ion and the concentration of the undissociated acid is negligible; the conjugate base must be weak.

Chemistry 130 Conjugate acid and base strengths

There are a variety of strong acids – any material that Acid pK A has a larger pKA than the hydronium ion will form the hydronium ion in solution. HI −9 Any material with an acid constant smaller than the HBr −8 hydronium ion will establish a measurable equilibrium HCl −6 between the acid and the dissociated hydronium ion and H2 SO4 −3 associated conjugate base – the anion. + H3 Oaq −1.7

Water therefore acts as a leveling solvent and restricts HNO3 −1.3 the degree of acidity possible in aqueous solution.

Chemistry 130 Autoprotolysis of water

Water can act as both an acid and a base – it can form the hydronium ion as well as the hydroxide ion in solution.

Pure water also undergoes a 'self­equilibrium':

+ ­ H2 Ol  H2 Ol  H3 Oaq  OHaq

+ ­ [Products] [H3 Oaq][OHaq] K = = W Reactants 2 [ ] [H2Ol] + ­ −14 K W = [H3Oaq][OHaq] = 1×10

pK W = 14

This is autoprotolysis or autionization.

Chemistry 130 pKW , pH and pOH

The autoprotolysis of water relates the hydronium ion concentration and the hydroxide ion concentration to the autoprotolysis constant of water:

+ ­ H2 Ol  H2 Ol  H3 Oaq  OHaq

+ ­ −14 K W = [H3Oaq][OHaq] = 1×10 pK lg H O+ OH­ lg H O+ lg OH­ 14 w = − 10 [ 3 aq][ aq] = − 10 [ 3 aq]− 10 [ aq] = pK lg H O+ lg OH­ p H pOH 14 w = − 10 [ 3 aq]− 10 [ aq] =  =

pK w = pH  pOH = 14 This relationship dictates the concentrations of hydroxide and hydronium ion in all aqueous solutions

Chemistry 130 Simple pH and pOH calculations

Q: What is the pH of a solution of 0.0002 M solution of HI?

Molarity of solution = 0.0002 mol L−1 The ionization equation of HI in water is

+ ­ HIaq  H2 Ol  H3 Oaq  Iaq

The K A of HI is very large and we know that HIaq is a strong acid + −1 Therefore, [H3 Oaq] = 0.0002 mol L lg H O+ 3.699 10 [ 3 aq ] = −

pH lg H O+ 3.699 = − 10 [ 3 aq] =

Chemistry 130 Simple pH and pOH calculations

Q: Calculate pOH, the pH and the concentration of hydroxide ion in a solution of 0.0002 M solution of HI? We have already calculated the p H of the solution p H lg H O+ 3.699 = − 10 [ 3 aq] =

The relationship between p H, pOH and pK W is

pK w = p H  pOH = 14

3.699  pOH = 14 pOH = 14 − 3.699 = 10.301

pOH lg OH­ 10.301 = − 10 [ aq ] = ­ −10.301 −11 −1 [OHaq] = 10 = 5 × 10 mol L Chemistry 130 Simple pH and pOH calculations

Q: Calculate pOH, the pH and the concentration of hydroxide ion in a solution of 0.00075 M solution of KOH? pOH calculation OH­ 0.00075 mol L−1 lg OH­ 3.125 [ aq] = [ aq] = − pOH lg OH­ 3.125 = − 10 [ aq ] = As

pK w = p H  pOH = 14 p H  3.125 = 14 p H = 14 − 3.125 = 10.875 As p H lg H O+ = − 10 [ 3 aq] lg H O+ 10.875 10 [ 3 aq] = − + −10.875 −11 −1 [H3Oaq] = 10 = 1.33 × 10 mol L Chemistry 130 Weak acids and bases

These calculations are straightforward as HI is a strong acid and KOH is a strong base.

For weak acids and bases, which do not fully ionize in solution, we have to use the rules for equilibria with which we are already familiar.

We also use

pK A = −lg K A pK B = −lg K B

which we have defined for the generic acid reaction and base reaction in water.

Chemistry 130 Weak acid/base calculations

Q: Calculate pOH, the pH and the concentration of hydroxide ion in a solution of 0.5 M solution of acetic acid?

+ ­ −5 CH3 CO2 Haq  H2 Ol  H3 Oaq  CH3 CO2aq K A = 1.8 × 10 Acetic acid Acetate ion

+ ­ CH3 CO2 Haq H3 Oaq CH3 CO2aq Initial concentrations 0.5 ≈0 0 Change −x x x Equilibrium concentrations 0.5−x x x

At equilibrium, the new concentrations are: [CH3 CO2 Haq] = 0.5−x ­ [CH3 CO2aq] = x + [H3 Oaq] = x

Chemistry 130 Weak acid/base calculations

−5 The acid constant for acetic acid is K A = 1.8 × 10

−14 whereas the autoprotolysis constant for water is K W = 1 × 10 The contribution of the self­ionization of water is of the order of 10­7 to the concentration of hydronium ion.

In general, if the acid constant for the weak acid is sufficiently large in comparison to the autoprotolysis constant for water, we can ignore the autoprotolysis of water with respect to the hydrogen ion concentration

Chemistry 130 Weak acid/base calculations

Q: Calculate pOH, the pH and the concentration of hydroxide ion in a solution of 0.5 M solution of acetic acid?

+ ­ −5 CH3CO2 Haq  H2 Ol  H3Oaq  CH3 CO2aq K A = 1.8 × 10 Acetic acid Acetate ion

We now set up the equilibrium constant for aqueous acetic acid: ­ + [CH3 CO2aq][H3 Oaq] K A = [CH3 CO2 Haq] and use the concentrations that we have already calculated:

[CH3 CO2 Haq] = 0.5−x ­ + ­ [CH3 CO2aq][H3 Oaq]  x x [CH3 CO2aq] = x K A = = + [CH3 CO2 Haq] 0.5−x [H3 Oaq] = x

Chemistry 130 Weak acid/base calculations

Q: Calculate pOH, the pH and the concentration of hydroxide ion in a solution of 0.5 M solution of acetic acid?

+ ­ −5 CH3 CO2 Haq  H2 Ol  H3 Oaq  CH3 CO2aq K A = 1.8 × 10 Acetic acid Acetate ion

We now solve the equilibrium equation that we have set up:

x  x −5 K = = 1.8 × 10 A 0.5−x x2 = 1.8 × 10−5 0.5−x

There are two ways of solving this equation

1. use the quadratic formula 2. make the assumption that x≪0.5 Chemistry 130 Weak acid/base calculations

Q: Calculate pOH, the pH and the concentration of hydroxide ion in a solution of 0.5 M solution of acetic acid?

+ ­ −5 CH3 CO2 Haq  H2 Ol  H3 Oaq  CH3 CO2aq K A = 1.8 × 10 Acetic acid Acetate ion

Using assumption 2, that x≪0.5

x2 x2 K = ≈ = 1.8 × 10−5 A 0.5−x 0.5 x2 ≈ 0.5 × 1.8 × 10−5 ≈ 9 × 10−6 1 2 x ≈  9 × 10−6 ≈ 0.003 + −1 + [H3 Oaq ] = 0.003 mol L pH = −lg [H3 Oaq] = 2.52 ­ −12 −1 pOH = 14−pH = 11.48 [OHaq] = 3.33 × 10 mol L Chemistry 130 Weak acid/base calculations

Q: Calculate pOH, pH and the concentrations of both hydroxide ion and hydronium in a solution of 1.5 M solution of hydroxylamine?

­ + −9 NH2 OHaq  H2Ol  OHaq  NH3 OHaq KB = 9.1 × 10 Hydroxylamine Hydroxylammonium

­ + NH2OHaq OHaq NH3 OHaq Initial concentrations 1.5 ≈0 0 Change −x x x Equilibrium concentrations 1.5−x x x

At equilibrium, the new concentrations are: [NH2 OHaq] = 1.5−x + [NH3 OHaq] = x ­ [OHaq] = x

Chemistry 130 Weak acid/base calculations

Q: Calculate pOH, pH and the concentrations of both hydroxide ion and hydronium in a solution of 1.5 M solution of hydroxylamine?

­ + −9 NH2 OHaq  H2Ol  OHaq  NH3OHaq K B = 9.1 × 10 Hydroxylamine Hydroxylammonium

We now set up the base equilibrium constant for aqueous hydroxylamine: + ­ [NH3 OHaq ][OHaq] K A = [NH2OHaq] and use the concentrations that we have already calculated: [NH OH ] = 1.5−x 2 aq + ­ 2 + [NH3 OHaq ][OHaq] x [NH3OHaq] = x K B = = ­ [NH2OHaq] 1.5−x [OHaq] = x

Chemistry 130 Weak acid/base calculations

Q: Calculate pOH, pH and the concentrations of both hydroxide ion and hydronium in a solution of 1.5 M solution of hydroxylamine?

­ + −9 NH2 OHaq  H2Ol  OHaq  NH3 OHaq KB = 9.1 × 10 Hydroxylamine Hydroxylammonium

The equation that we need to solve is: + ­ 2 [NH3 OHaq][OHaq] x −9 K B = = = 9.1 × 10 [NH2 OHaq] 1.5−x Using the quadratic method, x2 = 9.1 × 10−9 1.5−x so x2 = 1.5−x⋅9.1 × 10−9 = 1.365 × 10−8 − 9.1 × 10−9 x x2  9.1 × 10−9 x − 1.365 × 10−8 = 0

Chemistry 130 Weak acid/base calculations

Q: Calculate pOH, pH and the concentrations of both hydroxide ion and hydronium in a solution of 1.5 M solution of hydroxylamine?

­ + −9 NH2OHaq  H2 Ol  OHaq  NH3 OHaq K B = 9.1 × 10 Hydroxylamine Hydroxylammonium

1 2 2 2 −b± b − 4ac  For a quadratic of the form ax  bx  c = 0, x = 2a In this case, a = 1 b = 9.1 × 10−9 c = −1.365 × 10−8 and so 1 2 −9.1 × 10−9±9.1 × 10−92 − 4⋅1⋅−1.365 × 10−8 x = 2 1 −9 −8 2 −9.1 × 10 ± 5.460000008 × 10  −4 −1 x = = 1.168 × 10 mol L 2 Chemistry 130 Weak acid/base calculations

Q: Calculate pOH, pH and the concentrations of both hydroxide ion and hydronium in a solution of 1.5 M solution of hydroxylamine?

­ + −9 NH2 OHaq  H2 Ol  OHaq  NH3 OHaq K B = 9.1 × 10 Hydroxylamine Hydroxylammonium

The equation that we need to solve is: + ­ 2 [NH3 OHaq ][OHaq] x −9 K B = = = 9.1 × 10 [NH2OHaq] 1.5−x Using the assumption that x≪1.5 x2 x2 ≈ = 9.1 × 10−9 1.5−x 1.5 so x2 = 1.5⋅9.1 × 10−9 = 1.365 × 10−8 1 x ≈ 1.365 × 10−8 2 ≈ 1.168 × 10−4 mol L−1

Chemistry 130 Weak acid/base calculations

Q: Calculate pOH, pH and the concentrations of both hydroxide ion and hydronium in a solution of 1.5 M solution of hydroxylamine?

­ + −9 NH2OHaq  H2 Ol  OHaq  NH3 OHaq K B = 9.1 × 10 Hydroxylamine Hydroxylammonium

In this case there is no difference between the methods to the third place of ­ −4 −1 decimal and so x = [OHaq] = 1.168 × 10 mol L

pOH lg OH­ lg 1.168 10−4 3.93 = − 10 [ aq] = − 10  ×  = As pOH  p H = 14 p H = 14 − p OH = 14 − 3.93 = 10.07 and so as pH lg H O+ = − 10 [ 3 aq] + −10.07 −11 −1 [H3 Oaq] = 10 = 8.56 × 10 mol L

Chemistry 130 Weak acid/base calculations

Given that pH and pOH are related via the relationship

+ ­ −14 K W = [H3Oaq][OHaq] = 1×10

pK w = pH  pOH = 14

then in aqueous solution, only one of these quantities is required for all to be calculable.

Similarly, if KA is known for an acid or KB is known for a base, then using the change on the establishment of equilibrium will give the hydronium ion concentration or the hydroxide concentration and so all is known.

If we know the pH or pOH, we can also calculate KA or KB for the system, using similar methods.

Chemistry 130 Weak acid/base calculations

Q: Calculate pKB for dimethylamine given that a 0.164 M solution has a pH of 11.98. ­ + Me2 NHaq  H2 Ol  OHaq  Me2 NH2aq Dimethylamine Dimethylammonium

−1 When [Me2 NHaq] = 0.164 mol L , p H = 11.98 p H  p OH = 14 so 11.98  pOH = 14 so p OH = 14 − 11.98 = 2.02 Given that pOH lg OH­ = − 10 [ aq] ­ −2.02 −3 −1 [OHaq] = 10 = 9.55 × 10 mol L

Chemistry 130 Weak acid/base calculations

Q: Calculate pKB for dimethylamine given that a 0.164 M solution has a pH of 11.98.

­ + Me2 NHaq  H2 Ol  OHaq  Me2 NH2aq Dimethylamine Dimethylammonium

­ −3 −1 From the p H calculation, [OHaq ] = 9.55 × 10 mol L

­ + Me2 NHaq OHaq Me2 NH2aq Initial concentrations 0.164 ≈0 0 Change −9.55 × 10−3 9.55 × 10−3 9.55 × 10−3 Equilibrium concentrations 0.164−9.55 × 10−3 9.55 × 10−3 9.55 × 10−3

Chemistry 130 Weak acid/base calculations

Q: Calculate pKB for dimethylamine given that a 0.164 M solution has a pH of 11.98.

­ + Me2 NHaq  H2 Ol  OHaq  Me2 NH2aq Dimethylamine Dimethylammonium

We now know all the concentrations requred to calculate K B

−3 −1 −1 [Me2 NHaq] = 0.164−9.55 × 10 = 1.5445 × 10 mol L ­ −3 −1 [OHaq] = 9.55 × 10 mol L + −3 −1 [Me2 NH2aq] = 9.55 × 10 mol L + ­ −3 −3 [Me2 NH2aq][OHaq] 9.55 × 10 9.55 × 10  −4 K 5.9 10 B = = −1 = × [Me2 NHaq] 1.5445 × 10  pK lg K 3.23 B = − 10  B = Chemistry 130 Polyprotic acids

A polyprotic acid is one that can ionize more than once. Common examples include

Sulfuric acid H2 SO4

Phosphoric acid H3 PO4

Carbonic acid H2 CO3

Note that acids such as acetic acid are not polyprotic

+ ­ CH3 CO2 Haq  H2 Ol  H3 Oaq  CH3 CO2aq

­ + 2­ CH3 CO2aq  H2 Ol  H3 Oaq  CH2 CO2aq

Chemistry 130 Polyprotic acids

Each ionization of a polyprotic acid has an associated acid constant.

For phosphoric acid H3 PO4

+ ­ −3 H3 PO4aq  H2 Ol  H3 Oaq  H2 PO4aq K A 1 = 7.1 × 10

­ + 2­ −8 H2 PO4aq  H2 Ol  H3 Oaq  HPO4aq K A 2 = 6.3 × 10

2­ + 3­ −13 HPO4aq  H2 Ol  H3 Oaq  PO4aq K A 3 = 4.3 × 10

Note that each acid constant differs from the one before and, although the first ionization, in this case, is strong, the others are not. −3 H3 PO4aq K A 1 = 7.1 × 10 Strong acid ­ −8 H2 PO4aq K A 2 = 6.3 × 10 Weak acid 2­ −13 HPO4aq K A 3 = 4.3 × 10 Weak acid Chemistry 130 Polyprotic acids

If the first ionization of a polyprotic acid is described by a large acid constant, then equal concentrations of the hydronium ion and dihydrogenphosphate ons are produced.

The equilibrium for the second ionization is

­ + 2­ −8 H2 PO4aq  H2 Ol  H3 Oaq  HPO4aq K A 2 = 6.3 × 10 and the equilibrium constant is

2­ + 2­ ­ [HPO4aq ][H3 Oaq] [HPO4aq][H2 PO4aq ] −8 K A 2 = ­ = ­ = 6.3 × 10 [H2 PO4aq ] [H2 PO4aq] 2 ­ −8 K A 2 = [HPO4 aq ] = 6.3 × 10

and so the acidity of the second ionization is independent of the initial concentration of phosphoric acid Chemistry 130 Chemistry 130

Acid and Base equilibria

Dr. John F. C. Turner

409 Buehler Hall

[email protected]

Chemistry 130 Salts of strong and weak acids

When a salt is dissolved, the equilibria for the conjugate acid and base are established.

Dissolving the salt introduces the conjugate acid or the conjugate base into the solution and the normal equilibria occur

Salts of strong acids and strong bases form neutral solutions

Salts of strong acids and weak bases form acidic solutions

Salts of weak acids and strong bases form basic solutions

The pH of solutions of salts of weak acids and weak bases depend on the acid constant of the acid and the base constant of the base

Chemistry 130 Salts of strong and weak acids

The reaction that occurs when an anion associated with a weak acid is dissolved in water changes the pH of the solution.

This happens because the anion is the conjugate base of the associated acid and the acid­base equilibrium for that acid is established.

­ For nitrous acid, the associated anion is nitrite, NO2

­ ­ ­ [OHaq][HNO2aq] NO2aq  H2 Ol  OHaq  HNO2aq K B = ­ NO2aq + ­ + [H3 Oaq] [OHaq][HNO2aq ] [H3 Oaq] We can multiply K B by + to give K B = ­ ⋅ + [H3 Oaq] NO2aq [H3 Oaq] ­ + [OHaq][HNO2aq] [H3Oaq] K B = ­ ⋅ + NO2aq [H3Oaq] For nitrous acid, K = 7.2 × 10−4 A,HNO2 Chemistry 130 Salts of strong and weak acids

For nitrous acid, the associated anion is nitrite, ­ ­ ­ [OHaq][HNO2aq] NO2aq  H2 Ol  OHaq  HNO2aq K B = ­ NO2aq

+ ­ + [H3 Oaq] [OHaq][HNO2aq] [H3 Oaq] We can multiply K B by + to give K B = ­ ⋅ + [H3 Oaq] NO2aq [H3 Oaq] ­ + [OHaq][HNO2aq] [H3 Oaq] K W K B = ⋅ = NO­ H O+ K 2aq [ 3 aq] A,HNO2 For nitrous acid, K = 7.2 × 10−4 and so A,HNO2 −14 1 × 10 −11 K B = = 1.39 × 10 7.2 × 10−4

Chemistry 130 Salts of strong and weak acids

For nitrous acid, the associated anion is nitrite, ­ ­ ­ [OHaq][HNO2aq] NO2aq  H2 Ol  OHaq  HNO2aq K B = ­ NO2aq

+ ­ + [H3 Oaq] [OHaq][HNO2aq ] [H3 Oaq] We can multiply K B by + to give K B = ­ ⋅ + [H3 Oaq] NO2aq [H3 Oaq] ­ + [OHaq][HNO2aq] [H3Oaq] K W K B = ⋅ = NO­ H O+ K 2aq [ 3 aq] A,HNO2 For nitrous acid, K = 7.2 × 10−4 and so A,HNO2 −14 1 × 10 −11 K B = = 1.39 × 10 7.2 × 10−4

Chemistry 130 Strength of conjugate acids and bases

The acid and base strength of a conjugate acid­base pair, such as

­ nitrite­nitrous acid NO2 − HNO2

­ acetic acid­acetate CH3 CO2 H − CH3 CO2

+ ammonia­ammonium NH3 − NH4 are related by the relationship

K A K B = K W

pK A  pK B = pK W = 14

Chemistry 130 Strength of conjugate acids and bases

This situation occurs because the conjugate base of a weak acid and the conjugate acid of a weak base are both appreciably strong.

The acid or base strength of a conjugate acid or base of a strong base or acid is extremely weak and is negligible in most applications.

The appreciable strength of a conjugate acid or base and the presence of an equilibrium, because the base or acid is weak means that additions of acid or base to a solution that contains the acid­base conjugate pair will not effect the pH of the solution greatly.

These solutions are termed 'buffers'.

Chemistry 130 Buffer solutions

A buffer solution is one that contains a conjugate acid­base pair and is used to provide a relatively constant pH in chemical reactions, biological and medical systems and in industrial settings.

Because of the presence of the conjugate acid­base, we can write

­ + HAaq  H2 Ol  Aaq  H3 Oaq Weak acid Weak acid

[A ­ ][H O+ ] [HA ] K = aq 3 aq and so K aq = [H O+ ] A HA A ­ 3 aq [ aq] [A aq]

from which we can calculate the pH if we know the concentrations of the acid and conjugate base and the acid constant for the acid.

Chemistry 130 Buffer solutions

Given that

­ + HAaq  H2 Ol  Aaq  H3 Oaq Weak acid Weak acid

+ [HAaq] [H3Oaq] = K A ­ [ Aaq] [HA ] [HA ] pH lg H O+ lg K aq lg K lg aq = − 10 [ 3 aq] = − 10 A ­ = − 10  A  − 10 ­ [A ] [A ]  aq   aq  [A­ ] pH = −lg K  lg aq 10  A  10 [HA ]  aq 

Chemistry 130 Buffer solutions

For a buffer, the pH is given by

[A­ ] pH = p K  lg aq A 10 [HA ]  aq 

which is the Henderson­Hasselbalch equation.

Chemistry 130 Chemistry 130