Acid base equilibria

Acid and Base Equilibria

Let’s define acids and bases before we begin discussions about equilibria of weak acids and weak bases in aqueous media.

Acids There are three definitions that can be used to define acids. These definitions can be used to identify acids as they relate to their structural formulas within three modalities. An acid can be described by the Arrhenius definition, or the Brønsted-Lowery definition, or the Lewis definition of acids.

An Arrhenius acid is a substance that dissociates in aqueous media to produce hydronium ions, H3O+. Hydronium ions are water molecules that form a coordinate covalent bond (sometimes referred to as a dative bond) with a hydrogen ion. You can also referred to a hydronium ion as a protonated water molecule. Following is an illustration of the formation of a hydronium ion from a strong acid, i.e., an acid that can be 100% dissociated.

HCl (aq) + H2O (l)

H

H .. + O ..

+

-

Cl

H

A Brønsted-Lowery acid is a substance that donates protons. The following example illustrates how a substance can donate a proton:

1

+ NH4 (aq)

acid

+ H2O (l) base

H

H .. + O ..

H

conjugate acid

+

NH3 (aq)

conjugate base

A Lewis acid is a substance that is an electron pair acceptor. The following example illustrates how a substance can accept a pair of electrons:

acid

base

conjugate base

conjugate acid

The most comprehensive definition of an acid is the Lewis definition. It can fit all conditions without relying on the acid containing hydrogen.

Bases There are three definitions that can be used to define bases. These definitions can be used to identify bases as they relate to their structural formulas within three modalities. A base can be described by the Arrhenius definition, or the Brønsted-Lowery definition, or the Lewis definition of bases.

An Arrhenius base is a substance that ionizes in aqueous media to produce hydroxide ions. Following is an illustration of the formation of a hydroxide ion from a base, i.e., a base that can be 100% dissociated.

2

+

NaOH H2O

Na

(aq)

+

-

OH (aq)

A Brønsted-Lowery base is a substance that is a proton acceptor.

+ NH4 (aq)

acid

+ H2O (l)

H

base

H .. + O ..

H

conjugate acid

+

NH3 (aq)

conjugate base

A Lewis base is an electron pair donor. The following example illustrates how a substance can donate a pair of electrons:

base

acid

The most comprehensive definition of a base is the Lewis definition. It can fit all conditions without relying on the base containing hydroxide.

Conjugate Acids and Bases

3

Acids that function as Arrhenius, Brønsted-Lowery, or the Lewis acids form conjugate bases, and bases that function as Arrhenius, Brønsted-Lowery, or the Lewis bases form conjugate acids _ + HCO3 (aq)

H2O (aq) base

acid

HBr

(aq)

acid

+

NH3 (aq) base

+

CO3

+

H3O(aq)

+

(aq)

conjugate acid

(aq)

conjugate base

conjugate acid

NH4

2-

+

Br

-

(aq)

conjugate base

Relative Strengths of Acids and Bases Proton transfer reactions proceed from the stronger acid-base pair to the weaker acid-base pair.

Let’s write a balanced equation for the reaction that occurs between ammonium chloride and sodium carbonate. Then, we will decide if the equilibrium lies predominantly toward the reactants or the products.

We will use the following diagram to help us make the decision:

4

The net ionic reaction between the reaction of ammonium chloride and sodium carbonate is

2 NH +4 (aq) + CO 23 − (aq) → 2 NH 3 (aq) + H 2CO 3 (aq)

5

Now lets use the fact that “Proton transfer reactions proceed from the stronger acid-base pair to the weaker acid-base pair.” We need to make a decision between the acid strength of NH4+ versus H2CO3. Clearly H2CO3 is a stronger acid than NH4+; therefore, the equilibrium would favor the left, i.e., the formation of the ammonium cation and the carbonate anion. The equilibrium can be written in the following manner:

2 NH +4 (aq) + CO 23 − (aq) ! 2 NH 3 (aq) + H 2CO 3 (aq) acid weaker acid

base weaker base

conjugate base stronger base

conjugate acid stronger acid

The Equilibrium of Weak Acids Let HA represent a weak acid. The equilibrium reaction of this generalized weak acid would be

HA (aq) + H 2O (l)  H 3O +(aq) +A-(aq) and the equilibrium constant is

⎡⎣ H 3O +(aq) ⎤⎦ ⎡⎣ A -(aq) ⎤⎦ Ka = ⎡⎣ HA (aq) ⎤⎦ Water is the solvent medium in which dissociation (proton transfer) occurs; therefore, the concentration of water (which is essentially invariant) is a part of Ka. The Equilibrium of Weak Bases Let B represent a weak base. The equilibrium reaction of this generalized weak base would be

B (aq) + H 2O (l)  BH +(aq) + HO-(aq)

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and the equilibrium constant is

⎡⎣ BH +(aq) ⎤⎦ ⎡⎣OH -(aq) ⎤⎦ Kb = ⎡⎣ B (aq) ⎤⎦ Water is the solvent medium in which dissociation occurs; therefore, the concentration of water (which is essentially invariant) is a part of Kb. The Values of K 1. A large value of K means that the reaction favors the formation of product(s).

2. A small value of K means that the reaction favors the reactants.

3. If Ka is greater than 1, then the acid is strong and dissociates 100%.

4. If Kb is greater than 1, then the base is strong and dissociates 100%.

5. If Ka is 10-16 – 1, then the acid is weak.

6. If Kb is 10-16 – 1, then the base is weak.

7. If Ka is less than 10-16, then the acid is very weak.

8. If Kb is less than 10-16, then the base is very weak.

9. Refer to Appendix D in your textbook for Ka and Kb values for weak acids and weak bases.

Cationic Acid Cationic acids are acids that are conjugate acids of a weak base (i.e., they have donated a proton). NH4+ is an example of a cationic acid

NH (+4 (aq) + H 2 O (l)  NH 3 (aq) + H 3O +(aq)

7

The K for this reaction is Ka or Khydrolysis (Kh), and it is

K a (K h

[ NH ] ⎡⎣ H O )= 3

3

⎡⎣ NH +4 ⎤⎦

+

⎤⎦

The Ka (Kh) for this cationic acid is determined from two other K’s - the autoionization of water and the Kb for ammonia: (1) Auto-ionization of water

H 2 O (l)  OH -(aq) + H 3O +(aq)

⎡⎣H3O+(aq) ⎤⎦ ⎡⎣OH-(aq) ⎤⎦ = K w Kw = 1.0 x 10-14 (2) Kb for NH3

NH 3 (aq) +H 2O (l)  NH +4 (aq) +OH -(aq)

⎡⎣ NH +4 (aq) ⎤⎦ ⎡⎣OH -(aq) ⎤⎦ = Kb ⎡⎣ NH3 (aq) ⎤⎦ Kb = 1.8 x 10-5 Multiplying the auto-ionization of water by the reciprocal Kb will give you Ka(Kh):

8

⎡⎣ NH3 (aq) ⎤⎦ 1 + ⎡ ⎤ ⎡ ⎤ x H O OH = x Kw 3 (aq) ⎦ ⎣ (aq) ⎦ ⎣ + K ⎡⎣ NH 4 (aq) ⎤⎦ ⎡⎣OH (aq) ⎤⎦ b ⎡⎣ NH 3 (aq) ⎤⎦ ⎡⎣ H 3O +(aq) ⎤⎦ Kw = = Kh Kb ⎡⎣ NH +4 (aq) ⎤⎦ ⎡⎣ NH 3 (aq) ⎤⎦ ⎡⎣ H3O +(aq) ⎤⎦ 1.0 x 10-14 = + 1.8 x 10-5 ⎡⎣ NH 4 (aq) ⎤⎦ K h = 5.6 x 10-10 Ka (Kh) is equal to the Kw divided by Kb

Anionic Base Anionic bases are bases that are conjugate bases of a weak acid (i.e., they have acquired a proton). CN- is an example of an anionic base.

CN -(aq) + H 2O(l)  HCN ( (aq) + OH -(aq) The K for this reaction is Kb or Khydrolysis (Kh), and it is

K b (K h ) =

[ HCN ] ⎡⎣ − OH ⎤⎦ ⎡⎣CN − ⎤⎦

The Kb (Kh) for this anionic base is determined from two other K’s - the autoionization of water and the Ka for cyanic acid (HCN): (1) Auto-ionization of water

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H 2 O (l)  OH -(aq) + H 3O +(aq)

⎡⎣H3O+(aq) ⎤⎦ ⎡⎣OH-(aq) ⎤⎦ = K w Kw = 1.0 x 10-14 (2) Ka for HCN

HCN (aq) + H 2O(l)  CN -(aq) + H 3O+(aq) + ⎡⎣CN -(aq) ⎤⎦ ⎡⎣ H3O(aq) ⎤⎦ = Ka ⎡⎣ HCN (aq) ⎤⎦

Kb = 4.9 x 10-10 Multiplying the auto-ionization of water by the reciprocal Ka will give you Kb(Kh):

⎡⎣ HCN (aq) ⎤⎦ 1 + ⎡ ⎤ ⎡ ⎤ x H O OH = x Kw 3 (aq) ⎦ ⎣ (aq) ⎦ ⎣ + Ka ⎡⎣CN (aq) ⎤⎦ ⎡⎣ H 3O (aq) ⎤⎦ ⎤⎦ ⎡⎣ HCN (aq) ⎤⎦ ⎡⎣OH (aq) Kw = = Kh Ka ⎡⎣CN (aq) ⎤⎦

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⎤⎦ 1.0 x 10-14 ⎡⎣ HCN (aq) ⎤⎦ ⎡⎣OH (aq) = 4.9 x 10-10 ⎡⎣CN (aq) ⎤⎦

K h = 2.0 x 10-5 Kb (Kh) is equal to the Kw divided by Ka Application of Kh for a cationic base Calculate Kh for CH3CH(OH)COO- given that the Ka for lactic acid, CH3CH(OH)COOH, is 1.4 x 10-4. + ⎡⎣CH 3CH(OH)COO-(aq) ⎤⎦ ⎡⎣ H 3O(aq) ⎤⎦ = Ka ⎡⎣CH 3CH(OH)COOH (aq) ⎤⎦

⎡⎣CH 3CH(OH)COOH (aq) ⎤⎦ 1 = + Ka ⎡⎣CH 3CH(OH)COO -(aq) ⎤⎦ ⎡⎣ H 3O (aq) ⎤⎦ ⎡⎣CH 3CH(OH)COOH (aq) ⎤⎦ 1 + ⎡ ⎤ ⎡ ⎤ x H O OH = x Kw 3 (aq) (aq) ⎣ ⎦ ⎣ ⎦ + Ka ⎡⎣CH 3CH(OH)COO -(aq) ⎤⎦ ⎡⎣ H 3O (aq) ⎤⎦ ⎡⎣CH 3CH(OH)COOH (aq) ⎤⎦ ⎡⎣OH -(aq) ⎤⎦ Kw = = Kh Ka ⎡⎣CH 3CH(OH)COO -(aq) ⎤⎦

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⎡⎣CH3CH(OH)COOH (aq) ⎤⎦ + ⎡ ⎤ ⎡ ⎤ x H O OH 3 (aq) (aq) ⎦ ⎣ ⎦⎣ + ⎡⎣CH3CH(OH)COO-(aq) ⎤⎦ ⎡⎣ H3O(aq) ⎤⎦ 1 x 1.0 x 10−14 −4 1.4 x 10 = 7.1 x 10−11 = K h =

pH and pOH The pH and pOH are expressions of the strength of an acidic solution and the strength of a basic solution respectively. A pH value less than 7 means that the solution is acidic. A pH value greater than 7 means that the solution is basic. A pH value equal to seven means that the solution is neutral. The “potential for hydrogen (pH)” is equal to the negative common logarithm of the hydronium ion concentration.

pH = - log ⎡⎣H3O+(aq) ⎤⎦ The pOH, “potential for hydroxide,” is equal to the negative common logarithm of the hydroxide ion concentration.

pOH = - log ⎡⎣OH-(aq) ⎤⎦

The sum of pH and pOH is equal to 14 (pH + pOH = 14).

1. Solutions with pH less than 7.00 at 25oC are acidic.

2. Solutions with pH greater than 7.00 at 25oC are basic.

3. Solutions with pH equal to 7.00 at 25oC are neutral.

Application of pH Let’s calculate the pH of a solution made by dissolving 0.7000 g of NaOH in

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sufficient water to produce a volume of 500.0 mL. Sodium hydroxide, NaOH, is a strong base, and it is 100% ionized in aqueous medium to form OH- (aq).

The number of moles of sodium hydroxide can be calculated in the following manner:

mole of NaOH =

0.7000g = 0.01750 mol g 40.00 mol

Since sodium hydroxide is 100% dissociated, the number of moles of the hydroxide ion is

mole of OH(−aq ) = 0.01750 mol The concentration (moles/L) of –OH can be calculated in the following manner:

0.01750 mol 0.5000 L mol ⎡⎣OH -(aq) ⎤⎦ = 0.03500 L ⎡⎣OH -(aq) ⎤⎦ =

The pOH and pH can be calculated in the following manner:

pOH = - log ⎡⎣OH-(aq) ⎤⎦

13

pOH = - log [0.03500] pOH = - (-1.456) pOH = 1.456 pH + pOH = 14.000 pH = 14.000 - pOH pH=14.000 - 1.456 = 12.544 Another Application of pH If [H3O+] in vinegar is 1.6 x 10-3, calculate its pH.

⎡⎣ H3O+ ⎤⎦ = 1.6 x 10-3 M pH = - log (1.6 x 10-3 ) pH = - (-2.8) = 2.8 Another Application of pH The pH of seawater is 8.30. Calculate the [H3O+] and [-OH] of seawater.

pH = 8.30 pH = - log [H 3O + ] - pH = log [H 3O + ] 10- pH = [H 3O + ] 10- 8.30 = [H 3O + ] 5.01 x 10- 9 = [H 3O + ]

[H 3O + ][ - OH] = 1.0 x 10-14 5.01 x 10- 9 = [H3O + ] 5.01 x 10- 9 x [ - OH] = 1.0 x 10-14 1.00 x 10-14 [ OH] = = 2.00 x 10-6 -9 5.01 x 10 -

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Indicators An indicator is a substance that changes colors in some known pH range.

Phenolphthalein is a common indicator for basic solutions that have a pH range between 8.2 and 10.0. The indicator is colorless in an acidic medium, and in basic medium, the color is pink.

The structure of Phenolphthalein

The structure of Conjugate

in acidic medium

Phenolphthalein in basic medium

colorless

pink

Calculating Ka from pH and the Initial Concentration of an Acid

15

The following process can be used to calculate the equilibrium constant of a weak acid given its pH and initial concentration.

HA (aq) + H 2O (l)  H 3O+(aq) +A-(aq) + ⎡⎣ H 3O(aq) ⎤⎦ ⎡⎣ A-(aq) ⎤⎦ Ka = ⎡⎣ HA(aq) ⎤⎦ + ⎡⎣ H 3O(aq) ⎤⎦ = ⎡⎣ A-(aq) ⎤⎦ + ⎡⎣ H 3O(aq) ⎤⎦ Ka = ⎡⎣ HA(aq) ⎤⎦

2

[HA] = a o pH = - log ⎡⎣ H 3O +(aq) ⎤⎦ 10-pH = ⎡⎣ H 3O +(aq) ⎤⎦ 2

⎡⎣10-pH ⎤⎦ = ⎡⎣ H 3O +(aq) ⎤⎦ ⎡⎣10-pH ⎤⎦ Ka = ao

2

2

Where ao is the initial concentration Application: A 0.10 M solution of aqueous lactic acid, CH3CH(OH)COOH, has a pH equal to 2.43 at 25oC. Calculate the Ka for lactic acid at 25oC.

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⎡⎣10-pH ⎤⎦ Ka = ao

2

⎡⎣10-2.43 ⎤⎦ Ka = 0.10

2

2

⎡⎣3.72 x 10−3 ⎤⎦ Ka = 0.10 1.38 x 10−5 Ka = = 1.4 x 10−4 0.10

Calculating Equilibrium Concentrations and pH from Ka and the Initial Concentration of an Acid

HA (aq) + H 2O (l)  H 3O+(aq) +A-(aq) + ⎡⎣ H 3O(aq) ⎤⎦ ⎡⎣ A-(aq) ⎤⎦ Ka = ⎡⎣ HA(aq) ⎤⎦ + ⎡⎣ H 3O(aq) ⎤⎦ = ⎡⎣ A-(aq) ⎤⎦ + ⎡⎣ H 3O(aq) ⎤⎦ Ka = ⎡⎣ HA(aq) ⎤⎦

2

Manipulating the equation gives us an approximation equation that can be used to calculate the pH of an acid from its initial concentration and K a.

17

2

⎡⎣ H 3O +(aq) ⎤⎦ Ka = ⎡⎣ HA (aq) ⎤⎦

⎡⎣ HA (aq) ⎤⎦ K a = ⎡⎣ H 3O +(aq) ⎤⎦

2

⎡⎣ HA (aq) ⎤⎦ K a = ⎡⎣ H 3O +(aq) ⎤⎦ pH = - log ⎡⎣ H 3O +(aq) ⎤⎦ pH = - log ⎡⎣ a o K a ⎤⎦ Application: Calculate the equilibrium concentrations of the hydronium ion and the benzoate ion of a 0.020 M solution of benzoic acid. Calculate the pH a 0.020 M solution of benzoic.

COO

COOH +

H2O

_ +

H3O

+

Ka = 6.3 x 10-5 and ao = 0.20 M The equilibrium concentrations of the hydronium ion and the benzoate anion can be calculate in the following manner: a o K a = ⎡⎣ H 3O +(aq) ⎤⎦ 0.020 x 6.3 x 10 −5 = ⎡⎣H 3O +(aq) ⎤⎦ 1.1 x 10−3 M = ⎡⎣ H 3O +(aq) ⎤⎦

The pH can be calculated in the following manner:

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pH = - log ⎡⎣ a o K a ⎤⎦ pH = - log ⎡ 0.020 x 6.3 x 10−5 ⎤ ⎣ ⎦ pH = - (-2.9) = 2.9

Let’s calculate the hydronium ion concentration and pH of a 0.010 M formic acid solution. (Ka = 1.8 x 10-4).

HCOOH (aq) + H 2O (l)  H 3O +(aq) +HCOO-(aq) + ⎡⎣ H 3O(aq) ⎤⎦ ⎡⎣ HCOO-(aq) ⎤⎦ Ka = ⎡⎣ HCOOH (aq) ⎤⎦ + ⎡⎣ H 3O(aq) ⎤⎦ = ⎡⎣ HCOO-(aq) ⎤⎦ 2

+ ⎡⎣ H 3O(aq) ⎤⎦ Ka = ⎡⎣ HCOOH (aq) ⎤⎦

2

+ ⎡⎣ H 3O(aq) ⎤⎦ Ka = ⎡⎣ HCOOH (aq) ⎤⎦

x2 1.8 x 10 = 0.010-x 1.8x10 4 (0.010-x)=x 2 -4

x 2 + 1.8x10 -4 x - 1.8x10 -6 = 0 -b ± b 2 -4ac 2a a = 1; b = 1.8x10-4 ; and c = 1.8x10-6 x=

x=

-1.8x10-4 ± 3.24 x 10 −8 - 4 (1) (-1.8x10 -6 ) 2 (1)

-1.8x10-4 ± 2.7 x 10 −3 x= 2 x = 1.3 x 10-3 M = ⎡⎣ H 3O + ⎤⎦

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⎡⎣ H 3O +(aq) ⎤⎦ ⎡⎣ HCOO-(aq) ⎤⎦ Ka = ⎡⎣ HCOOH (aq) ⎤⎦ (x) (x) 0.010 - x x2 1.8 x 10-4 = 0.010 1.8 x 10-4 =

1.8 x 10-6 = x 1.3 x 10-3 = x

We could use the approximation equation ⎡⎣ H 3O + (aq) ⎤⎦ = a o K a

However, we must demonstrate that (ao – x) is approximately ao is valid, i.e., ao is much larger than x.

If the acid is less than 15% ionized, then the approximation is valid In the above problem x x 100 = % ionization ao 1.3 x 10-3 x 100 = 13 % 0.010

So, instead of using the long method, we could have used the approximation equation ⎡⎣ H 3O + (aq) ⎤⎦ = a o K a

Therefore, we could have made the following Approximation: 0.010 is greater than x; consequently 0.010 M – x = 0.010 M

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⎡⎣ H 3O + (aq) ⎤⎦ = a o K a = 0.010 x 1.8 x 10 −4 ⎡⎣ H 3O + (aq) ⎤⎦ = 1.8 x 10 −6 = 1.3 x 10 −3

What if we change the initial concentration of the formic acid to 0.0010 M? Calculate the hydronium ion concentration and pH of a 0.0010 M formic acid solution. (Ka = 1.8 x 10-4) Let’s make an approximation: ⎡⎣ H 3O +(aq) ⎤⎦ ⎡⎣ HCOO-(aq) ⎤⎦ Ka = ⎡⎣ HCOOH (aq) ⎤⎦ (x) (x) 0.0010 - x x2 1.8 x 10-4 = 0.0010 1.8 x 10-4 =

1.8 x 10-7 = x 4.2 x 10-4 = x

Assume 0.0010 is greater than x

x x 100 = % ionization ao

4.2 x 10-4 x 100 = 42 % 0.0010 Therefore, the assumption cannot be made, and the quadratic equation must be used.

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HCOOH (aq) + H 2O (l)  H 3O+(aq) +HCOO-(aq) + ⎡⎣ H 3O(aq) ⎤⎦ ⎡⎣ HCOO-(aq) ⎤⎦ Ka = ⎡⎣ HCOOH (aq) ⎤⎦ + ⎡⎣ H 3O(aq) ⎤⎦ = ⎡⎣ HCOO-(aq) ⎤⎦ 2

+ ⎡⎣ H 3O(aq) ⎤⎦ Ka = ⎡⎣ HCOOH (aq) ⎤⎦

2

+ ⎡⎣ H 3O(aq) ⎤⎦ Ka = ⎡⎣ HCOOH (aq) ⎤⎦

x2 1.8 x 10 = 0.0010-x 1.8x10 4 (0.0010-x) = x 2 -4

x 2 + 1.8x10 -4 x - 1.8x10 -7 = 0

-b ± b 2 -4ac 2a a = 1; b = 1.8x10-4 ; and c = 1.8x10-7 x=

-1.8x10-4 ± 3.24 x 10−8 - 4 (1) (-1.8x10-7 ) x= 2 (1) -1.8x10-4 ± 8.7 x 10−4 x= 2 x = 3.4 x 10-4 M = ⎡⎣ H3O + ⎤⎦

x x 100 = % ionization ao

3.4 x 10-4 x 100 = 34 % 0.0010

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Calculating the pH of an Aqueous Solution of a Weak Base First, let’s set up the equilibrium equations for a weak base:

B (aq) +H 2O (l)  BH +(aq) +OH -(aq) ⎡⎣ BH +(aq) ⎤⎦ ⎡⎣OH -(aq) ⎤⎦ Kb = ⎡⎣ B (aq) ⎤⎦ ⎡⎣ BH +(aq) ⎤⎦ = ⎡⎣OH -(aq) ⎤⎦ ⎡⎣OH -(aq) ⎤⎦ Kb = ⎡⎣ B (aq) ⎤⎦

2

The Kb for the base The following equation results from making the approximation that the initial concentration is greater than the amount dissociated:

⎡⎣OH -(aq) ⎤⎦ Kb = ⎡⎣ B (aq) ⎤⎦

2

K b ⎡⎣ B (aq) ⎤⎦ = ⎡⎣OH -(aq) ⎤⎦

2

K b ⎡⎣ B (aq) ⎤⎦ = ⎡⎣OH -(aq) ⎤⎦

23

pH + pOH = 14.00 K b ⎡⎣ B

(aq)

⎤⎦ = ⎡⎣OH -(aq) ⎤⎦

− log K b ⎡⎣ B

(aq)

⎤⎦ = pOH

pH − log K b ⎡⎣ B

(aq)

⎤⎦ = 14.00

pH = 14.00 + log K b ⎡⎣ B

(aq)

⎤⎦

pH = 14.00 + log K b b o where bo is the initial concentration of the base Application using the approximation equation: Calculate the pH of a 0.010 M aqueous solution of pyridine, C5H5N. Pyridine is a weak base with Kb = 1.7 x 10-9. The initial concentration of the base (pyridine) is 0.010 M.

pH = 14.00 + log K b b o pH = 14.00 + log 1.7 x 10 −9 x 0.010 pH = 14.00 + log 1.7 x 10 −9 x 0.010 pH = 14.00 -5.4 pH = 8.6 Calculating the pH of an Aqueous Solution of the Salt of a Weak Acid Treating a weak acid with a strong base (-OH) results in the formation an anionic base, A-, that functions as a BrØnsted-Lowery base. The resulting anionic base, A-, reacts with water to form hydroxide ions.

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HA (aq) +OH -(aq)  A- (aq) +H 2O(l) A-(aq) +H 2O(l)  HA(aq) +OH -(aq) K w ⎡⎣ HA (aq) ⎤⎦ ⎡⎣OH (aq) ⎤⎦ Kh = = Ka ⎡⎣ A-(aq) ⎤⎦ K w ⎡⎣OH (aq) ⎤⎦ Kh = = Ka ⎡⎣ A-(aq) ⎤⎦

2

⎡⎣ OH -(aq) ⎤⎦ = ⎡⎣ A -(aq) ⎤⎦

2

Kw Ka

2 K ⎡⎣ A -(aq) ⎤⎦ w = ⎡⎣ OH -(aq) ⎤⎦ Ka

ao

Kw = ⎡⎣OH -(aq) ⎤⎦ Ka

pH + pOH = 14.00 ao

Kw = ⎡⎣OH -(aq) ⎤⎦ Ka

− log a o

Kw = pOH Ka

pH − log a o

Kw = 14.00 Ka

pH = 14.00 + log a o

Kw Ka

The resulting equation uses the approximation that the concentration of the anionic acid is much larger that the amount that forms the hydroxide ion.

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Application: Calculate the pH of a 0.015 M solution of sodium hypochlorite, NaClO. The Ka for hypochlorous acid is 3.5 x 10-8.

ClO-(aq) + H 2O(l)  HClO (aq) + OH -(aq) pH = 14.00 + log a o

Kw Ka

1.00 x10−14 pH = 14.00 + log 0.015 x 3.5x10-8 1.00 x10−14 pH = 14.00 + log 0.015 x 3.5x10-8 pH = 14.00 -4.2 pH = 9.8

The pH of an Aqueous Solution of the Salt of a Weak Base Treating a weak base with a strong acid results in the formation a cationic acid, BH+, that functions as a BrØnsted-Lowery acid. The resulting cationic acid, BH+, reacts with water to form hydronium ions.

B

(aq)

+ + + H 3O(aq) → BH (aq) + H 2O(l)

+ + BH (aq) + H 2O(l) ! B(aq) + H 3O(aq)

+ K w ⎡⎣ B(aq) ⎤⎦ ⎡⎣ H 3O(aq) ⎤⎦ Kh = = + Kb ⎡⎣ BH (aq) ⎤⎦

+ ⎡ H 3O(aq) ⎤⎦ K Kh = w = ⎣ + Kb ⎡⎣ BH (aq) ⎤⎦

2

26

2

+ ⎡⎣ H 3O(aq) ⎤⎦ Kw = Kb ⎡⎣ BH +(aq) ⎤⎦ 2 K + ⎡⎣ BH +(aq) ⎤⎦ w = ⎡⎣ H 3O(aq) ⎤⎦ Ka

bo

Kw + ⎤⎦ = ⎡⎣ H 3O(aq) Kb

+ ⎤⎦ = pH -log ⎡⎣ H 3O(aq)

⎡ K ⎤ -log ⎢ b o w ⎥ = pH Kb ⎦ ⎣ Application: Calculate the pH of a 0.50 M solution of ammonium chloride, NH4Cl. The Ka for ammonia is 1.8 x 10-5. + NH +4 (aq) + H 2O(l) ! NH 3 (aq) + H 3O(aq)

Kw Kb

+ ⎡⎣ H 3O(aq) ⎤⎦ = ⎡⎣ BH +(aq) ⎤⎦

2

2 K + ⎡⎣ BH +(aq) ⎤⎦ w = ⎡⎣ H 3O(aq) ⎤⎦ Ka

bo

Kw + ⎤⎦ = ⎡⎣ H 3O(aq) Kb

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bo

Kw + ⎤⎦ = ⎡⎣ H 3O (aq) Kb

1.00 x 10−14 + ⎤⎦ 0.50 x = ⎡⎣ H 3O (aq) −5 1.8 x 10 1.00 x 10−14 − log 0.50 x = pH 1.8 x 10−5 - (-4.8) = pH 4.8 = pH

Common Ion Effect

The common ion effect is the addition of a common ion to the equilibrium.

An example is the addition of NaA to an acid solution of HA, a weak acid.

The initial concentration of HA is ao. The equilibrium involves a small amount (x) of HA dissociating to “x” amount of A- and “x” amount of H3O+ with (ao – x) HA remaining at equilibrium. ao

Initially

ao - x

x

x

at equilibrium

+ HA (aq) + H 2O(l) ! A-(aq) + H 3O(aq)

y

y

y

+ NaA (aq) → Na (aq) + A-(aq)

The addition of the common ion

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+ HA(aq) + H 2O(l) ! H 3O(aq) + A(-aq )

The concentration of A- is x + y

⎡⎣ H 3O +(aq) ⎤⎦ ⎡⎣ A -(aq) ⎤⎦ Ka = ⎡⎣ HA (aq) ⎤⎦ Ka =

(x) (x+y) a o -x

Assumptions: y is greater than x; a ois greater than x Ka =

(x) (y) ao

Ka o =x y + ⎤⎦ = x and y = A(−aq) , the common ion Where ⎡⎣H3O(aq)

Application: Let’s calculate the equilibrium concentration of the hydronium ion in a 0.100 M solution of acetic acid, CH3COOH. The equilibrium constant of acetic acid at 25 oC is 1.8 x 10-5.

The equation for the dissociation of acetic acid at 25 oC is + CH 3 COOH (aq) + H 2O(l) ! H 3O(aq) + CH 3COO(-aq )

+ ⎡⎣ H 3O(aq) ⎤⎦ ⎡⎣CH 3COO- ⎤⎦ Ka = ⎡⎣CH 3COOH (aq) ⎤⎦

without adding a common ion, [H3O+] = [CH3COO-]; therefore,

1.8 x 10

−5

x2 = 0.100

1.8 x 10 −5 x 0.100 = x 2

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1.8 x 10 −5 =

x (x + 0.1) + ⎤⎦ 1.8 x 10 -5 x 0.100 = x = ⎡⎣ H 3O(aq) 0.1 -x

+ ⎤⎦ 1.8 x 10 -6 = ⎡⎣ H 3O(aq)

+ ⎤⎦ 1.3 x 10 −3 = ⎡⎣ H 3O(aq)

Now, let’s add a common ion. Let’s add sufficient sodium acetate so that the solution is 0.100 M in CH3COONa. The equilibrium is 0.100 – x

x

x

+ CH 3 COOH (aq) + H 2O(l) ! H 3O(aq) + CH 3COO(-aq )

0.100

0.100

0.100

+ CH 3 COONa (aq) → Na (aq) + CH 3COO(-aq )

1.8 x 10 −5 =

x (x + 0.1) 0.1 -x

0.100 is greater than x Therefore, x = [H3O+] = 1.8 x 10-5 M Consequently, the common ion decreases the concentration of the hydronium ion. In short, the common ion shifts the equilibrium to the reactants or suppresses the dissociation of the weak acid.

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Calculating the pH and the Equilibrium Concentrations of Species in Solution formed from the Reactions of Anionic Bases that Produce Multi-Species in Solution Following is a generalized example of a species that forms multiple species in solution:

(1) A 3-(aq) + H 2O (l) ! HA 2-(aq) + OH -(aq) (2) HA 2-(aq) + H 2O (l) ! H 2 A -(aq) + OH -(aq) (3) H 2 A -(aq) + H 2O (l) ! H 3A (aq) + OH -(aq) Following are the equations that govern the behavior of multiple species in solution: K h1 = ⎡⎣ H 3O

+ (aq)

⎤⎦ ⎡⎣OH

(aq)

⎡⎣ HA 2-(aq) ⎤⎦ ⎡⎣ HA 2-(aq) ⎤⎦ ⎡⎣ OH -(aq) ⎤⎦ ⎤⎦ x = ⎡⎣ A 3-(aq) ⎤⎦ ⎡⎣ H 3O +(aq) ⎤⎦ ⎡⎣ A 3-(aq) ⎤⎦

⎡⎣ HA 2-(aq) ⎤⎦ ⎡⎣OH -(aq) ⎤⎦ 1 K h1 = K w x = K a3 ⎡⎣ A 3-(aq) ⎤⎦ ⎡⎣ HA 2-(aq) ⎤⎦ = ⎡⎣OH -(aq) ⎤⎦ If ⎡⎣ HA 2-(aq) ⎤⎦ and ⎡⎣OH -(aq) ⎤⎦ equal x and the initial concentration of ⎡⎣ A 3-(aq) ⎤⎦ = a o then Kw x2 = K a3 ao

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Kw x2 = K a3 ao aoKw =x K a3 where x = ⎡⎣ HA 2-(aq) ⎤⎦ and ⎡⎣OH -(aq) ⎤⎦

x-y

y y

2HA(aq) + H 2O ! H 2 A-(aq) + - OH (aq)

⎡⎣OH -(aq) ⎤⎦ = y+x ⎡⎣ H 2 A -(aq) ⎤⎦ = y ⎡⎣ HA 2-(aq) ⎤⎦ = x − y but x is greater than y; therefore,

y+x is essentially x and x-y is essentially x

K h 2 = ⎡⎣ H 3O K h2

+ (aq)

⎤⎦ ⎡⎣OH

(aq)

⎡⎣ H 2 A -(aq) ⎤⎦ ⎡⎣ H 2 A -(aq) ⎤⎦ ⎡⎣OH -(aq) ⎤⎦ ⎤⎦ x = ⎡⎣ HA 2-(aq) ⎤⎦ ⎡⎣ H3O +(aq) ⎤⎦ ⎡⎣ HA 2-(aq) ⎤⎦

⎡⎣ H 2 A -(aq) ⎤⎦ ⎡⎣OH -(aq) ⎤⎦ 1 = Kw x = K a2 ⎡⎣ HA 2-(aq) ⎤⎦

⎡⎣OH -(aq) ⎤⎦ = x as a consequence of an approximation ⎡⎣ HA 2-(aq) ⎤⎦ = x as a consequence of an approximation then y = ⎡⎣ H 2 A -(aq) ⎤⎦ Kw =y K a2

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y

z

z

H 2 A-(aq) + H 2O ! H 3A + - OH (aq)

⎡⎣OH -(aq) ⎤⎦ = z+x ⎡⎣ H 3A (aq) ⎤⎦ = z ⎡⎣ HA 2-(aq) ⎤⎦ = y-z but x is greater than z and y is greater than z; therefore, z+x is essentially x and y-z is essentially y K h3 = ⎡⎣ H 3O K h3

+ (aq)

⎤⎦ ⎡⎣OH

(aq)

⎡⎣ H 3 A (aq) ⎤⎦ ⎣⎡OH -(aq) ⎦⎤ ⎡⎣ H 3 A (aq) ⎤⎦ ⎤⎦ x = ⎡⎣ H 2 A -(aq) ⎤⎦ ⎡⎣ H 3O +(aq) ⎤⎦ ⎡⎣ H 2 A -(aq) ⎤⎦

⎡⎣ H 3 A (aq) ⎤⎦ ⎡⎣OH -(aq) ⎤⎦ 1 = Kw x = K a1 ⎡⎣ H 2 A -(aq) ⎤⎦

⎡⎣OH -(aq) ⎤⎦ = x as a consequence of an approximation ⎡⎣ H 2 A -(aq) ⎤⎦ = y as a consequence of an approximation then z = ⎡⎣ H 3 A (aq) ⎤⎦ ⎡ H 3A (aq) ⎤⎦ x Kw = ⎣ K a1 y y Kw = ⎡⎣ H 3A (aq) ⎤⎦ x K a1

Application: Calculate the equilibrium concentrations of all species in solution

for 0.10 M solution of Na2CO3.

H 2CO 3 (aq) +H 2O (l)  HCO-3 (aq) +H 3O +(aq) K a1 = 4.3 x 10 −7 HCO-3 (aq) + H 2O (l)  CO2-3 (aq) + H 3O +(aq) K a2 = 5.6 x 10 -11

33

(1) CO 2-3 (aq) + H 2O (l)  HCO-3 (aq) + OH -(aq) (2) HCO-3 (aq) + H 2O(l)  H 2CO 3 +OH -(aq)

Kw x2 = K a2 ao aoKw =x K a2 where x = ⎡⎣ HCO3- (aq) ⎤⎦ and ⎡⎣OH -(aq) ⎤⎦

1 x 10-14 x2 = 4.3 x 10-7 0.10 M 0.10 M x 1 x 10-14 =x 5.6 x 10-11 4.2 x 10-3 = x

The concentration of ⎡⎣HCO3- (aq) ⎤⎦ and ⎡⎣OH-(aq) ⎤⎦

x

HCO

3 (aq)

y

y

+ H 2 O(l)  H 2 CO 3 +OH -(aq)

⎡⎣OH -(aq) ⎤⎦ = y+x ⎡⎣ H 2 CO3 (aq) ⎤⎦ = y ⎡⎣ HCO3- (aq) ⎤⎦ = x − y but x is greater than y; therefore, y+x is essentially x and x-y is essentially x

34

Kw =y K a1 1 x 10-14 =y 4.30 x 10-7 2.3 x 10-8 = y

+ ⎡⎣ Na (aq) ⎤⎦ = 0.20 M ⎡⎣OH -(aq) ⎤⎦ = 4.2 x 10-3 M

⎡⎣ HCO3- (aq) ⎤⎦ = 4.2 x 10-3 M ⎡⎣ H 2CO3 (aq_ ⎤⎦ = 4.2 x 10-3 M

Calculate the pH of the solution pH = 14.00 + log

aoKw K a2

0.10 x 1 x 10-14 pH = 14.00 + log 5.61 x 10-11 pH = 14.00 + log (4.22 x 10-3 ) = 14.00 -2.4 = 11.6

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