1

Relationship between pH and pOHRelationship between pH and pOH

the sum of the pH and pOH of a solution = 14.00 at 25°C

can use pOH to find pH of a solution

the sum of the pH and pOH of a solution = 14.00 at 25°C

can use pOH to find pH of a solution

2

pKpK

a way of expressing the strength of an acid or base is pK

pKa = -log(Ka), Ka = 10-pKa

pKb = -log(Kb), Kb = 10-pKb

the stronger the acid, the smaller the pKa

larger Ka = smaller pKa

because it is the –log

a way of expressing the strength of an acid or base is pK

pKa = -log(Ka), Ka = 10-pKa

pKb = -log(Kb), Kb = 10-pKb

the stronger the acid, the smaller the pKa

larger Ka = smaller pKa

because it is the –log

3

Finding the pH of a Strong AcidFinding the pH of a Strong Acid

there are two sources of H3O+ in an aqueous solution of a strong acid – the acid and the water

for the strong acid, the contribution of the water to the total [H3O+] is negligible

for a monoprotic strong acid [H3O+] = [HA]

0.10 M HCl has [H3O+] = 0.10 M and pH = 1.00

there are two sources of H3O+ in an aqueous solution of a strong acid – the acid and the water

for the strong acid, the contribution of the water to the total [H3O+] is negligible

for a monoprotic strong acid [H3O+] = [HA]

0.10 M HCl has [H3O+] = 0.10 M and pH = 1.00

4

Finding the pH of a Weak AcidFinding the pH of a Weak Acid

there are also two sources of H3O+ in and aqueous solution of a weak acid – the acid and the water

however, finding the [H3O+] is complicated by the fact that the acid only undergoes partial ionization

calculating the [H3O+] requires solving an ICE problem for the reaction that defines the acidity of the acid (the top equation)

For a weak acid Ka << 1 so we may be able to apply approximations to avoid solving a quadratic

there are also two sources of H3O+ in and aqueous solution of a weak acid – the acid and the water

however, finding the [H3O+] is complicated by the fact that the acid only undergoes partial ionization

calculating the [H3O+] requires solving an ICE problem for the reaction that defines the acidity of the acid (the top equation)

For a weak acid Ka << 1 so we may be able to apply approximations to avoid solving a quadratic

5

[HNO2] [NO2-] [H3O+]

initial

change

equilibrium

Find the pH of 0.200 M HNO2(aq) solution @ 25°CFind the pH of 0.200 M HNO2(aq) solution @ 25°C

Write the reaction for the acid with water

Construct an ICE table for the reaction

Enter the initial concentrations – assuming the [H3O+] from water is ≈ 0

since no products initially, Qc = 0, and the reaction is proceeding forward

HNO2 + H2O NO2- + H3O+

[HNO2] [NO2-] [H3O+]

initial 0.200 0 ≈ 0

change

equilibrium

6

[HNO2] [NO2-] [H3O+]

initial 0.200 0 0

change

equilibrium

represent the change in the concentrations in terms of x

sum the columns to find the equilibrium concentrations in terms of x

substitute into the equilibrium constant expression

+x+x-x

0.200 -x x x

Find the pH of 0.200 M HNO2(aq) solution @ 25°CFind the pH of 0.200 M HNO2(aq) solution @ 25°C

7

determine the value of Ka from Table 15.5

since Ka is very small, approximate the [HNO2]eq = [HNO2]init and solve for x

Ka for HNO2 = 4.6 x 10-4

[HNO2] [NO2-] [H3O+]

initial 0.200 0 ≈ 0

change -x +x +x

equilibrium 0.200 x x0.200 -x

Find the pH of 0.200 M HNO2(aq) solution @ 25°CFind the pH of 0.200 M HNO2(aq) solution @ 25°C

8

Ka for HNO2 = 4.6 x 10-4

[HNO2] [NO2-] [H3O+]

initial 0.200 0 ≈ 0

change -x +x +x

equilibrium 0.200 x x

check if the approximation is valid by seeing if x < 5% of [HNO2]init

the approximation is valid

x = 9.6 x 10-3

Find the pH of 0.200 M HNO2(aq) solution @ 25°CFind the pH of 0.200 M HNO2(aq) solution @ 25°C

9

Ka for HNO2 = 4.6 x 10-4

[HNO2] [NO2-] [H3O+]

initial 0.200 0 ≈ 0

change -x +x +x

equilibrium 0.200-x x x

x = 9.6 x 10-3

substitute x into the equilibrium concentration definitions and solve

[HNO2] [NO2-] [H3O+]

initial 0.200 0 ≈ 0

change -x +x +x

equilibrium 0.190 0.0096 0.0096

Find the pH of 0.200 M HNO2(aq) solution @ 25°CFind the pH of 0.200 M HNO2(aq) solution @ 25°C

10

Ka for HNO2 = 4.6 x 10-4

substitute [H3O+] into the formula for pH and solve

[HNO2] [NO2-] [H3O+]

initial 0.200 0 ≈ 0

change -x +x +x

equilibrium 0.190 0.0096 0.0096

Find the pH of 0.200 M HNO2(aq) solution @ 25°CFind the pH of 0.200 M HNO2(aq) solution @ 25°C

11

Ka for HNO2 = 4.6 x 10-4

[HNO2] [NO2-] [H3O+]

initial 0.200 0 ≈ 0

change -x +x +x

equilibrium 0.190 0.0096 0.0096

check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka

though not exact, the answer is reasonably close

Find the pH of 0.200 M HNO2(aq) solution @ 25°CFind the pH of 0.200 M HNO2(aq) solution @ 25°C

Tro, Chemistry: A Molecular Approach

12

What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?

13

Write the reaction for the acid with water

Construct an ICE table for the reaction

Enter the initial concentrations – assuming the [H3O+] from water is ≈ 0

HC6H4NO2 + H2O C6H4NO2- + H3O+

[HA] [A-] [H3O+]

initial 0.012 0 ≈ 0

change

equilibrium

What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?

14

[HA] [A-] [H3O+]

initial 0.012 0 0

change

equilibrium

represent the change in the concentrations in terms of x

sum the columns to find the equilibrium concentrations in terms of x

substitute into the equilibrium constant expression

+x+x-x

0.012 -x x x

What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?

HC6H4NO2 + H2O C6H4NO2- + H3O+

15

determine the value of Ka

since Ka is very small, approximate the [HA]eq = [HA]init and solve for x

[HA] [A2-] [H3O+]

initial 0.012 0 ≈ 0

change -x +x +x

equilibrium 0.012 x x0.012 -x

What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?

HC6H4NO2 + H2O C6H4NO2- + H3O+

16

Ka for HC6H4NO2 = 1.4 x 10-5

check if the approximation is valid by seeing if x < 5% of [HC6H4NO2]init

the approximation is valid

x = 4.1 x 10-4

[HA] [A2-] [H3O+]

initial 0.012 0 ≈ 0

change -x +x +x

equilibrium 0.012 x x

What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?

17

x = 4.1 x 10-4

substitute x into the equilibrium concentration definitions and solve

[HA] [A2-] [H3O+]

initial 0.012 0 ≈ 0

change -x +x +x

equilibrium 0.012-x x x

What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?

18

substitute [H3O+] into the formula for pH and solve

[HA] [A2-] [H3O+]

initial 0.012 0 ≈ 0

change -x +x +x

equilibrium 0.012 0.00041 0.00041

What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?

19

check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka

the values match

[HA] [A2-] [H3O+]

initial 0.012 0 ≈ 0

change -x +x +x

equilibrium 0.012 0.00041 0.00041

What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?

20

[HClO2] [ClO2-] [H3O+]

initial 0.100 0 ≈ 0

change

equilibrium

Write the reaction for the acid with water

Construct an ICE table for the reaction

Enter the initial concentrations – assuming the [H3O+] from water is ≈ 0

HClO2 + H2O ClO2- + H3O+

Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2

Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2

21

represent the change in the concentrations in terms of x

sum the columns to find the equilibrium concentrations in terms of x

substitute into the equilibrium constant expression

[HClO2] [ClO2-] [H3O+]

initial 0.100 0 ≈ 0

change -x +x +x

equilibrium 0.100-x x x

Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2

Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2

Tro, Chemistry: A Molecular Approach

22

Wirte an expression for Ka and use the ICE table to obtain an equation for x

since Ka is very small, approximate the [HClO2]eq = [HClO2]init and solve for x

[HClO2] [ClO2-] [H3O+]

initial 0.100 0 ≈ 0

change -x +x +x

equilibrium 0.100-x x x

Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2

Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2

23

check if the approximation is valid by seeing if x < 5% of [HNO2]init

the approximation is invalid

x = 3.3 x 10-2

[HClO2] [ClO2-] [H3O+]

initial 0.100 0 ≈ 0

change -x +x +x

equilibrium 0.100-x x x

Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2

Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2

24

if the approximation is invalid, solve for x using the quadratic formula

Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2

Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2

25

x = 0.028

substitute x into the equilibrium concentration definitions and solve

[HClO2] [ClO2-] [H3O+]

initial 0.100 0 ≈ 0

change -x +x +x

equilibrium 0.100-x x x

[HClO2] [ClO2-] [H3O+]

initial 0.100 0 ≈ 0

change -x +x +x

equilibrium 0.072 0.028 0.028

Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2

Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2

26

substitute [H3O+] into the formula for pH and solve

[HClO2] [ClO2-] [H3O+]

initial 0.100 0 ≈ 0

change -x +x +x

equilibrium 0.072 0.028 0.028

Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2

Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2

27

check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka

the answer matches

[HClO2] [ClO2-] [H3O+]

initial 0.100 0 ≈ 0

change -x +x +x

equilibrium 0.072 0.028 0.028

Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2

Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2

28

What is the Ka of a weak acid if a 0.100 M solution has a pH of 4.25?

What is the Ka of a weak acid if a 0.100 M solution has a pH of 4.25?

Use the pH to find the equilibrium [H3O+]

Write the reaction for the acid with water

Construct an ICE table for the reaction

Enter the initial concentrations and [H3O+]equil

HA + H2O A- + H3O+

[HA] [A-] [H3O+]

initial

change

equilibrium

[HA] [A-] [H3O+]

initial 0.100 0 ≈ 0

change

equilibrium 5.6E-05

29

[HA] [A-] [H3O+]

initial 0.100 0 0

change

equilibrium

fill in the rest of the table using the [H3O+] as a guide

if the difference is insignificant, [HA]equil = [HA]initial

substitute into the Ka expression and compute Ka

+5.6E-05+5.6E-05−5.6E-05

0.100 5.6E-05

5.6E-05 5.6E-050.100

What is the Ka of a weak acid if a 0.100 M solution has a pH of 4.25?

What is the Ka of a weak acid if a 0.100 M solution has a pH of 4.25?

HA + H2O A- + H3O+

30

Percent IonizationPercent Ionization

another way to measure the strength of an acid is to determine the percentage of acid molecules that ionize when dissolved in water – this is called the percent ionization the higher the percent ionization, the stronger the acid

another way to measure the strength of an acid is to determine the percentage of acid molecules that ionize when dissolved in water – this is called the percent ionization the higher the percent ionization, the stronger the acid

• since [ionized acid]equil = [H3O+]equil

31

What is the percent ionization of a 2.5 M HNO2 solution if Ka for HNO2 = 4.6 x 10-4 ?

What is the percent ionization of a 2.5 M HNO2 solution if Ka for HNO2 = 4.6 x 10-4 ?

Write the reaction for the acid with water

Construct an ICE table for the reaction

Enter the Initial Concentrations

Define the Change in Concentration in terms of x

Sum the columns to define the Equilibrium Concentrations

HNO2 + H2O NO2- + H3O+

[HNO2] [NO2-] [H3O+]

initial

change

equilibrium

[HNO2] [NO2-] [H3O+]

initial 2.5 0 ≈ 0

change

equilibrium

+x+x-x

2.5 - x x x

32

determine the value of Ka from Table 15.5

since Ka is very small, approximate the [HNO2]eq = [HNO2]init and solve for x

[HNO2] [NO2-] [H3O+]

initial 2.5 0 ≈ 0

change -x +x +x

equilibrium 2.5-x ≈2.5 x x

What is the percent ionization of a 2.5 M HNO2 solution if Ka for HNO2 = 4.6 x 10-4 ?

What is the percent ionization of a 2.5 M HNO2 solution if Ka for HNO2 = 4.6 x 10-4 ?

33

[HNO2] [NO2-] [H3O+]

initial 2.5 0 ≈ 0

change -x +x +x

equilibrium 2.5 0.034 0.0342.5 - x x x

substitute x into the Equilibrium Concentration definitions and solve

x = 3.4 x 10-2

What is the percent ionization of a 2.5 M HNO2 solution if Ka for HNO2 = 4.6 x 10-4 ?

What is the percent ionization of a 2.5 M HNO2 solution if Ka for HNO2 = 4.6 x 10-4 ?

HNO2 + H2O NO2- + H3O+

34

[HNO2] [NO2-] [H3O+]

initial 2.5 0 ≈ 0

change -x +x +x

equilibrium 2.5 0.034 0.034

Apply the Definition and Compute the Percent Ionization

since the percent ionization is < 5%, the “x is small” approximation is valid

What is the percent ionization of a 2.5 M HNO2 solution if Ka for HNO2 = 4.6 x 10-4 ?

What is the percent ionization of a 2.5 M HNO2 solution if Ka for HNO2 = 4.6 x 10-4 ?

HNO2 + H2O NO2- + H3O+

35

Relationship Between [H3O+]equilibrium and [HA]initialRelationship Between [H3O+]equilibrium and [HA]initial

What happens to [H3O+] at equilibrium if we dilute [HA]?

decreasing/increasing the initial concentration of acid results in increased/decreased percent ionization

this means that the increase in H3O+ concentration is slower than the increase in acid concentration

What happens to [H3O+] at equilibrium if we dilute [HA]?

decreasing/increasing the initial concentration of acid results in increased/decreased percent ionization

this means that the increase in H3O+ concentration is slower than the increase in acid concentration

36

Finding the pH of Mixtures of AcidsFinding the pH of Mixtures of Acids

generally, you can ignore the contribution of the weaker acid to the [H3O+]eq

for a mixture of a strong acid with a weak acid, the complete ionization of the strong acid provides more than enough [H3O+] to shift the weak acid equilibrium to the left

so far that the weak acid’s added [H3O+] is negligible

for mixtures of weak acids, generally only need to consider the stronger for the same reasons

as long as one is significantly stronger than the other,

and their concentrations are similar

generally, you can ignore the contribution of the weaker acid to the [H3O+]eq

for a mixture of a strong acid with a weak acid, the complete ionization of the strong acid provides more than enough [H3O+] to shift the weak acid equilibrium to the left

so far that the weak acid’s added [H3O+] is negligible

for mixtures of weak acids, generally only need to consider the stronger for the same reasons

as long as one is significantly stronger than the other,

and their concentrations are similar

37

Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)

Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)

Write the reactions for the acids with water and determine their Kas

If the Kas are sufficiently different, use the strongest acid to construct an ICE table for the reaction

Enter the initial concentrations – assuming the [H3O+] from water is ≈ 0

HF + H2O F- + H3O+ Ka = 3.5 x 10-4

[HF] [F-] [H3O+]

initial 0.150 0 ≈ 0

change

equilibrium

HClO + H2O ClO- + H3O+ Ka = 2.9 x 10-8

H2O + H2O OH- + H3O+ Kw = 1.0 x 10-14

38

[HF] [F-] [H3O+]

initial 0.150 0 0

change

equilibrium

Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)

Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)

represent the change in the concentrations in terms of x

sum the columns to find the equilibrium concentrations in terms of x

substitute into the equilibrium constant expression

+x+x-x

0.150 -x x x

39

determine the value of Ka for HF

since Ka is very small, approximate the [HF]eq = [HF]init and solve for x

Ka for HF = 3.5 x 10-4

[HF] [F-] [H3O+]

initial 0.150 0 ≈ 0

change -x +x +x

equilibrium 0.150 x x0.150 -x

Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)

Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)

40

Ka for HF = 3.5 x 10-4

[HF] [F-] [H3O+]

initial 0.150 0 ≈ 0

change -x +x +x

equilibrium 0.150 x x

check if the approximation is valid by seeing if x < 5% of [HF]init

the approximation is valid

x = 7.2 x 10-3

Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)

Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)

41

Ka for HF = 3.5 x 10-4

[HF] [F-] [H3O+]

initial 0.150 0 ≈ 0

change -x +x +x

equilibrium 0.150-x x x

x = 7.2 x 10-3

substitute x into the equilibrium concentration definitions and solve

[HF] [F-] [H3O+]

initial 0.150 0 ≈ 0

change -x +x +x

equilibrium 0.143 0.0072 0.0072

Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)

Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)

42

Ka for HF = 3.5 x 10-4

substitute [H3O+] into the formula for pH and solve

[HF] [F-] [H3O+]

initial 0.150 0 ≈ 0

change -x +x +x

equilibrium 0.143 0.0072 0.0072

Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)

Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)

43

check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka

though not exact, the answer is reasonably close

[HF] [F-] [H3O+]

initial 0.150 0 ≈ 0

change -x +x +x

equilibrium 0.143 0.0072 0.0072

Ka for HF = 3.5 x 10-4

Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)

Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)

44

NaOH Na+ + OH-

Strong Arrhenius BasesStrong Arrhenius Bases

the stronger the base, the more willing it is to accept H+

use water as the standard acid

for strong bases, practically all molecules are dissociated into OH– or accept H’s

multi-OH strong bases completely dissociated

[HO–] = [strong base] x (# OH)

the stronger the base, the more willing it is to accept H+

use water as the standard acid

for strong bases, practically all molecules are dissociated into OH– or accept H’s

multi-OH strong bases completely dissociated

[HO–] = [strong base] x (# OH)

Calculate the pH at 25°C of a 0.0015 M Sr(OH)2 solution and determine if the solution is acidic, basic, or neutral

Calculate the pH at 25°C of a 0.0015 M Sr(OH)2 solution and determine if the solution is acidic, basic, or neutral

pH is unitless. The fact that the pH > 7 means the solution is basic

[Sr(OH)2] = 1.5 x 10-3 M

pH

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

[H3O+][OH-] pH[Sr(OH)2]

[OH-]=2[Sr(OH)2]

[OH-] = 2(0.0015)= 0.0030 M

46

Practice: Calculate the pH of a 0.0010 M Ba(OH)2

solution and determine if it is acidic, basic, or neutral

47

[H3O+] = 1.00 x 10-14

2.0 x 10-3 = 5.0 x 10-12M

pH > 7 therefore basic

Ba(OH)2 = Ba2+ + 2 OH- therefore [OH-] = 2 x 0.0010M = 0.0020M = 2.0 x 10-3 M

pH = -log [H3O+] = -log (5.0 x 10-12)pH = 11.30

Kw = [H3O+][OH-]

Practice: Calculate the pH of a 0.0010 M Ba(OH)2

solution and determine if it is acidic, basic, or neutral