1. Relationship 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

2. pK 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 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

4. Finding 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

5. Find 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
HNO2 + H2O NO2- + H3O+
[HNO2]
[NO2-]
[H3O+]
initial
change
equilibrium
[HNO2]
[NO2-]
[H3O+]
initial
0.200
≈ 0
change
equilibrium
since no products initially, Qc = 0, and the reaction is proceeding forward

6. Find the pH of 0.200 M HNO2(aq) solution @ 25°C
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
[HNO2]
[NO2-]
[H3O+]
initial
0.200
change
equilibrium -x +x +x x x x

7. Find the pH of 0.200 M HNO2(aq) solution @ 25°C
Ka for HNO2 = 4.6 x 10-4
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
0.200
≈ 0
change -x +x
equilibrium x x

8. Find the pH of 0.200 M HNO2(aq) solution @ 25°C
Ka for HNO2 = 4.6 x 10-4
check if the approximation is valid by seeing if x < 5% of [HNO2]init
[HNO2]
[NO2-]
[H3O+]
initial
0.200
≈ 0
change -x +x
equilibrium x
x = 9.6 x 10-3
the approximation is valid

9. Find the pH of 0.200 M HNO2(aq) solution @ 25°C
Ka for HNO2 = 4.6 x 10-4
substitute x into the equilibrium concentration definitions and solve
[HNO2]
[NO2-]
[H3O+]
initial
0.200
≈ 0
change -x +x
equilibrium
0.190
0.0096
[HNO2]
[NO2-]
[H3O+]
initial
0.200
≈ 0
change -x +x
equilibrium
0.200-x x
x = 9.6 x 10-3

10. Find the pH of 0.200 M HNO2(aq) solution @ 25°C
Ka for HNO2 = 4.6 x 10-4
substitute [H3O+] into the formula for pH and solve
[HNO2]
[NO2-]
[H3O+]
initial
0.200
≈ 0
change -x +x
equilibrium
0.190
0.0096

11. Find the pH of 0.200 M HNO2(aq) solution @ 25°C
Ka for HNO2 = 4.6 x 10-4
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka
[HNO2]
[NO2-]
[H3O+]
initial
0.200
≈ 0
change -x +x
equilibrium
0.190
0.0096
though not exact, the answer is reasonably close

12. What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?
Tro, Chemistry: A Molecular Approach

13. 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+
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
[HA]
[A-]
[H3O+]
initial
0.012
≈ 0
change
equilibrium

14. 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+
[HA]
[A-]
[H3O+]
initial
0.012
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 x x x

15. 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+
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
change -x +x
equilibrium x x

16. What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?
Ka for HC6H4NO2 = 1.4 x 10-5
check if the approximation is valid by seeing if x < 5% of [HC6H4NO2]init
[HA]
[A2-]
[H3O+]
initial
0.012
≈ 0
change -x +x
equilibrium x
x = 4.1 x 10-4
the approximation is valid

17. What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?
initial
0.012
≈ 0
change -x +x
equilibrium
0.012-x x
substitute x into the equilibrium concentration definitions and solve
x = 4.1 x 10-4

18. What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?
substitute [H3O+] into the formula for pH and solve
[HA]
[A2-]
[H3O+]
initial
0.012
≈ 0
change -x +x
equilibrium

19. What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka
[HA]
[A2-]
[H3O+]
initial
0.012
≈ 0
change -x +x
equilibrium
the values match

20. Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2
HClO2 + H2O ClO2- + H3O+
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]
[ClO2-]
[H3O+]
initial
0.100
≈ 0
change
equilibrium

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

22. Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2
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
change -x +x
equilibrium
0.100-x x
Tro, Chemistry: A Molecular Approach

23. Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2
check if the approximation is valid by seeing if x < 5% of [HNO2]init
[HClO2]
[ClO2-]
[H3O+]
initial
0.100
≈ 0
change -x +x
equilibrium
0.100-x x
x = 3.3 x 10-2
the approximation is invalid

24. Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2
if the approximation is invalid, solve for x using the quadratic formula

25. Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2
substitute x into the equilibrium concentration definitions and solve
[HClO2]
[ClO2-]
[H3O+]
initial
0.100
≈ 0
change -x +x
equilibrium
0.072
0.028
[HClO2]
[ClO2-]
[H3O+]
initial
0.100
≈ 0
change -x +x
equilibrium
0.100-x x
x = 0.028

26. Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2
substitute [H3O+] into the formula for pH and solve
[HClO2]
[ClO2-]
[H3O+]
initial
0.100
≈ 0
change -x +x
equilibrium
0.072
0.028

27. Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka
[HClO2]
[ClO2-]
[H3O+]
initial
0.100
≈ 0
change -x +x
equilibrium
0.072
0.028
the answer matches

28. 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
0.100
≈ 0
change
equilibrium
5.6E-05
[HA]
[A-]
[H3O+]
initial
change
equilibrium

29. What is the Ka of a weak acid if a 0.100 M solution has a pH of 4.25?
HA + H2O A- + H3O+
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
[HA]
[A-]
[H3O+]
initial
0.100
change
equilibrium
−5.6E-05
+5.6E-05
+5.6E-05
0.100 
5.6E-05
0.100
5.6E-05
5.6E-05

30. Percent Ionization since [ionized acid]equil = [H3O+]equil
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
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
2.5
≈ 0
change
equilibrium
[HNO2]
[NO2-]
[H3O+]
initial
change
equilibrium -x +x +x
2.5 - x x x

32. What is the percent ionization of a 2
What is the percent ionization of a 2.5 M HNO2 solution if Ka for HNO2 = 4.6 x 10-4 ?
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
change -x +x
equilibrium
2.5-x ≈2.5 x

33. What is the percent ionization of a 2
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+
substitute x into the Equilibrium Concentration definitions and solve
[HNO2]
[NO2-]
[H3O+]
initial
2.5
≈ 0
change -x +x
equilibrium
0.034
2.5 - x x x
x = 3.4 x 10-2

34. What is the percent ionization of a 2
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+
Apply the Definition and Compute the Percent Ionization
[HNO2]
[NO2-]
[H3O+]
initial
2.5
≈ 0
change -x +x
equilibrium
0.034
since the percent ionization is < 5%, the “x is small” approximation is valid

35. Relationship 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

36. Finding 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

37. Find the pH of a mixture of 0. 150 M HF(aq) solution and 0
Find the pH of a mixture of M HF(aq) solution and M HClO2(aq)
HF + H2O F- + H3O+ Ka = 3.5 x 10-4
HClO + H2O ClO- + H3O+ Ka = 2.9 x 10-8
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
H2O + H2O OH- + H3O+ Kw = 1.0 x 10-14
[HF]
[F-]
[H3O+]
initial
0.150
≈ 0
change
equilibrium

38. Find the pH of a mixture of 0. 150 M HF(aq) solution and 0
Find the pH of a mixture of M HF(aq) solution and 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
[HF]
[F-]
[H3O+]
initial
0.150
change
equilibrium -x +x +x x x x

39. Find the pH of a mixture of 0. 150 M HF(aq) solution and 0
Find the pH of a mixture of M HF(aq) solution and M HClO2(aq)
Ka for HF = 3.5 x 10-4
determine the value of Ka for HF
since Ka is very small, approximate the [HF]eq = [HF]init and solve for x
[HF]
[F-]
[H3O+]
initial
0.150
≈ 0
change -x +x
equilibrium x x

40. Find the pH of a mixture of 0. 150 M HF(aq) solution and 0
Find the pH of a mixture of M HF(aq) solution and M HClO2(aq)
Ka for HF = 3.5 x 10-4
check if the approximation is valid by seeing if x < 5% of [HF]init
[HF]
[F-]
[H3O+]
initial
0.150
≈ 0
change -x +x
equilibrium x
x = 7.2 x 10-3
the approximation is valid

41. Find the pH of a mixture of 0. 150 M HF(aq) solution and 0
Find the pH of a mixture of M HF(aq) solution and M HClO2(aq)
Ka for HF = 3.5 x 10-4
substitute x into the equilibrium concentration definitions and solve
[HF]
[F-]
[H3O+]
initial
0.150
≈ 0
change -x +x
equilibrium
0.143
0.0072
[HF]
[F-]
[H3O+]
initial
0.150
≈ 0
change -x +x
equilibrium
0.150-x x
x = 7.2 x 10-3

42. Find the pH of a mixture of 0. 150 M HF(aq) solution and 0
Find the pH of a mixture of M HF(aq) solution and M HClO2(aq)
Ka for HF = 3.5 x 10-4
substitute [H3O+] into the formula for pH and solve
[HF]
[F-]
[H3O+]
initial
0.150
≈ 0
change -x +x
equilibrium
0.143
0.0072

43. Find the pH of a mixture of 0. 150 M HF(aq) solution and 0
Find the pH of a mixture of M HF(aq) solution and M HClO2(aq)
Ka for HF = 3.5 x 10-4
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka
[HF]
[F-]
[H3O+]
initial
0.150
≈ 0
change -x +x
equilibrium
0.143
0.0072
though not exact, the answer is reasonably close

44. Strong 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)
NaOH  Na+ + OH-

45. Calculate the pH at 25°C of a 0
Calculate the pH at 25°C of a M Sr(OH)2 solution and determine if the solution is acidic, basic, or neutral
Given:
Find:
[Sr(OH)2] = 1.5 x 10-3 M pH
Concept Plan:
Relationships:
[H3O+]
[OH-] pH
[Sr(OH)2]
[OH-]=2[Sr(OH)2]
Solution:
[OH-]
= 2(0.0015)
= M
Check:
pH is unitless. The fact that the pH > 7 means the solution is basic

46. Practice: Calculate the pH of a 0
Practice: Calculate the pH of a M Ba(OH)2 solution and determine if it is acidic, basic, or neutral

47. Practice: Calculate the pH of a 0
Practice: Calculate the pH of a M Ba(OH)2 solution and determine if it is acidic, basic, or neutral
Ba(OH)2 = Ba OH- therefore
[OH-] = 2 x M = M = 2.0 x 10-3 M
Kw = [H3O+][OH-]
[H3O+] =
1.00 x 10-14
2.0 x 10-3
= 5.0 x 10-12M
pH = -log [H3O+] = -log (5.0 x 10-12)
pH = 11.30
pH > 7 therefore basic