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# Weak Acids and Bases

### Skills to develop

• Define a weak acid or base.
• Calculate pH and pOH of a weak acid or base solution using simple formula, quadratic equation, and including autoionization of water.
• Calculate the pH or pOH quickly.

# Weak Acids and Bases

Weak acids and bases are only partially ionized in their solutions, whereas strong acids and bases are completely ionized when dissolve in water.

Common Weak Acids
AcidFormula
FormicHCOOH
AceticCH3COOH
TrichloroaceticCCl3COOH
HydrofluoricHF
Hydrocyanic HCN
Hydrogen
sulfide
H2S
Water H2O
Conjugate acids
of weak bases
NH4+
Common Weak Bases
BaseFormula
ammoniaNH3
trimethyl
ammonia
N(CH3)3
pyridineC5H5N
ammonium
hydroxide
NH4OH
waterH2O
HS- ionHS-
conjugate bases
of weak acids
e.g.:
HCOO-
Some common weak acids and bases are given here. Furthermore, weak acids and bases are very common, and we encounter them often both in the academic problems and in everyday life.

The ionization of weak acids and bases is a chemical equilibrium phenomenon. The equilibrium principles are essential for the understanding of equilibria of weak acids and weak bases.

The conjugate acid-base pairs have been discussed in Acids and Bases. In this connection, you probably realize that conjugate acids of weak bases are weak acids and conjugate bases of weak acids are weak bases.

## Ionization of Week Acids

H         O
|       //
H--C--C
|       \\
H         O-
Acetic acid, CH3COOH, is a typical weak acid, and it is the ingradient of vinegar. It is partially ionized in its solution. CH3COOH = CH3COO- + H+ The structure of the acetate ion, CH3COO-, is shown on the right.

Example 1

In a solution of acetic acid, the equilibrium concentrations are found to be [CH3COOH] = 1.000; [CH3COO-] = 0.0042. Evaluate the pH of this solution and the equilibrium constant of ionization of acetic acid.

Solution
From the ionization of acetic acid,

CH3COOH = CH3COO- + H+
0.100             0.0042     0.0042
we conclude that
[H+] = [CH3COO-]
= 0.0042.
Thus, pH = -log0.0042 = 2.376.

The equilibrium constant of ionzation,

(0.0042)2
K = ------------- = 1.78x10-5
1.000

Discussion
The equilibrium constant of an acid is represented by Ka; and similar to the pH scale, a pKa scale is defined by

pKa = - log Ka

and for acetic acid, pKa = 4.75. Note that Ka = 10-pKa

### The pH and pKa of Weak acid

There are many weak acids, which do not completely dissociate in aqueous solution. As a general discussion of weak acids, let HA represent a typical weak acid. Then its ionization can be written as: HA = H+ + A- In a solution whose label concentration is C (= [HA] + [A-]), let us assume that x is the concentration that has undergone ionization. Thus, at equilibrium, the concentrations are [HA] = C - x
[H+] = [A-] = x
Make sure you understand why they are so, because you will have to setup these relationship in your problem solving. In summary, we formulate them as HA = H+ + A-
C
- initial concentration, assume x M ionized
C-x    x       x - equilibrium concentration

x2
Ka = -------
C - x

 Ka = 10-pKa
pKa = - log Ka

The pKa values of many weak acids are listed in table form in handbooks, and some of these values are given in the Handbook of CAcT, pKa of Acids.

Example 2

The pKa of acetic acid is 4.75. Find the pH of acetic acid solutions of labeled concentrations of 1.0 M, 0.010 M, and 0.00010 M.

Solution
Assume the label concentration as C and x mole ionized, then the ionization and the equilibrium concentrations can be represented by the formulation below.

CH3COOH = CH3COO- + H+
C - x                     x             x

x2
Ka = -------
C - x
The equation is then
x2 + Ka x = C Ka = 0

The solution of x is then
- Ka + (Ka2 + 4 C Ka)(1/2)
x  =  ---------------------------
2
Recall that Ka - 1.78e-5, the values of x for various
C are given below:
 C = 1 0.01 0.00010 M x = 0.0042 0.00041 0.0000342 M pH = 2.38 3.39 4.47

Discussion
In the above calculations, the following cases may be considered:

1. If x is small (< 1% of) compared to C, then C-x is approximately C. Thus, x = (Ka * C)(1/2) Note that you are comparing x with C here. If C > 100*Ka, the above method gives satisfactory results.
2. If x is not small by comparison with C, or C is not large in comparison to Ka, then the equation takes this form: x2 + Ka x - C Ka = 0, and the solution for x, which must not be negative, has been given above.
3. Both cases 1 and 2 neglect the contribution of [H+] from the ionization of water. However, if the pH calculated from cases 1 and 2 falls in the range between 6 and 7, the concentration from self-ionization of water cannot be neglected.

When the contribution of pH due to self-ionization of water cannot be neglected, there are two equilibria to be considered.

HA  =  H+  +  A-
C-x     x     x

H2O = H+ + OH- 55.6 y y <- - - ([H2O] = 55.6)

Thus, [H+] = (x+y), [A-] = x, [OH-] = y,

and the two equilibria are

(x+y) x Ka = --------- ........ (1) C - x

and Kw = (x+y) y, ........... (2) (Kw = 1E-14)

There are two unknown quantities, x and y in two equations, and (1) may be rearranged to give

x2 + (y + Ka) x - C Ka = 0
-(y+Ka) + ((y+Ka)2 + 4 C Ka)(1/2)
x  =  -------------------------------------
2

One of the many methods to find a suitable solution for this problem is to use iterations, or successive approximations.

1. Assume that y = 1E-7

2. Calculate an x value using the quadratic form

-(y+Ka) + ((y+Ka)2 + 4 CKa)(1/2)
x  =  -------------------------------------
2

3. Calculate a new y value (yn) from the x just obtained using

yn = 1E-14/(x+y)

4. Replace y in step (2) by yn, and recalculate x.

5. Repeat steps (2) and (3) until the new values and the old values differ insignificantly.
The above procedure is actually a general method that always gives a satisfactory solution. This technique have to be used to calculate the pH of dilute weak acid solutions. Further discussion is given in the
Exact Calculation of pH.

## Calculate pOH of Basic Solutions

The discussion on weak acids provide a paradigm for the discussion of weak bases. For weak base B, the ionization is B- + H2O = HB + OH-
and
[HB] [OH-]
Kb = -------------
[B-]

The pOH can be calculated for a basic solution if Kb is given. In this case, the discussion is similar and parallel to that given above for the calculation of pH of weak acids when Ka is know.

### Confidence Building Questions

• A weak acid is a compound that
a. is completely ionized in solution,
b. is not completely ionized in solution,
c. gives a high pH in a solution,
d. gives a low pH in its solution,

Consider...

The pH of a solution depends on both the concentration and the degree of ionization, (or using Ka as an indicator). In contrast, a strong acid is completely ionized in solution.

• The acidity constant, Ka, for a strong acid is
a. infinity,
b. very large,
c. very small,
d. zero.

Consider...

Infinity is a concept, it does not represent a definite value. Derive your answer from the definition of equilibrium constant. A strong acid is "completely" ionized in its solution, but the concentration of the conjugate acid is not zero. Thus, a very large Ka is more realistic than infinity.

• Household vinegar is usually 5% acetic acid by volume. Calculate the molarity of this solution.
Assume density of solution to be 1 g/mL. The formula weight of CH3COOH is 60.

Consider...

Assume 1 L solution. You have 50 mL acetic acid in 1 L vinegar. The density is 1 g/mL, thus, you have 50 g acetic acid. There is 50 mL vinegar in 1.0 L of vinegar, 50 g/60 g per mol = 0.8 mol/L

• Acetic acid is a typical and familiar compound that provides a good example for numerical problems. Its acidity constant Ka is 1.85e-5. What is the pH of a concentrated vinegar, which is a 1.0 M acetic acid solution?

Consider...

Use the approximation of [H+] = square root of (Ka * C). [H+] = (1.85e-5)1/2 = 4.3e-3. The approximation is justified because 0.0043/1.0 = 0.4%. Note that most text books give Ka = 1.75e-5, but we assume a slightly different value in this and the following problems.

• If you dilute the vinegar 100 times in a soup that you are cooking, the concentration of your soup is 0.010 M in acetic acid.
Other ingredients are ignored. What is the pH of this solution?

Consider...

Use the approximation of [H+] = sqrt (Ka * C). The concentration of H+ goes from 3.3E-3 in a 1 M solution down to 4.3E-4 M in a 0.01 M solution. The concentration of H+ decreases 10 times when the concentration of the acid decreases 100 times.

• What is the pH of a 0.010 M HCl solution?

Consider...

[H+] = 0.010 M. The concentration of H+ is from HCl, which is a strong acid. The pH of a 0.01 M HCl solution is lower than that of a 1 M acetic acid solution, compare with the previous problem.

• What is the pH of a 1.0E-4 acetic acid solution (Ka = 1.85E-5)?

Hint...

Note that [H+] = 4.3E-5 is 4% of [HAc] (= 1.0E-4). Use the formula

-Ka + (Ka2 + 4 C Ka)(1/2)
x  =  -----------------------
2

and see what you get. You should use the quadratic formula to calculate [H+]. The value using the quadratic formula is 4.5 rather than 4.4 from sqrt(C*Ka)

• What is the pH of a 1.0E-3 M chloroacetic acid solution (Ka = 1.4E-3)? This is an interesting numerical problem. Make a good effort to solve it.

Hint...

Even a strong acid with concentration of 1.0E-3 M gives a pH of 3. When C and Ka are comparable, you have to use the quadratic formula.

• What is the pH of a 1.0e-6 M chloroacetic acid solution (Ka = 1.4E-3)?

Consider...

At this concentration, the acid is almost completely ionized.

• What is the pH of a 1.0e-7 M chloroacetic acid solution (Ka = 1.4e-3)?