Common Weak Acids | |
---|---|
Acid | Formula |
Formic | HCOOH |
Acetic | CH_{3}COOH |
Trichloroacetic | CCl_{3}COOH |
Hydrofluoric | HF |
Hydrocyanic | HCN |
Hydrogen sulfide | H_{2}S |
Water | H_{2}O |
Conjugate acids of weak bases | NH_{4}^{+} |
Common Weak Bases | |
---|---|
Base | Formula |
ammonia | NH_{3} |
trimethyl ammonia | N(CH_{3})_{3} |
pyridine | C_{5}H_{5}N |
ammonium hydroxide | NH_{4}OH |
water | H_{2}O |
HS^{-} ion | HS^{-} |
conjugate bases of weak acids | e.g.: HCOO^{-} |
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.
H O | // H--C--C | \\ H O- |
---|
Example 1
Solution
From the ionization of acetic acid,
The equilibrium constant of ionzation,
Discussion
The equilibrium constant of an acid is represented by Ka; and similar
to the pH scale, a pKa scale is defined by
and for acetic acid, pKa = 4.75. Note that Ka = 10^{-pKa}
x^{2}
Ka = -------
C - x
pKa = - log KaKa = 10^{-pKa}
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
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.
x^{2}
Ka = -------
C - x
The equation is then
x^{2} + Ka x = C Ka = 0
- K_{a} + (K_{a}^{2} + 4 C K_{a})^{(1/2)} x = --------------------------- 2 Recall that Ka - 1.78e-5, the values of x for various C are given below:
C = | 1.0 | 0.010 | 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:
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 xThere are two unknown quantities, x and y in two equations, and (1) may be rearranged to giveH_{2}O = H^{+} + OH^{-} 55.6 y y <- - - ([H_{2}O] = 55.6)
Thus, [H^{+}] = (x+y), [A^{-}] = x, [OH^{-}] = y,
and the two equilibria are
(x+y) x K_{a} = --------- ........ (1) C - x
and K_{w} = (x+y) y, ........... (2) (K_{w} = 1E-14)
-(y+K_{a}) + ((y+K_{a})^{2} + 4 C K_{a})^{(1/2)} x = ------------------------------------- 2
-(y+K_{a}) + ((y+K_{a})^{2} + 4 CK_{a})^{(1/2)} x = ------------------------------------- 2
The pOH can be calculated for a basic solution if K_{b} is given. In this case, the discussion is similar and parallel to that given above for the calculation of pH of weak acids when K_{a} is know.
Answer b
Consider...
The pH of a solution depends on both the concentration and the
degree of ionization, (or using K_{a} as an indicator).
In contrast, a strong acid is completely ionized in solution.
Answer b
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
K_{a} is more realistic than infinity.
Answer 0.8 M
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
Answer 4.3e-3
Consider...
Use the approximation of [H^{+}] = square root of (K_{a} * 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.
Answer 3.4
Consider...
Use the approximation of [H^{+}] = sqrt (K_{a} * 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.
Answer 2
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.
Answer 4.4
Hint...
Note that [H^{+}] = 4.3E-5 is 4% of [HAc] (= 1.0E-4).
Use the formula
-K_{a} + (K_{a}^{2} + 4 C K_{a})^{(1/2)} x = ----------------------- 2and 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*K_{a})
Answer 3.2
Hint...
Even a strong acid with concentration of 1.0E-3 M gives a pH of 3.
When C and K_{a} are comparable, you have to use the quadratic formula.
Answer 6.0
Consider...
At this concentration, the acid is almost completely ionized.
Answer 6.7
Consider...
From the previous question, you know that the chloroacetic acid should
have been completely ionized. Thus, [H^{+}] is about 2e-7, half of that
is contributed by the self-ionization of water. This corresponds to
a pH of 6.7.
Answer 7
Consider...
Critical judgment is required.
The pH is entirely due to the self-ionization of water at this
concentration.