|Common Weak Acids|
of weak bases
|Common Weak Bases|
of weak acids
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.
From the ionization of acetic acid,
The equilibrium constant of ionzation,
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
Ka = -------
C - x
|Ka = 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.
Assume the label concentration as C and x mole ionized, then the ionization and the equilibrium concentrations can be represented by the formulation below.
Ka = -------
C - x
The equation is then
x2 + Ka x = C Ka = 0
- 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||0.010||0.00010 M|
|x =||0.0042||0.00041||0.0000342 M|
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 give
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)
-(y+Ka) + ((y+Ka)2 + 4 C Ka)(1/2) x = ------------------------------------- 2
-(y+Ka) + ((y+Ka)2 + 4 CKa)(1/2) x = ------------------------------------- 2
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.
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.
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.
Answer 0.8 M
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
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.
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.
[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.
Note that [H+] = 4.3E-5 is 4% of [HAc] (= 1.0E-4). Use the formula
-Ka + (Ka2 + 4 C Ka)(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*Ka)
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.
At this concentration, the acid is almost completely ionized.
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.
Critical judgment is required. The pH is entirely due to the self-ionization of water at this concentration.