IV. Weak Acids and Weak Bases | ||||
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IV-1. Introduction and Background | IV-2. Sample Problems | IV-3. Problem List | IV-4. Weak Acid Problems | IV-5. Weak Base Problems |
Below are a weak acid and a weak base problem. These problems are straightforward in that there are no initial reactions that affect the concentration of the weak acid or base.
What is the pH of a 0.100 M solution of acetic acid (K_{a} = 1.8x10^{-5})?
1. First identify what species are present, and any reactions that could occur. (The title of this problem states that it is a weak acid problem, but not all problems will tell you the type of equilibrium that will be involved.) The only specie present is acetic acid. The [H^{+}] of water (1.0x10^{-7} M) can almost always be neglected. If the concentration of the weak acid is low, then you must solve a complex equilibria problem that includes the dissociation of water. The pre-equilibrium concentration of acetic acid is 0.100 M.
2. The balanced equilibrium reaction is: CH_{3}COOH_{(aq)} H^{+}_{(aq)} + CH_{3}COO^{-}_{(aq)}
and the equilibrium constant expression is:
[H^{+}][CH_{3}COO^{-}] K_{a} = ------------ [CH_{3}COOH]
3. Now calculate the reaction quotient, Q, to determine the direction in which the reaction will proceed to reach equilibrium. In acid-base problems, it is seldom necessary to calculate Q exactly. In this sample problem, the pre-equilibrium concentration of acetate ion, COO^{-}, is zero, so Q is zero and the reaction goes in the forward direction.
4. For each mol of CH_{3}COOH that dissociates, 1 mole each of H^{+} and CH_{3}COO^{-} forms. The pre-equilibrium concentrations, the concentration changes, and the equilibrium concentrations are summarized in the following table:
CH_{3}COOH | H^{+} | COO^{-} | |
---|---|---|---|
[ ]_{o} | 0.100 M | ~0 | 0 |
[ ] | -x M | +x M | +x M |
[ ]_{eq} | (0.100 - x) M | x M | x M |
Where [ ]_{o} are the pre-equilibrium concentrations, [ ] are the changes in concentrations, and [ ]_{eq} are expressions for the equilibrium concentrations.
5. We can now calculate the equilibrium concentrations using the equilibrium constant expression:
[H^{+}][CH_{3}COO^{-}] K_{a} = ------------ [CH_{3}COOH] (x)(x) K_{a} = -------- (0.100-x)K_{a} = x^{2} / (0.100-x)
We could solve this expression exactly, but since we are dealing with a weak acid, x is much smaller than 0.100 and can be neglected in the denominator.
1.8x10^{-5} = x^{2} / 0.100
x^{2} = 1.8x10^{-6}
x = 1.3x10^{-3} M
Note that x << 0.100 M, and the approximation was valid.
This problem asked for the pH of the solution.
[H^{+}] = x = 1.3x10^{-3} M
pH = -log[H^{+}] = -log(1.3x10^{-3})
pH = 2.87 |
We can calculate Q to check that we are at equilibrium:
Q = (1.3x10^{-3})(1.3x10^{-3}) / 0.100 = 1.810^{-5}. Q = K_{a}, so the system is at equilibrium.
What is the pH of a 0.100 M solution of ammonia (K_{b} = 1.8x10^{-5})?
1. First identify what species are present, and any reactions that could occur. (The title of this problem states that it is a weak base problem.) The only specie present is ammonia. The [OH^{-}] of water (1.0x10^{-7} M) can almost always be neglected. If the concentration of the weak base is low, then you must solve a complex equilibria problem that includes the dissociation of water. The pre-equilibrium concentration of ammonia is 0.100 M.
2. The balanced equilibrium reaction is: NH_{3}_{(aq)} + H_{2}O NH_{4}^{+}_{(aq)} + OH^{-}_{(aq)}
and the equilibrium constant expression is:
[NH_{4}^{+}][OH^{-}] K_{b} = ----------- [NH_{3}]
3. Now calculate the reaction quotient, Q, to determine the direction in which the reaction will proceed to reach equilibrium. In acid-base problems, it is seldom necessary to calculate Q exactly. In this sample problem, the pre-equilibrium concentration of ammonium ion, NH_{4}^{+}, is zero, so Q is zero and the reaction goes in the forward direction.
4. For each mol of NH_{3} that reacts with water, 1 mole each of NH_{4}^{+} and OH^{-} forms. The pre-equilibrium concentrations, the concentration changes, and the equilibrium concentrations are summarized in the following table:
NH_{3} | NH_{4}^{+} | OH^{-} | |
---|---|---|---|
[ ]_{o} | 0.100 M | 0 | ~0 |
[ ] | -x M | +x M | +x M |
[ ]_{eq} | (0.100 - x) M | x M | x M |
Where [ ]_{o} are the pre-equilibrium concentrations, [ ] are the changes in concentrations, and [ ]_{eq} are expressions for the equilibrium concentrations.
5. We can now calculate the equilibrium concentrations using the equilibrium constant expression:
[NH_{4}^{+}][OH^{-}] K_{b} = ----------- [NH_{3}] (x)(x) K_{b} = --------- (0.100-x)K_{b} = x^{2} / (0.100-x)
We could solve this expression exactly, but since we are dealing with a weak base, x is much smaller than 0.100 and can be neglected in the denominator.
1.8x10^{-5} = x^{2} / 0.100
x^{2} = 1.8x10^{-6}
x = 1.3x10^{-3} M
Note that x << 0.100 M, and the approximation was valid.
This problem asked for the pH of the solution.
[OH^{-}] = x = 1.3x10^{-3} M
pOH = -log[OH^{-}] = -log(1.3x10^{-3})
pOH = 2.87
pH = 14.00 - POH = 14.00 - 2.87
pH = 11.13 |
We can calculate Q to check that we are at equilibrium:
Q = (1.3x10^{-3})(1.3x10^{-3}) / 0.100 = 1.810^{-5}. Q = K_{a}, so the system is at equilibrium.
Equilibrium Practice Problems | |
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