V. Advanced Weak Acid and Weak Base Equilibria | |||
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V-1. Introduction to Acidic and Basic Salts |
V-2. Sample Problem | V-3. List of Problems | V-4. Practice Problems |
The following problem is an example of determining the pH, and therefore the equilibrium [H^{+}], when an acidic salt is added to water. For a reminder of how to begin, see the document on the general solution of equilibria problems.
What is the pH of a 0.500 M solution of NH_{4}Cl in water?
1. In any equilibrium problem, the first step is to identify the nature of the specie dissolving in water. NH_{4}Cl (ammonium chloride) contains the spectator ion Cl^{-}, so NH_{4}Cl is a salt and completely dissociates:
NH_{4}Cl_{(s)} NH_{4}^{+}_{(aq)} + Cl^{-}_{(aq)}
Cl^{-}_{(aq)} is a spectator ion and will not be involved in any equilibria. NH_{4}^{+}_{(aq)} can dissociate to produce NH_{3 (aq)} and H^{+}_{(aq)}, so the solution will be acidic.
NH_{4}^{+}_{(aq)} NH_{3 (aq)} + H^{+}_{(aq)}
K_{b} for ammonia is 1.8x10^{-5}. Using K_{a} * K_{b} = 1.00x10^{-14}, then K_{a} for NH_{4}^{+}_{(aq)} is 5.6x10^{-10}.
The pre-equilibrium concentration of NH_{4}^{+}_{(aq)} is given, and is 0.500 M. The pre-equilibrium concentration of H^{+}_{(aq)} is 1.0x10^{-7} due to the dissociation of pure water. We can consider the pre-equilibrium [H^{+}] to be 0.00 M since it is much lower than 0.500 M. The pre-equilibrium concentration of NH_{3 (aq)} is 0.00 M.
2. First balance the reaction: NH_{4}^{+}_{(aq)} NH_{3 (aq)} + H^{+}_{(aq)}
Then write the equilibrium constant expression for this reaction:
[H^{+}][NH_{3}] K_{a} = ---------- [NH_{4}^{+}]
3.
[H^{+}][NH_{3}] Q = ---------- [NH_{4}^{+}]
Since the pre-equilibrium concentrations of the products are zero, Q is zero. Thus, Q < K_{eq} and the reaction will proceed in the forward direction.
4. For each mol of NH_{4}^{+} that dissociates, 1 mole each of H^{+} and NH_{3} forms:
NH_{4}^{+} | H^{+} | NH_{3} | |
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[ ]_{o} | 0.500 M | ~0.00 M | 0.00 M |
[ ] | -x M | +x M | +x M |
[ ]_{eq} | (0.500-x) M | (0.00+x) M | (0.00+x) M |
5. Once you have the expressions for the equilibrium concentrations, you can use the equilibrium constant expression to solve for x:
[H^{+}][NH_{3}] K_{a} = ---------- [NH_{4}^{+}] x^{2} 5.6x10^{-10} = -------- (0.500-x)There are two ways to approach this problem from here. You can rearrange the equation above and solve for x using the quadratic equation, or you can make the approximation that x will be much less than 0.500 M and drop the x in the denominator.
Making this approximation gives:
x^{2} 5.6x10^{-10} = ------ 0.500
x^{2} = 2.8x10^{-10}
x = 1.66x10^{-5}
Before continuing, check that the approximation was valid:
x^{2} 5.6x10^{-10} = ------------------- (0.500 - 1.66x10^{-5})
x^{2} = 2.8x10^{-10}
x = 1.66x10^{-5}
We get the same answer for x, 1.66x10^{-5}, so the approximation was valid.
Now find what the problem requested:
[H^{+}] = x = 1.66x10^{-5} M
pH = -log[H^{+}]
pH = -log(1.66x10^{-5})
pH = 4.78 |
Note that K_{a} was known only to 2 significant figures, so the answer after taking the log has 2 digits past the decimal point.
Check results:
(1.66x10^{-5})(1.66x10^{-5}) Q = ---------------------- (0.500-1.66x10^{-5})Q = 5.5x10^{-10}. Q = K_{a}, so the system is at equilibrium.
Equilibrium Practice Problems | |
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