Liberato Cardellini
Dipartimento di Scienze dei Materialie della Terra
Universita, Via Brecce Bianche60131 Ancona, Italy
INTRODUCTION
An important part of every general chemistry curriculum is the treatment of ionic equilibrium problems in aqueous solution. In this topic, two important aspects of stoichiometric calculations are treated: 1) the conceptual and formative chemical aspect; 2) precision of the result. These two aspects are difficult to come to terms with when dealing with ionic equilibrium calculations. Chaston (1) has devised an interesting method for calculating the exact pH for dilute solutions of acids and bases. However, this method has two drawbacks: 1)it is not of general use; 2) it relies heavily on algebraic manipulations. Using this or a similar method, the student forgets the chemical aspect of the problem to be solved. On the other hand, using a series of rote-learned formulae is not a better didactic approach. The student has to ask itself questions in order to decide what formula to use: is this a buffer solution? Are we at the equivalence point? Is the hydrolysis appreciable? Can the dissociation of this weak acid be considered negligible? And so on. These questions are bewildering for the student not yet an expert chemist. He has to decide the chemical problem from the evaluation of terms as "negligible", "perceptible" or "significant". Another approach can be used. As instructors, we can agree that the difficulty faced by students in solving ionic equilibrium problems is the recognition of the chemical approach required in different situations. So, the ideal method has to:
(([H^{+}]approx / [H^{+}]) - 1) x 100 = E (1)
a good method of calculation is one that produces results with E no greater than 1%.
HOW THIS METHOD WORKS1. Water is always ionized according to the equilibrium:
H_{2}O ⇌ H^{+} + OH^{-}
and the ion product of water must always be verified: [H^{+}][OH^{-}] = K_{W}. Sometimes the H^{+} coming from water ionization must be taken into account.
2. Strong acids, bases and ionic salts are completely ionized:
HCl → H^{+} + Cl^{-}
NaOH → Na^{+} + OH^{-}
NH_{4}Cl → NH_{4}^{+} + Cl^{-}
3. Weak acids and bases are partially ionized; conjugate species undergo hydrolysis:
NH_{4}^{+} + H_{2}O ⇌ NH_{4}OH + H^{+}
4. Normally, we use the mass balance equation and the electro-neutrality condition for verifying the correctness of the result.
5. The only approximations allowed are of a numerical nature: 1.00x10^{-2} + 3.45x10^{-8} = 1.00x10^{-2}. The tolerable error is 1%. In dealing with multiple equilibria it is wise, but not essential from a calculation point of view, to consider the equilibrium with the greatest K first.
In the ionic equilibrium calculations, [H^{+}] varies between very different values: thus we can use a numerical criteria for approximations. In an aqueous solution 1.00x10^{-2}M of hydrochloric acid, the contribution from water ionization is 1.00x10^{-12}M; so, we have:
[H^{+}]= 1.00x10^{-2} M + 1.00x10^{-12} M = 1.00x10^{-2} MThe numerical importance of the water dissociation on the pH of the acid solution has been discussed elsewhere (3). Now, we use some examples for showing how this method works. Let us consider this problem: calculate the pH of a solution 1.00x10^{-8} M of hydrochloric acid. The following reactions take place:
H_{2}O ⇌ H^{+} + OH^{-}
HCl → H^{+} + Cl^{-}
From the theory and from the information present in the text of the problem, we establish that an acid solution has to have a pH less than 7; so we can write:
[H^{+}] > [OH^{-}]K_{W} = 1.00x10^{-14} = [H^{+}][OH^{-}]
HCl | → | H^{+} | + | Cl^{-} | |
b | 1.00x10^{-8} | 0 | 0 | ||
a | 0 | 1.00x10^{-8} | 1.00x10^{-8} |
Where b means before (the reaction takes place) and a means after. So we have:
[H^{+}] = 1.00x10^{-8}M [Cl^{-}] = 1.00x10^{-8}MIf r mol L^{-1} of H_{2}O ionizes at equilibrium,
H_{2}O | ⇌ | H^{+} | + | OH^{-} | |
Initial concentrations | 1.00x10^{-8} | 0 | |||
Equilibrium concentrations | 1.00x10^{-8} + r | r |
r must be a positive number in the interval 0 < r < (K_{W})^{1/2}M. Now, we combine the various information:
1.00x10^{-14}M^{2} = (1.00x10^{-8} + r)M x rM
r = 9.51x10^{-8}M
[OH^{-}] = r = 9.51x10^{-8}M [H+] = 1.00x10^{-8}M + 9.51x10^{-8}M = 1.05x10^{-7}MpH = 6.978
Verification. It is necessary to verify if the solution is acceptable.
1 . Electro-neutrality condition
[OH^{-}] + [Cl^{-}] = 1.05x10^{-7}M2. Dissociation constant
[H^{+}][OH^{-}] = 9.99x10^{-15}M^{2}The conditions are fulfilled taking significant figures into account. Thus, students know that they have solved the problem in the correct way.
Let us consider the titration of a weak acid by a strong base. The titration curve is reported in all textbooks; but just few select points are calculated. For example, Petrucci (4) performs the titration of 25.00 mL 0.1000 M CH_{3}COOH with 0.1000 M NaOH: a) Before the addition of any NaOH; b) after addition of 15.00 mL 0.1000 M NaOH; c) at the equivalence point; d) after addition of 26.00 mL 0.1000 M NaOH. But, what if we add 0.25 mL 0.1000 M NaOH or 24.90 mL 0.1000 M NaOH or 25.02 mL 0.1000 M NaOH?
For example, let us consider the problem: To 25.00 mL of 0.1000 M CH_{3}COOH are added 24.95 mL of 0.1000 M NaOH. Calculate the pH of the resulting solution. With this method, the smartest students can solve the problem:
(25.00 mL) x (0.1000 mol CH_{3}COOH/L) x (10^{-3} L/mL) = 2.500 x 10^{-3} mol CH_{3}COOH
(24.95 ml) x (0.100 mol NaOH/L) x (10^{-3} L/mL) = 2.495x10^{-3} mol NaOH = 2.495x10^{-3} mol OH^{-}
V = 25.00 mL + 24.95 mL = 49.95 mL
CH_{3}COOH | + | OH^{-} | → | CH_{3}COO^{-} | + | H_{2}O | |
b | 2.500x10^{-3} | 2.495x10^{-3} | 0 | 0 | |||
a | 5x10^{-6} | 0 | 2.495x10^{-3} | 2.495x10^{-3} |
So, we have:
[CH_{3}COOH] = 5.0x10^{-6} mol CH_{3}COOH/4.995x10^{-2} L = 1.001x10^{-4} M [CH_{3}COO^{-}] = 2.495x10^{-3} mol CH_{3}COO^{-}/4.995x10^{-2} L = 4.995x10^{-2} MWe consider the CH_{3}COOH equilibrium,
CH_{3}COOH | ⇌ | H^{+} | + | CH_{3}COO^{-} | |
Initial concentrations | 1.001x10^{-4} | 1.00x10^{-7} | 4.995x10^{-2} |
For establishing the direction toward which the net reaction proceeds, we calculate the reaction quotient, Q:
Q = [H^{+}][CH_{3}COO^{-}][CH_{3}COOH] = 1.00x10^{-7} x 4.995x10^{-2} / 1.001x10^{-4} = 4.995x10^{-5}
Because Q > K_{a} the reaction proceeds from right to left: if r_{1} of CH_{3}COOH are formed at equilibrium,
CH_{3}COOH | ⇌ | H^{+} | + | CH_{3}COO^{-} | |
Initial concentrations | 1.001x10^{-4} + r_{1} | 1.00x10^{-7} - r_{1} | 4.995x10^{-2} - r_{1} |
r_{1} must be a positive number in the interval 0 < r < 1.00x10^{-7} M.
K_{a}[CH_{3}COOH] = [H^{+}][CH_{3}COO^{-}]
(1.753x10^{-5}M)x(1.001x10^{-4} + r_{1})M = (1.00x10^{-7} - r_{1})M x (4.995x10^{-2} - r_{1})M
r_{1} = 6.484x10^{-8} mol/L
[H^{+}] coming from CH_{3}COOH is: 1.00x10^{-7} - r_{1} = 3.516x10^{-8} M. Now, exactly how much is [H^{+}] coming from the water ionization is calculated: if r_{2} mol L^{-1} of water ionizes in a solution of CH_{3}COOH/CH_{3}COONa, at equilibrium,H_{2}O | ⇌ | H^{+} | + | OH^{-} | |
Initial concentrations | 3.516x10^{-8} | 0 | |||
Equilibrium concentrations | 3.516x10^{-8} - r_{2} | r_{2} |
K_{W}=r_{2}(3.516x10^{-8} + r_{2})
r_{2} = 8.395x10^{-8} M. Then, these values are iterated: we consider again the acetic acid equilibrium,
CH_{3}COOH | ⇌ | H^{+} | + | CH_{3}COO^{-} | |
Initial concentrations | 1.001x10^{-4} | 8.395x10^{-8} | 4.995x10^{-2} |
Again (Q = 4.189x10^{-5} > K_{a}) the forward reaction takes place: if r_{1} mol L^{-1} of CH_{3}COOH are formed at equilibrium,
CH_{3}COOH | ⇌ | H^{+} | + | CH_{3}COO^{-} | |
Initial concentrations | 1.001x10^{-4} + r_{1} | 8.395x10^{-8} - r_{1} | 4.995x10^{-2} - r_{1} |
(1.753x10^{-5}) x (1.001x10^{-4} + r_{1})M = (8.395x10^{-8} - r_{1})M x (4.995x10^{-2 - r1)M}
r_{1} = 4.880x10^{-8} mol/L
and [H^{+}] = 8.395x10^{-8} - r_{1} = 3.515x10^{-8}M
pH = 7.454
Verification:
[CH_{3}COOH] = 1.001x10^{-4} + r_{1} = 1.0x10^{-4}MWe have:
[Na^{+}] + [H^{+}] = 4.995x10^{-2}MIn this way students know that they have performed the task correctly.
CONCLUSIONSAs Freiser (12) warned us, when dealing with this topic, we "must avoid the Scylla of oversimplification to achieve clarity and the Charybdis of cumbersome rigorous equations". From our experience, this way to systematize as well as to improve the ease of teaching these important aspects of general chemistry works: the large majority of students are successful in solving ionic equilibrium problems and in dealing with the chemistry involved.
A FINAL REMARKHe has research publications in chemical education, study of charge-transfer complexes and chemistry of free radicals. Liberato is author of two books, one on how to solve chemical problems and another on using a computer in chemical education. He has spent three periods of study leave working with the late Professor Lord John Tedder and with Dr John C. Walton at the Chemical Centre of the University of St. Andrews in Scotland, doing radical chemistry research. Now, problems related with chemical education are of high priority.