A PROCEDURE FOR SOLVING EQUILIBRIUM PROBLEMS'
8
ROBERT NEILSON BOYD Washington Square College, New York University, New York, New York
As GENERALLY
presented, the solution of simple equilibrium problems is straightforward and uncomplicated. First, one writes the chemical equation, listing all the substances involved and their respective concentrations; the equilibrium expressionis written next; finally, one introduces particular values for the concentrations and the equilibrium constant, K., for the given system, and solves the resulting mathematical equation for the desired quantity. When problems arise which involve hydrolysis of salts of weak acids or bases, as in Example I, the student is generally taught to recognize them as a specific type, and in solving them to use a special "short-cut" method since values for K , (or K.) are not listed in tables. Ezample I : What is the concentration of OH- in 0.01 M sodium acetate? Substances , Ac- H1O = HAc OHY Y Concentrations 0.01 Y
+ -
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At this point, the student may be shown that by a simple mathematical artifice, that is, by inserting [H+] in both numerator and denominator of the righthand term of the expression for KH (equivalent to multiplying the right-hand side of the equation by unity), and then collecting terms, [HAcl[OH-I !H+l Ka = [Ac-I [H+l 1 lHAc1 = x [OH-][H+]= x K w = K" [Ac-l[H+l KA Kn he can obtain an expression, K E = KwIKA, that is easily remembered, easily evaluated, and easily used. - YXY , :Y K E K. = ~ [AAcl[OH-I = Ac(0.01 - y) - 0.01 1 x lo-"
, 2
1.85 x 10" - 0.01 ar [OH-] 2.3 X 10-8 M Unfortunately, what the student is not shown is (1) that what he has done involves the fundamental principles of equilibrium systems; and (2) that what he has done here for hydrolysis systems can be extended widely to more complicated types of problems involving several co-existing equilibrium systems. 'Presented at the 118th Meeting of the American Chemical Society, Chicago, September, 1950.
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In brief, the subscript in K His emphasized too strongly; the uniqueness of its character is nonexistent, for it is still essentially K,, for the reaction as written. It is fundamental that if several systems are involved in one over-all system, and that if the total system is in equilibrium, then the several systems are a t equilibrium, individually and collectively. The mathematical trick of introducing [H+] was far more than a convenient device; it implied that two individual systems, water and acetic acid, are coabined into one equilibrium. Furthermore, it can be made into a rational procedure; for it ought to appear reasonable to the student that if the only equilibrium expression involving hydroxide ion that he can think of as being applicable a t this point is K w , then he probably ought to bring it into the calculation. This he can do by introducing the symbol, [H+l, to give [H+] [OH-]; but having done so in the numerator, he is required to introduce it also into the denominator, and the term for KAis thereby automatically obtained. This procedure has the advantage, moreover, of direct application t o problems which can prove extremely tedious in solution. with many chances for errors, or which may even appear under the best of conditions to be unsolvable without extreme effort. It is the chief purpose of this paper to demonstrate the validity of this extension by the use of a variety of illustrations. For instance, a common problem that is enoountered in elementary courses involves a complex ion and an insoluble solid, as in Example 11. Ezample II: What concentration of ueous ammonia will be required to dissolve 0.001 mol of ~ a p e liter r of solutionl Usually the student is told first of all to write the equation for the reaction: AgCl
+ 2NHa = Ag(NH&++ C10.001
0.001
Then he is expected to see that two equilibria are involved : AgCl = Ag+
+ C1-
Ag(NHah+ = AgC
sincehe knows that
+ 2NHr
KnP= [Ag+l[CI-I Klne~b
[&+I [NHJ'
[rnNH7,
the [CI-1 is to be 0,001 M , he is then instmcted to calculate the [Ag+] from the bt of these; K , = [Ag+][CI-] 1.56 X 10-10 = [Ag+] (0.001) [Ag+l = 1.56 X lo-' M
198
APRIL. 1952
199
and then to insert this value into the equilibrium expression for the complex ion in order to obtain the value for [NHa], since the [Ag+] will be the same for both equilibria. The value so obtained is 2.1 X.10-'M.
lo-'
=
Mg++ O.O1
+ 2NROH
= M~(OH)+ E 2N%+
0.05
[NIL+]: = [NH,OH]'[Mg++]
Here K., is evaluated by introducing [OH-12, which enables the separations of the two factors, KBZand l/K,,,, that represent the major equilibria of the system.
lAg+1[NRJ2 [Ag(NHahf1 (1.56 X 1O-')[NH,Ia
Kuut%b=
6.8 X
Substances Concentrations
This procedure of an intermediate calculation involving an ion common to both equilibria was avoided in the hydrolysis case. If it could be avoided there, i t should he unnecessary in this instance, as well. Therefore, let us begin over again. First, write the chemical equation, together with the concentrations of the various comnonents: Substances Concentrations
AgCI
+ 2NHs - Ag(NHA+ f CI0.001
0.001
then the equilibrium expression for the reaction
I t is no longer necessary to calculate the [OH-] needed to precipitate Mg(OH)%,and then use this to calculate the INHIOH], according to the generally employed method; instead, since the [OH-] is the same for the separate equilibria, it can he employed as shown to enable the immediate use of both equilibrium constants, simultaneously rather than one a t a time. Two insoluble solids can he handled easily, as Example IV shows.
Now, it is necessary to evaluate K., (just as it was previously necessary to evaluate K,), and this can he done by considering what equilibria are involved and what equilibrium expressions are therefore available Ezample IV: Solid AgCl is added to a 0.1 M KBr solution. What is the ratio of the [Cl-I to the [Br-] in the solution for use. Obviously [CI-] must be connected to the when equilibrium is reached7 K., for AgC1, so the term [Ag+] may be introduced in ApCl = ApBr + ClBrthe numerator in order to make explicit this relationship of [Ag+] and [CI-1; but this requires that [Ag+] be put into the denominator, as well, and examination of the resulting equation shows that the other equilibrium system, as represented by KiDat.b,has now hecome apparent, too. There is no need to calculate [Ag+], and then [CI-1, finally securing the ratio of [Cl-] to [Br-] by another step, when one operation is sufficient. Furthermore, it becomes obvious that the ratio will be the same whatever the starting concentration of Br- (or of C1-, for that matter), and the figure of 203 as an equilibrium constant signifies a great deal more than a value for a mere ratio of two ionic concentrations. The solution of the problem is a routine matter from Another important group of problems deals with this point: precipitations by HB. Examples V and VI are typical of these, and they also demonstrate the ease with which such problems can be solved.
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Only one computation is necessary, limiting the chances for errors; the two equilibria are involved in a very direct fashion; and, what is more important, they are brought in together, as they should be. Example 111 shows how the procedure may he applied to a problem involving two bases, one'of which is an insoluble substance.
Emmple V: What concentration of Cd'+ can exist in 0.3 M HCI saturated with HsS? Cd++ + H I S = CdS 2H+ LH+I' Keq = [Cd++l[H2Sl [Hf1' I S 3 1 [H+I2[S--1 K"q = [Cd++][H,S] [S--1 - [Cd++][S--1 [H,S] 1 [H+la K., = - X K,r = 1.1 X lo-" = 1.1 x 10' = KP 1 X lo-" ICd++lIH~Sl
Ezample 111: A solution is 0.01 M in Mg++ and 0.05 M in of N H , will ~ ~ Mg(OH)* NHc~~, begin to precipitate?
The choice of [S--1 as the operating factor is an obvious one because of the need for involving [Cd++] IS--]
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JOURNAL OF CHEMICAL EDUCATION
U)O
as well as the hydrogen sulfide equilibrium. K12 can be used here for HeS as in all cases where [Hf] and [S--1 are not dependent solely upon the ionization of the H2S. Remembering that a saturated solution of H2S is about 0.1 M, and that practically none of the hydrogen ion comes from its ionization,
proper equilibrium constants. In the case at hand, CuS, S, and H:O would be included. Ka' =
[Cu++lS[Sla[NOla[H~O14 [CuS]~[NOs-]s[H+]a
The equilibrium constant then can he evaluated by involving the separate equilibrium constants for CuS, (0.3)P the oxidation of S-- to S, and the reduction of NOalo' = [ c ~ + +(0.1) I to NO. This is done by introducing [%-I3 in the [Cdt+] = 8.2 X 10-'M numerator to com~letethe ex~ressionfor the K.. of Ezample VI: Calculate the minimum [H+]necessary to prevent CuS; the necessary introduction of the same term in precipitation of ZnS when a solution which is 0.01 M in Zn++ the denominator completes the constant, [S]/[S--1, for is saturated with HB. the oxidation of S-- to S. The expression for the Zu++ + HsS = ZnS + 2H+ reduction of NOa- to NO is already complete. 0.01 0.1
~-
Inserting the known values for the various terms gives or in exponentials which are easier to handle, Many problems concerning amphoteric substances can be handled in the same fashion. Example VII is a good - illustration. Ezample VII: How many mols of Zn(0HX can be dissolved by one liter of 0.1 M NaOH? Zn(OH), 20H- = Zn0,-2H20 Substances z Concentrationa 0.1-22
+
K.,
=
lo-'
=
+
=
=
(10-87.6)~(10".4)~1018)2
=
10S5.7 g 1086
This high value for K.. indicates that the equilibrium point of the reaction as witten lies far to the right in the direction of the oxidation of the sulfide. If a concentration of unity is assigned to NO, NOs-, and H+, the equilibrium concentration of cupric ion would be 1012M. Of course, two of the equilibria, namely those involving CuS and the oxidation of s--, can be combined by means of available oxidation-reduction potential values, and when this is added to the NO3-/NO half-reaction,
Z
(0.1
- 22)Z
Neglecting - 22 as compared to 0.1 gives z
K.,
[ZnOa--1
lo-'
lo-' mols of Zn(0H)a per liter. The Power of the proposed n ~ t h o dbecomes greater the larger the number of separate equilibria which make UD the over-all reaction. Some of the most striking examples include oxidation and reduction systems, such as are found in Examples VIlI and IX. or
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an e. m. f. of +0.36 volt for the over-all reaction of oxidation of CuS is obtained. This gives a K., of approximately 10" in agreement with the previous value. (For HgS, the potential value of -1.05 volt leads to a K., of 104, an indication that HgS will not be nearly so oxidieed by dilute nitric acid as ;* c , ,1~ VYY., LY
&le VIII: Use equilibrium principles to show that cus be separated from HgS by dilute nitric acid. The products are Cu++, S, and NO. 3CuS 2NOs8H+ = 3Cuf+ f 35 2NO GLO [Cu++I~NO12 K s q = [NO,-12[H+l~
~t must be recognized that the calculations are valid only in the range of concentrations where the law of Mass Action holds, that the values are given only for 25"C., and that no account can be taken of the speed with which equilibrium will be reached. But for qualitative purposes, the calculations are helpful in indicating the possibilities inherent in a chemical procedure.
When oxidation-reduction equilibria are of importance, it is frequently convenient to mention all substances that are part of the system, whether solid, liquid, or gas, in order to facilitate the choice of the
Ezample ZX: Show by e uilibrium oansiderations the feasibility of a qualitative prooe!ure which calls for the dissolving of HgS by bromine, the products being S, Br-, and a complex ion of mercury, H&--. HgS Bh 2Br- = HgEr-S
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