Universiiv
James N. Butler British Columbia
of
\ancouver, 8. C.
1
I
An Approath to Complex Equilibrium Problems
In t,he typical introductory analytical chemistry course, it is conventional to spend considerable time t,reat,ingproblems in ionic equilibrium. Such problems have long been used to help students understand the acid base t,itrations and precipitations encountered in t,he analytical laboratory. Equilibrium problems can be made to serve an addit,ional purpose if a certain type of approach to them is used. The more complex problems provide a convenient introduction to the use of approximations in simplifying complicated mathematical expressions, a technique which the student will find very useful in more advanced courses in both physics and chemistry. One possible approach to equilibrium problems is described in this paper. The general method used is to set up enough equations relating the various conr~ntrationspresent to define the system completely and then to use the student's experimental knowledge to make approximations which simplify the equations. A series of examples of increasing difficulty is used to illustrate this approach. If the solution under consideration contains only a slightly dissociated solvent (water) and fully dissociated solutes (strong acids or bases and their soluble salts) only one equilibrium need be considered, namely (As is convent,ional, t.he hydrated proton is designated H+.) There are, however, two concentrations to be considered, even in pure water, and so we must obtain another independent equation relating them in order to solve for their values. Such a relation is obtained by specifying that the solution be electrically neutral; t,hat is, that the number of positive charges be equal to t,he number of negative charges [Ht] = [OH-] Combining these two equations gives simply [Ht]
=
[OH-] =
dG
=
10'
which is a well known result. The emphasis here is placed not on the result, but on the use of the electroneutrality principle to obtain a second equation relating the two concentrations of interest. This principle is easily extended to include solutions of completely dissociated acids or bases, as is illustrated by the following example: Find [H+] and [OH-] in a M and a lo-' M solution of HzS04. Assume romplete dissociation of the acid.* There are three concentrations, [H+], [OH-], and *Actually, &SO4 dissociates in two ~ t e p s ;,the dissociation constant for HS0.c being about 10-9. For ~lmplicityin this example complete dissociation is assumed.
(SO4=),so we must have three relations between them: solvent equilibrium: [Hf] [OH-] = KW = =1" eleetroneutrality: [ H I ] = [OH-] 2 [SO4-] mass halance: [SO,q = 4 = or lo-'
+
Not,e that the sulfate ion has t.wo charges, so that in t,he electroneutrality relation (which is a charge b a h c e ) we must multiply its concentration by the number of charges. Note also that the mass balance says simply that all the sulfate ion present came from the sulfuric acid, whose "analytical concentration" is Co. Consider now the case where Co = lo-' M . The solution will be quite acid, so t,hat ( H i ) is much larger than [OH-]. This means that we can approximate the electroneutrality condition by neglecting [OH-] compared ni th [H+] : IH +I 2 [SO,-I Substituting from the mass balance, [H+] = 2 X 10Wa, and using the solvent equilibrium we ralculate [OH-] = 5 X 10-12. We immediately see that our approximation was excellent. I t is important to emphasize that we can neglect a small concentration mly when it is added to or subtracted from a large concentration, never when it occurs in multiplication or division. Turning to the case where Ca = lo-' M we see that we cannot make the simplification that we did before, since the solut,ion is nearly neutral. The electroneutrality condition becomes on substitution for (SOT), [Ht]
=
[OH-]
+2 x
lo-'
substituting in the solvent. equilibrium for [Hi], me ohtain a quadratic equation in [OH-]: [OH-]' + 2 x 10-7 [OH-] - lo-" = 0 solving this by the quadratic formula gives [OH-] = and substitut,ion in the water equilibrium gives [H+] = 2.4 X lo-'. I t can be readily seen that all three concentrat,ions are approximately 10-7 molar and so none can be neglected. A crucial question now arises: suppose we had made the approximat,ion that [H+] is large compared to [OH-] and used the approximate electroneutrality relation obtained in the first part of the problem. How would we know whether it was applicable or not? The simplest answer is to use the approximation: 4.1 X
[H+]
"2 [SOO-]
=2 X
10-7
and note that [OH-] = 5 X 10-'when calculated using this result. The approximation can be seen to be poor, and in fact on comparison with the exact solut,ion can be seen to int,roducean error of about 20%. Volume 38, Number 3, Morch 196 1
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141
All of the essentials of the method have been illustrated hy t,his simple example. They may be summarized in general t,erms as follows: 1. Set op expressions for all the independent equilibria involved in the system concerned. 2. List all the conrentrations of interest. These must include 811 concentrations in the equilibrium expressions, and mas also contain additional speci~s(such as Nat, NOa-, ete.) which are present but do not renrt. 3, If the ions of the mlvent are involved in the equilibria (arid-base systems) set up the condition for electroneutrelity. 4. Uar maas halnnre relations to obtain enough independent relations to define the system completely. The total numher oi cnnecnt,rationsshould be eqmal to tho total numher of relations. 5. Mnkr as man? approximations in tho mass-balance and electro-ncut,mlity relations a8 is eon~istentwith your quslitativr chemirnl knowledge of t,ho system. In particular: Neglert [OH-] mmpamd to [HI] in acid solutions; neglect [Ht1 rompared t,o [OH-] in hasir solutiona; assume preripitation or complex formntion rrartions go to completion, etc. €. Salre the npproxim:rte equations. 7. Chrrk tho ~.csoltsobtained hv the aooroximnte method to arts not valid and if nrrcssary rr-calculate the results.
Illustrative Applications
Calculate [Pb++], the concentration of lead ion i n gram-ions per liter, for a. saturated solution of PbCL in water and for a saturated solution of I'bCIZ in 0.05 112' NaCl. The equilibrium considered here is the solubility of lead chloride, xx-hich is described hy the solubilit,y produrt (at room t,cmperature) [Ph++][Cl-la
=
1.8 X 10-'
Since the ions of water do not affect the equilibrium they need not be considered. I n the pure water solution, only I%++ aud CI- ions need he considered; t,he rlect,ro~~rut.rality relat,ion gives
,f(X) = 0 is known, a briter approximation is given by
where f(Xo) is the function evaluated at Xo and f'(Xo) is the derivative evaluated a t Xo. I t is convenient to make X approximately unity, so let us set [Pb++] = 0.01X; then our cubic equat,ion is f(X) = X 3 + 5 X P + 6 . 2 5 X - 4 5 = O
Its derivat,ive is Evaluating a t Xo = 3.6, f(Xn) = 89.0, f'(Xn) = 81.2, giving X I = 2.50. A second approximation, using ,f(X1) = 17.4, f'(X1) = 50.0, gives X* = 2.15. A third approximation, using f(X2) = 1.4, fl(X2) = 41.6, gives X1 = 2.12, which is close enough. Thus the saturated solution of PbCL in 0.05 M NaCl has [Ph++] = 0.021. Please note that if the usual approach to a "common ion" problem had been used, that is to set [CI-] = 0.05, we would have calculated [Pb++] = 0.072, which is larger t,han the value for the saturat,ed solution in pure water. Although the mathematics is more difficult, the inclusiou of an occasional problem in which the usual approximations are not valid is often sufficient to make t,he student realize that not all problems can he classed into standard types. Such problems as this, whieh involve the solution of higher degree equations where an approximate root is known, serve to introduce the wdent to Scwton's method, which may prove very useful to him in later work. 2 ml qf 0.04 d NH4CI are mi.xed with 1 ml qf 0.01 dl NH?. Calculate [H+]. If the solution is diluted to .SO liters, what is [fI+]? Two equilibria are involved here; t,he ionization of wat,er and the dissociation of aqueous ammonia: [H+] [OH-]
Altrrnatively, this relation can be considered to he a mass halancc: I:or earh mole of precipitak that dissolves, onc mole of lcad ion and 2 moles of chloride ion arc formed. Hence the conceutration of chloride is t,wice that of lcad. Substitut.ion in the soluhilit,y or [Pb++] = produrt gives 4 [Pb++]' = 1.8 X 0.03G. Whpn sodium (:hloride is added, [Cl-] is inrreased, and hrnce [I%++] derreaxx A precise statement of this is provided by inrluding t,he concentration of sodium ion in the elect,roneut~ralityrondition
Alternatively, t,his can be considered to be a mass balance on chloride ion: All t,he chloride iou in snlut,ion comeseither from S a C l or from PbClp. The concentratiou of sodium ion is simply the concentration of SaCl, 0.05, and the result may he subst,ituted in the solubility product rclat,ion to give
[I%++] mill be smallcr thau the value obtained for pure wat,er, 0.036, which would indicat,e that the two t,erms in t,hc brackets are nearly the same size. We can s o h t,his cubic equat,ion by using Xewton's method of appmximat,ing roots: If an approximate root. X o of 142
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=
K , = 10-I
