Solution of problems involving equilibrium constants

right.) HXeH++X-. (If acid is strong, equilibrium is displaced completely to the right.) Step 4: ... (The above is true since the numbers of mols of p...
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ASSUMING that the

Step 2: The solution under consideration contains completely answers to problems are l e s important in a student's chemical education than an ionized HCI in a solvent which is only verysligbtly ionized. Step 3: H,O = H + + OH-. understanding of the means used to obtain the answers, Step 4: a review of journals and texts seems to indicate that 1 pH = log the solution of problems has not been satisfactorily (1) [H+l treated. Some methods suggest mere substitution of [Hi] [OH-] = lo-" (2) known values into an explicit expression to obtain [ H f ] = [OH-] + [CI-] answers; this obviates thinking on the part of the (3) student who then proceeds all too often to apply this [cl-1= lo-" (4) substitution in a formula to a case which the formula Step 5: From (3) and (4), was never designed to cover. Still other methods [ H t ] = [OH-] + lo-* involve the making of approximations that are not a t all obvious, nor are they shown to be valid. The From this and (2), usual alternative to these unsatisfactory methods invariably suggests the difficulty, often almost insurmountable, of solving a number of unwieldy simultaneous equations. I t is the purpose of this paper to suggest a procedure for the solution of problems-and particularly of problems having to do with equilibrium constants-which starts from fundamentals, is as is much less than 4(10-"), Since simple as possible, avoids approximations that are not obvious or shown to be valid, and requires the student to be urecise in his thinking. In Halving prohlems of t h e types mentioned above From this and (I), 10' t.he student may be advised to take the following steps: pH e log - = 6.998 E 7.0 1. Read the problem. 2. Visualize the situation being discussed, considering such things a s the probable range of the pH, the solution's volume, etc. 3. Write chemical equilibrium equations for any equilibria prevailing in the solution. 4. Write mathematicsl equations as they become apparent in terms of the equilibrium constants, data. given in the problem, the consideration that solutions have no net charge, etc. If the pH, pOH, pC1, per cent hydrolysis or anything else is desired, it is, of course, necessary to write an equation involvi~g the desired variable, such as, for example, the d e f i ~ n gequation for pH. 5. After n equations having only n unknowns have been written, make any obvious approximations and solve, keeping on the alert to make any further approximstions that may be justibble from either chemical or mathematical considerations. To avoid the solution of complicated equations guesses may be made that certain terms of these equations are negligible Provided that these guesses are later shown to be correct.

the manner in which these steps may he applied, two problems will be solved. Problem 1 : A factory's waste materials consist of inerk and HCI, ~h~~~ are discharged into a stream at such rate that the HC1 concentration of the stream is lo-' molar. The Fish and Wildlife Service wishes to know the pH of the stream.


Problem 2: Develop an equation expressing the [H+] in a solution &iCh might have been obtained by mixing b milliequivalents of-a monoacidic base with a milliequivalents of a monohasic acid in terms of equilihrium constants, a, b, and the volume u of the solution in millilitels. Using the derived equation determine the [H+] and the of a 0.1 M NH~CNsolution. Step 2: The solution under consideration may be a solution of an acid, a base, a salt, or a a d t with excess acid or base. The base and/or the acid involved may be weak, causing hydrolysis of the salt's cations and/or anions. It might be observed that the solution being discussed could be a solution of a strong base, a stmng acid, a. weak base, a weak acid, a salt of a strong base acid, a salt of a stronz base and a weak acid. a salt and a of a weak base and a, strong acid,-a salt of a weak baseand a. weak acid or a. mixture of any of these kinds of salts with either the acid or the base from which it was produced, with the limitation that only monobasic acids and monoacidic bases may be considered. Some of the types of problems dealing with tbese solutions are ionization constant problems, common ion effect problems, titration curve calculations, buffer solution problems, and problems dealing with pH's of various kinds of salt solutions. The fact that a general equation far the H' ion concentration in all of tbese solutions can be derived throws doubt on the advisability of the common practice of treeting d l of the problems mentioned as though they were entirely unrelated.




Cross-multiplying, and grouping terms:

Step 3:

+ OHMOH = M + + OHHIO e H +


(If bass is strong, equilibrium is displaced completely to the right.) HXeH++X(If acid is strong, equilibrium is displaced completely to the right.) Step 4: [H+I[OH-I



Dropping out negligible terms,

lo-' [ H f ] - 10-lQ 107" [H+j5 + lo9[H+]l - 7 1.8





The base is stronger than the acid. Hence the solution is basic Therefore and lo-' > [H+] > 10-".

[H+l[X-l = K. (for strong acid li. = m) IHXl

[M+] [X-]


+ [MOH] = b/v


+ [HX] = a/u

+ [Hf1 = [OH-] + [X-I

(5) (6)

(The above is true since the numbers of mols of positive charges snd of negative charges per liter are equal.)


pH = log (l/[H+I)


[Ht1 =


- [1\1+1)





Elimmnting [M +I from (6) and (8). - Kb(b/u) [OH -1 KI


+ [H+] =


+ [X-l


10-10 7 = 1.8 100

4% 1010


e 4(Our guess was good.) 1010

From this and equation (7) above, 10'0 pH = log - = 10 - 0.8 = 9.2 6.24

Eliminating [HX] from (3) and (51, [H+][X-] =

10-¶a 1.8[HCl -

Guessing that the last term is negligible from a consideration of probable values of [Ht],

Step 5: Eliminating [MOHl from (2) and (4), [M+][OH-1 = Ka


lo* [H+]1 - lo-'* 7 1.8



Iclirninating [X-I from (9) and (lo), (11)

Many problems ask for the "per cent hydrolysis of the salt" in addition to the pH. For many salts the meaning of this is quite clear. With regard to salts of weak acids and weak bases, however, it is well to point out that the anion and the cation may hydrolyze to different extents. In this event the per cent hydrolysis of each ion must be considered and the larger of these taken as the per cent hydrolysis of the salt. If the per cent hydrolysis of the R H 8 N in the solution discussed above were desired one might obtain the [H+] in the same general way as ahove. Then v e have from equation (1)

Eliminating [OH-] from (1) and ( l l ) ,

[OH-] = [H




I'wm equ~tion(2) thr rnrm d conrmlmtlm oi carim hydrdvztrl 11sv eoncentrstion o i cnrion unl.ydroIy~~d ii


Similarly, from equation (3), the ratio of the concentration of anion hydrolyzed to the concentration of anion unhydrolyzed

With regard to the above expression, it is interesting is to note that if our solution is, for example, one of NaC1, b = a, 1/K, = 0, 1/K. = 0 and we have directly that [H+l = fiw. In determining the [H+] in a 0.1 M NH&N solution, we have from the above From this and equation (5) we have equation, substituting 7/10" for K., 1.8/105 for Kb, 10-l4 for K,, 0.1 for b/u, and 0.1 for a/v

[HX] = 0.047 The extent of hydrolysis =

0.00.1 47



In this case the salt's ions are seen to be hydrolyzed to almost exactly the same extent. This is not always

MAY, 1950


true. For example, the cation in 0.1 M NHH,103 terms in which the final equation's only unknown hydrolyzes to the extent of 0.0092 per cent while the occurs to the same power. If, for example, !ye are anion hydrolyzes t o the extent of 0.0032 per cent interested in solving for x in the simple equation (taking K. = 0.19). A word of caution may be in order with respect to the solution of systems of simultaneous equations. and we drop the second term since it is negligible with It may be noted above that after n equations having respect to the first term we obtain the impossibility only n unknowns were written, an unknown was that 0 = 1. If, however, we inspect the terms more eliminated which occurred in only two equations. This closely and observe that the difference between the reduces the system to one of n - 1 equations having two largest terms in .z2 is small compared to the second n - 1 unknowns in one operation, thus saving work. term, me see that the second term must be retained to With regard to the maliing of approximations, it will obtain the correct solution, x = 1000. Similar situabe noted that errors will sometimes be introduced if tions sometimes occur in the solution of problems inapproximations are made before we have grouped all volving equilibrium constants.