12-9 Ionic Equilibria
A. Solubility Equilibria
The solubility product expression for a 1-1 electrolyte is MX(5) = M'+ + X*~, KSO = ( Μβ +) ( Μ ' - ) , and, for the general case,
MvXv_(s) = v+Mz+ +
v_x
z-
y Kap = (MZ+)V+(XZ-)V- [Eq. (12-60)].The solid salt is taken to be in its standard state and hence to have unit activity, but we must remember that for A^p to apply, the solid must in fact be in equilibrium with the solution.
The solubility S of an electrolyte is defined as the number of gram formula weights that dissolve in the particular medium. If there is no added common ion, then S = (Mz+)/v+ = (X*-)/v_ , so in the general case
12-9 IONIC EQUILIBRIA 469
As an example, Κ8Ώ = 9 χ 1 0- 1 2 for A g2C r 04 at 25°C, so we have 9 χ 1 0 -1 2 = ( 2 )2( 1 ) S3, S = 1.3 χ 10"4.
If, however, an electrolyte is present which furnishes a common ion, say X*-, then the solubility is given by (Mz+)/v+ . The expression for Α^ρ becomes
*sp = (v+sy+(c + v_s)v-, where C is the concentration of added Xz- .
Continuing this example, if 0.1 m N a2C r 04 is also present, then 9 χ i o -1 2 = (2S)2(0.1 + 5 ) .
The resulting cubic equation is best solved by successive approximations:
/ 9 χ IO"1 2 W2
The first approximation, S i , gives a result small compared to 0.1, and is therefore adequate.
A more serious calculation would n o w proceed t o the estimation of Ky for a solution S molal in A g2C r 04 and 0.1 m in N a2C r 04 as the medium, s o as to obtain a more nearly correct KBj>
using Eq. (12-93).
Another type of complication is that the dissolved salt may not be fully disso
ciated. For example, appreciable amounts of CoOx (where Ox denotes oxalate ion) are present in solution as the undissociated molecule. If the solution is saturated with respect to CoOx(5), the concentration S0 of undissociated CoOx in solution is a constant. The observed solubility S is then .S0 + ( C o2 +) , and if N a2O x is added, S decreases due to the common ion effect. With further addition of N a2O x , however, the complex Co(Ox)i" begins t o form and the solubility, that is, total dissolved CoOx(s), increases. Other cases of complex formation with slightly soluble salts include the well-known example of A g ( C N )2" , as well as that of A g C l2" and other silver halide complexes.
Finally, if the anion of the slightly soluble salt is one of a weak acid, then the solubility will depend on the pH of the solution. Thus in the case of silver acetate the following simultaneous equilibria would hold:
AgAc(*) = A g+ + A c - , K8V = 1.8 Χ 10"3, H A c = H+ + A c - , K = 1.75 Χ 10"5.
If the pH is known, this determines the degree of dissociation α of the acid, and
Kai> = (Ag+XA<r) = S(otS).
In this case the total acetic acid substance is given by the solubility S, but the actual A c- concentration is only ocS.
B. Equilibrium across a Semipermeable Membrane
An important type of ionic equilibrium is that across a membrane which is per
meable only to certain of the ions present. The situation is known as one of Donnan
M e m b r a n e
1 JL
Nn + A - A"
M + ! M +
S S
!
F I G . 12-12. Donnan equilibrium.
equilibrium. As an example, consider the arrangement shown in Fig. 12-12. The membrane is permeable to solvent and to M+ and X" ions, but not to N+ ions.
Such selectivity is shown, for example, by the membranes of living nerve cells.
The condition for ionic equilibrium is that the activity a2 [see Eq. (12-52)] be the same on both sides of the membrane for that electrolyte to which it is permeable, in this case M+, X~. Thus
<4+«x-
= « - · (12-96)If the solutions are dilute, activities may be replaced by concentrations. Also, electroneutrality requires that (M+)1 = (X")1 = C and (Μ+)π + (N+)1 1 = (X")n. Equation (12-96) thus reduces to
C2 = (M+)1 1 [(Μ+)π + (N+)1 1]. (12-97) Suppose that C = 0.01 m and that (N+) = 0.1 m. We find (M+)1 1 = 9.9 χ 10~4 m
and see that the Donnan effect acts to exclude M+ from side II. If C is 1 0- 3 m, ( M+)n drops to 1 χ 10~5 m, and the exclusion ratio ( M+ ) 7( M+)n increases from 10.1 to 100.
The physical basis for the exclusion effect is that a potential difference, the Donnan potential φ, makes side II positive relative to side I. It can be shown that
Ψ ~ -ψ l n (M+)M ' (12-98)
where & is Faraday's number [see Eq. (13-12)]. In the numerical example, φ is 0.0594 V and 0.118 V for C values of 0.01 m and 0.001 m9 respectively, for 25°C.
Donnan potentials are important in biology. For example, nerve cells or axons appear to be permeable to K+ but not to N a+ ions and application of Eq. (12-88) gives about the observed potential across the resting cell membrane. Reduction of this potential by more than a certain threshold amount makes the membrane permeable to N a+ ions and a wave of local depolarization races along from cell to cell. Such nerve impulses travel at some 100 ft s e c- 1. The contraction of muscle cells also involves changes in cell membrane potential and in ratio of permeability to N a+ versus K+ ions.
C. Weak Acids and Bases
The treatment of dissociation equilibria involving weak acids and bases will be limited here to the cases of a simple acid HA and a simple base BOH. Even so, the general solutions can involve rather complex algebraic manipulations. A power
ful alternative approach is given in the Special Topics section.
12-9 IONIC EQUILIBRIA 471
We consider first the weak monobasic acid H A :
H A = H+ + A - , Ka = (H^p · ( 1 2 - 9 9 )
The general situation is one of a solution prepared by dissolving amounts of H A and of the salt M A so as to give the formalities fHA and fMA . The cation M + is assumed not to hydrolyze. With the solvent taken to be water, we have
H2o = H+ + O H - , Kw = ( H + X O H - ) . ( 1 2 - 1 0 0 )
The solution must be electrically neutral, and so
( M + ) + ( H + ) = ( A - ) + ( O H - ) , ( 1 2 - 1 0 1 )
where ( M + ) = /M A . Finally, by material balance
( H A ) + ( A " ) = / HA + / M A · ( 1 2 - 1 0 2 )
Note that ( H A ) and ( A-) denote the actual concentrations of these species, whereas /H A and fMA are the amounts weighed out per 1 0 0 0 g of water when the solution is made up.
The preceding four equations must be solved simultaneously, and it is helpful to reduce them as follows. From Eq. ( 1 2 - 1 0 1 ) we have
( A " ) = / M A + ( H+) - ( O H - )
and insertion of this result into Eq. ( 1 2 - 1 0 2 ) gives
( H A ) = /HA- ( H + ) + ( O H - ) .
Equation ( 1 2 - 9 9 ) then becomes
V _ ( # W M A + ( H + ) + ( O H - ) ]
K* ~ / H A - ( H + ) + ( O H " ) * ( 1 2"1 0 3 )
The simultaneous solution of Eqs. ( 1 2 - 1 0 0 ) and ( 1 2 - 1 0 3 ) then gives ( H+) and ( O H-) for any formal composition.
Equation (12-103) simplifies considerably under various special conditions.
Case 1. /MA = 0. Then
If ( O H ) < ( H + ) , then ΚΛ = ( H + )2/ [ /HA - (H+)]. This condition holds if ( H + )2 > 1 0 "1 2 and hence if K&fHA > 1 0 ~1 2. If also /H A > (H+), then K& = ( H + )2/ /HA . This condition holds if /HA > 100(H+) and hence if /H A > 1 04Aa.
A s examples, for 0.1 m Η Ac, K&fHA = 1.75 χ 1 0 ~β and is much greater than 1 0 "1 2. Since fHA is about 1 04Aa , the last approximation can just be used (to 1 % error) and ( H+)2 = 1.75 χ 10~β, ( H+) = 1.32 χ ΙΟ"3. However, i f /H A were 1 0 -8m , t h e n KJHA would be 1.75 x 1 0 "1 8, a n d the full equation (12-103) would be needed.
Case 2. fHA = 0. Then
v ( H+) [ /M A + CH+) - ( O H - ) ]
If w e divide ΚΛ by Kw t o give Kh , the hydrolysis constant, w e obtain
* . _ ( O H - ) K O H - ) - (H+)]
κ
"~ί;-
/MA + ( H+) - ( O H - ) · ( 1 2-1 0 5 ) This is of the same form as Eq. (12-104), and the same t w o types of approximation follow:( O H - )2
If KhfMA > 1 0 -1 2, then Kh =
If /MA > 104tfh , then Kh =
/MA - (OH") ( O H - )2
F o r example, = 1.01 χ 10"1 4/1.75 χ 1 0 ~5 = 5.77 χ 1 0- 1 0. For 0.1 m N a A c , KhfMA = 5.77 χ 1 0 "1 1, or more than 1 0- 1 2, and fMA = 0.1, or more than 104#h . T h e simplest form m a y then be used and ( O H ~ )2 = (0.1)(5.77 χ 10"1 0), (OH~) = 7.6 χ 10~β, (H+) = 1.3 χ Ι Ο- 9.
Case 3. If /MA and fHA are each greater than ( H+) or (OH~), then Eq. (12-103) reduces t o
^ = ( H ^ A ( 1 2_1 0 6 )
/HA
The solution is n o w said t o be buffered. That is, in order t o change (H+) appreciably, w e must add sufficient acid or base to change /MA or fHA appreciably. Thus in a solution 0.1 / in Η A c and 0 . 1 / i n N a A c , (H+) will be 1.75 χ 10"5. Addition of 0 . 0 1 / H C l c h a n g e s /H A c t o 0.11 a n d /N a A c to 0.09, and hence ( H+) only changes to 2.1 χ 10"5.
A parallel set of relationships holds for the weak base BOH:
The analog of Eq. (12-103) is
„ ( O H - ) [ / ,X + ( O H - ) - ( H + ) ] „
KB ~ / B O H - ( O H - ) + ( H + ) ' °2-107)
where X~ is a nonhydrolyzing anion. The various special cases are similarly analo
gous to those for the weak acid.
We can find the equilibrium concentrations in a solution containing H A , M A , BOH, and BX, by solving Eqs. (12-100), (12-103), and (12-107) simultaneously. If all of the concentrations (HA), (A~), (BOH), and ( B+) are much larger than that of ( H+) or (OH~), then these ions may be ignored in the charge balance, and the simultaneous equations to be solved are
B O H + H A = B+ + A ~ + Η , Ο , *ab = = ψ ,
( B + ) + / M A = ( A - ) + /BX , / H A + / M A = ( H A ) + (A~),
/BOH + / B X = ( B O H ) + ( B + ) .