12-9 Ionic Equilibria
A. Solubility Equilibria
The solubility product expression for a 1-1 electrolyte is MX(J) = M*+ + X2- , KSO = (M2+)(M2-), and, for the general case,
Mv+Xv_(s) = ,+Mz+ + v_x2-, ^sp = (Μκ+γ+(Χκ-γ- [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 Ksv> 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+ = (Xz-)/v_ , so in the general case
* s p = (
v+y+(v-)
v-sr.
12-9 IONIC EQUILIBRIA 469 As an example, K8V = 9 χ 1 0- 12 for A g2C r 04 at 25°C, so we have
9 χ 1 0 -12 - ( 2 )2( 1 ) S3, S = 1.3 X 10-4.
If, however, an electrolyte is present which furnishes a common ion, say Xz-, then the solubility is given by (Mz+)/v+ . The expression for Ksv> becomes
^
Sp = (v
+sy+(c + v.sy-
9where C is the concentration of added Xz-.
Continuing this example, if 0.1 m N a2C r 04 is also present, then 9 χ ΙΟ"12 = (25*)2(0.1 +
The resulting cubic equation is best solved by successive approximations:
Sl ~ ( = 4.7 χ ΙΟ"6.
\ (4)(0.1) /
The first approximation, S1, gives a result small compared to 0.1, and is therefore adequate.
A more serious calculation would now proceed to the estimation of Ky for a solution S molal in Ag2Cr04 and 0.1 m in Na2Cr04 as the medium, so as to obtain a more nearly correct KBp
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(s), 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)f ~ begins to form and the solubility, that is, total dissolved CoOx(^), 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:
AgAcO) = Ag+ + Ac-, KST> = 1.8 χ ΙΟ"3, HAc = H+ + Ac-, Κ = 1.75 Χ ΙΟ"5.
If the pH is known, this determines the degree of dissociation α of the acid, and Ksv = (Ag+)(Ac-) = S(*S).
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
Membrane
Nn+ A - A "
M+ ! M +
s
j
S!
F I G . 1 2 - 1 2 . 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
(12-96) If the solutions are dilute, activities may be replaced by concentrations. Also, electroneutrality requires that (M+)1 = (X-)1 = C and (M+)11 + (N+)11 = ( X ~ )n. Equation (12-96) thus reduces to
C2 = ( Μ + )π [(M+)11 + (N+)1 1]. (12-97) Suppose that C = 0.01 m and that (N+) = 0.1 m. We find (M+)11 = 9.9 x 10~4 m and see that the Donnan effect acts to exclude M+ from side II. If C is IO"3 m, (M+)11 drops to 1 x IO"5 m, and the exclusion ratio ( M+) V ( 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
(12-98) where J*" 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 H A 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 :
HA = H+ + A-
f=
(Hr+2^ · (
12"")
( H A )
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
H20 = H+ + OH-, Kw = ( H+) ( O H r ) . (12-100) The solution must be electrically neutral, and so
(M+) + (H+) = (A-) + (OH-), (12-101) where ( M+) = fMA . Finally, by material balance
(HA) + (A") = / H A + / M A · (12-102) Note that (HA) and ( A-) denote the actual concentrations of these species,
whereas fHA and fMA are the amounts weighed out per 1000 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. (12-101) we have
( A - ) = /MA + ( H + ) - ( O H - ) and insertion of this result into Eq. (12-102) gives
( H A ) = /H A- ( H + ) + (OH-).
Equation (12-99) then becomes
K ( #+) [ / M A + ( H+) + (OH-)]
*a = a - (H+) + ( O H - ) «/ h ( 1 2"1 0 3)
The simultaneous solution of Eqs. (12-100) and (12-103) then gives ( H+) and ( O H-) for any formal composition.
Equation (12-103) simplifies considerably under various special conditions.
Case 1. /MA = 0. Then
a (H+)[(H+) - (OH-)]
/HA - CH+) + (OH-)
If (OH ) < (H+), then K& = (H+)2/[/HA - (H+)]. This condition holds if (H+)2 > 10"12 and hence if ΚΆ/ΗΑ > 10"1 2. If also /HA > (H+), then K* = ( H + )2/ /H A. This condition holds if /HA > 100(H+) and hence if /HA > 1 04^a .
As examples, for 0.1 m HAc, ΚΆ/ΗΑ = 1.75 χ 10"6 and is much greater than 10~1 2. Since/HA is about 1 04#a , the last approximation can just be used (to 1 % error) and ( H+)2 = 1.75 χ 1 0_ β, (H+) = 1.32 χ 10-3. However, i f /HA were 10~8 m, then KJ^ would be 1.75 χ 10"1 3, and the full equation (12-103) would be needed.
Case 2. /HA = 0. Then
v ( H+) [ /MA + (H+) - (OH-)]
If we divide K& by ATW to give Kh , the hydrolysis constant, we obtain Kb = — = (QH~)[(OH~) - (H+)]
# a /MA + (H+) - (OH")
This is of the same form as Eq. (12-104), and the same two types of approximation follow:
(OH-)2
If KhfuA > IO"12, then Kh =
If /MA > 104#h , then Kh =
/MA - (OH-) (OH-)2
/MA
For example, Kh = 1.01 χ 10"14/1.75 χ 10~5 - 5.77 χ 1 0- 1 0. For 0.1 m NaAc, KhfMA = 5.77 χ 1 0- 1 1, or more than 10~1 2, and /MA = 0.1, or more than 10*Kb . The simplest form may then be used and (OH")2 = (0.1)(5.77 χ 10"1 0), (OH") = 7.6 χ 10~6, (H+) = 1.3 x IO"9.
Case 3. If fMA and /HA are each greater than (H+) or ( O H ) , then Eq. (12-103) reduces to
^ ( H J V M A ( 1 2 1 0 6)
/HA
The solution is now said to be buffered. That is, in order to change (H+) appreciably, we must add sufficient acid or base to change /MA or fHA appreciably. Thus in a solution 0.1 / in HAc and 0.1 / i n NaAc, (H+) will be 1.75 χ IO- 5. Addition of 0.01/HCl changes/HAC to 0.11 a n d /N a Ac
to 0.09, and hence (H+) only changes to 2.1 χ IO- 5.
A parallel set of relationships holds for the weak base BOH:
The analog of Eq. (12-103) is
( O H - ) [ /BX + (OH-) - (H+)]
K» = /BOH - (OH") + ( H+) · ( 1 2-1 0 7)
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 HA, MA, 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 ( O H-) , then these ions may be ignored in the charge balance, and the simultaneous equations to be solved are
BOH + HA - B+ + A - + H20 , *ab = = ^ ,
(B+) + / M A = (Α-) + Λ χ , / H A + / M A = (HA) + (A"), /BOH + / B X = (BOH) + (B+).
D. Titration Curves
An acid-base titration consists of adding successive amounts of a base to a solution of an acid, or of an acid to a solution of a base, and noting how the
12-9 IONIC EQUILIBRIA 473
hydrogen ion concentration [or pH = — log(H+)] varies. We will consider only the first situation here.
The calculation of a titration curve is clarified if one recognizes that a solution which initially has a formality fHA in acid is converted into one having formalities /HA and fMA by addition of a strong base, where fMA is the formality of the added base and /HA + fMA = /HA · As a specific example, consider a solution that initially is 0.1 fin acetic acid. Addition of sodium hydroxide sufficient to make the solution 0.05 / in N a O H must yield an equilibrium mixture identical to that which would be obtained if the solution were 0.05 fin H Ac and 0.05 fin NaAc. In eifect, the formal composition of the solution may be expressed in two alternative ways:
0.1 fin H Ac) ( 0.05 fin H Ac, 0.05 fin N a O H ) ~ (0.05 fin NaAc.
Both statements specify the same amount of acetic acid substance and sodium ion.
The general procedure for finding such alternative ways of expressing a formal composition is to consider species that would tend to react and then to suppose that the reaction goes to completion. The new formal composition is expressed in terms of the products. Thus H Ac and N a O H tend to react:
H Ac + Na+ + OH" = Na+ + Ac" + HaO. (12-108) The alternative expression of formal composition follows if we suppose that the
neutralization of the O . l m H A c by 0.05 m N a O H goes to completion, giving 0.05 m NaAc and residual 0.05 m HAc. The reaction does not in fact go entirely to completion, but the formal composition can be given as 0 . 0 5 / N a A c and 0 . 0 5 / H A c .
With this preamble, the calculation of a titration curve reduces to the calculation of (H+) from Eq. (12-103) for various ratios of /MA to /HA , with their total kept constant. The equivalence point is that for which /HA = 0. The result of such a calculation for a 0.002 m acetic acid solution is shown in Fig. 12-13, where FN is the degree of neutralization. (It has been assumed that the added sodium
hydrox-i ι I I I I I I I I I 1 0 0.2 0.4 0.6 0.8 1.0
FIG. 12-13. Titration curve for acetic acid using a strong base.
ide solution is sufficiently concentrated that dilution effects can be neglected.) Calculations of this type can be quite tedious, and as in the preceding subsection, the reader is reminded that a more general and powerful approach is given in the Special Topics section.
A titration may be followed conductimetrically. In a neutralization reaction such as that of Eq. (12-108), each portion of N a O H added converts an equivalent amount of HAc into Na+ + A c-. The conductance of the solution therefore increases since a weak electrolyte is being replaced by a strong one. At the end point, further addition of N a O H adds N a+ and O H- ions to those of the NaAc, and the conductance increases more rapidly, as illustrated in Fig. 12-14. If the acid being titrated is a strong acid, then the neutralization converts the ions H+ + A~ into N a+ + A~ or, in effect, substitutes sodium ion for hydrogen ion.
Since the equivalent conductivity of sodium ions is much less than that of hydrogen ions, the result is that the conductance drops during the titration until the end point is reached, and then rises as before.
Example. Suppose that 0.1 m HCl is being titrated with concentrated NaOH (we therefore neglect volume changes) and that the conductance is followed using an immersion-type con
ductivity cell of cell constant 0.2. The initial conductance is L0 = (CA)/1000k = (0.1)(426)/
(1000X0.2) = 0.213 ohm"1. When enough NaOH has been added to make the solution 0 . 0 5 / in NaOH or, alternatively, 0.05 m in NaCl and 0.05 m in residual HCl, the conductance Lx is
(0.05X426) + (0.05X126) χ
Lx = — 0.138 o h m- 1.
(1000X0.2)
At the end point the formality of the added NaOH is 0.1, and the solution consists of just 0.1 m NaCl. The conductance L2 = (0.1)(126)/(1000)(0.2) = 0.063 ohm-1. When the end point is overshot with 0 . 1 5 / added NaOH, the system is 0.1 m in NaCl and 0.05 m in NaOH, and L3 is now
(0.1X126) + (0.05X248)
(1000X0.2) 0.125 ohm"1.
Thus the series of L values 0.213, 0.138, 0.063, and 0.125 goes through a minimum at the end point.
FIG. 12-14. Conductimetric titration curves for a strong acid (upper left line) and a weak acid {lower left line).