Equilibrium © 2009, Prentice-Hall, Inc. Chemical Equilibrium.
Chapter 12 A Deeper Look at Chemical Equilibrium.
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Transcript of Chapter 12 A Deeper Look at Chemical Equilibrium.
Chapter 12
A Deeper Look at Chemical Equilibrium
why Ka, Kw, Ksp↑while adding [NaCl]↑
Effect of electrolyte concentration on concentration-based equilibrium constants
12-1 The effect of ionic strength on solubility of salts
The explanation
dissociation ↑by↑ionic strength of soln.
What do you mean by “ionic strength”?
ionic strength = μ= ½ΣCιZι2
[I] ionic charges
example at p. 244
Effect of charge on μ
Electrolyte Morality μ ex.
1 : 1 M M NaCl
2 : 1 M 3 M Na2SO4
3 : 1 M 6 M Al(NO3)3
2 : 2 M 4 M MgSO4
(1) rxn: aA + bB → cC + dD
Without considering μ, k =
(2) The effect of μ; the [C] activities (Ac)
[C]c[D]d
[A]a[B]b
∴ k = AA
aABb
ACcAD
d
γAaγB
b[A]a[B]b
γCcγD
d[C]c[D]d
=
12-2 Activity Coefficients
The Debye – Hückel eqn.
Fine γX of the ions from ZX (ionic charge) & αX (ave
rage size)
-log γX = 1 + 3.3 αX
0.51 ZX2
αX = effective diameter of the hydrated ion X (10-9m)
works fairly well for 0.1 M
for most single charged ions : αX ≈ 0.3 nm
-log γX ≈
the larger charged & smaller ions ;
the larger αX (Table 12-1)
For μ< 0.01 3.3 αX << 1
1 + 3.3 αX ≈ 1
∴ -log γX = 0.51 ZX2
1 +
0.51
Effect of ionic strength
increase decrease. at μ→ 0, γ → 1, Ax → [x]
(2) at μ= const. ion with larger charge, the bigger △γ
ex. △ γBaSO4 > △γAgCl
at μ = k
(3) at any μ, for same charge ions: γX ≈ γY
their difference (minor) could be from at any effective diameter of hydrated ion was formed
How to interpolate
can be also found for that is between values in Table 12-
1 by using linear interpolation.
High ionic strengths
at high μ(μ> 0.1 M), γ↑& could > 1
Ex. at p. 249 for a better estimate of solubility of PbI2
PbI2 (s) Pb 2+ + 2I-
x 2x
(2) From , interpolation in Table 12-1 to get Pb2+ &
I-, and Ksp = γPb2+γI
-2[Pb2+][I-]2 find x2 &
(1) Ksp = x(2x)2 = 7.9 x 10-9 find x1 &
(3) From 2 & Ksp = γPb2+γI
-2[Pb2+][I-]2 find x3
Compare x1, x2, and x3
Real definition of pH
• pH = - logAH+ = - log([H+]H+)
• Ex: at P. 251
Ex:
• What weight of Na2HPO4 and KH2PO4 would be required to prepare 200 mL of a buffer solution of pH 7.40 that has an ionic strength of 0.20?
Sol:
(1) pH = pKa2 + log [HPO4-2]/[ H2PO4
-]
7.40 = 7.20 + log X/Y
(2) µ = 1/2ΣCiZi2
0.20 = 1/2 {[Na+](1)2 + [X](2)2 + [K+](1)2 + [Y](1)2} = 3X + Y
• 解聯立方程式 (1) and (2)
12-3 Charge & Mass Balances
• Ex. • Consider the charge balance for
a solution prepared by weighing out 0.0250 mol of KH2PO4 plus 0.0300 mol of KOH and diluting to 1.00 L.
• Solution:• H2PO4
- H+ + HPO42-
• HPO42- H+ + PO4
3-
• [K+] = 0.0300 + 0.0250 = 0.0550M• [HPO4
2- ] = 0.020M• [PO4
3-] = 0.005Mcharge balance: [H+] + [K+] =[OH-] + [H2PO4
-] + 2[HPO42- ]+ 3[PO4
3-]
Mass balance
• K2HPO4 in H2O
• Mass balance:• K2HPO4 2 K+ + HPO4
-
HPO42- H+ + PO4
3-
HPO42- + H+ H2PO4
-
H2PO4- H+ H3PO4
[K+] = 2 {[H3PO4] + [H2PO4-] + [HPO4
2-] + [PO43-]}
12-4 Systematic Treatment of Equilibrium
• Ex. 1 at p.254• pH of 10-8 M KOH KOH K+ + OH-
H2O H+ + OH-
• Sol:• Charge balance: [K+] + [H+] = [OH-]• Mass balance: [K+] = 1.0 ×10-8 M
• K balance: Kw = [H+]γH+ x [OH-]γOH
-
• ∵ μ is very low in this soln., so γ≈ 1
• Kw = [H+](1.0 ×10-8 + [H+])
• [H+]2 + 1.0 ×10-8 + [H+] - 1×10-14 = 0
• [H+] = = 9.6 ×10-8 or -1.1 ×10-7 M
• pH = -log (9.6 ×10-8 ) = 7.02
2
101410100.1 14168
Ex. 2 at P.255 Solubility of CaF2
• CaF2(s) Ca2+ + 2F- Ksp = 3.9 ×10-11
• F- + H2O HF + OH- Kb = 1.5 ×10-11
• H2O H+ + OH- Kw = 1 ×10-14
• Sol:• Charge balance: 2[Ca2+] + [H+] = [F-] + [OH-]• Mass balance: 2[Ca2+] = [F-] + [HF]
• K balance: Ksp = [Ca2+]γCa2+ x [F-]2(γF
-)2
= 3.9 ×10-11
Kb = = 1.5 ×10-11
Kw = [H+]γH+ x [OH-]γOH
- = 1.0 ×10-14
F
OHHF
F
OHHF
][
][][
• Ignore μ & allγ= 1, (check later), & use buffer pH=3
[H+] = 1 ×10-3 ∴[OH-] = 1 ×10-11
• From Kb: 1.5 ×10-11 = = 1.5
• From mass balance: 2[Ca2+] = [F-] + 1.5[F-] = 2.5[F-]
[Ca2+] = 1.25[F-]• From Ksp: 1.25[F-]3 = 3.9 ×10-11
[F-] = = 3.15 ×10-4
∴ [Ca2+] = 3.9 ×10-4
][
10][ 11
F
HF
][
][F
HF
3
11
25.1
109.3
12-5 Fraction Composition Equilibrium
• HA H+ + A-
F
• F = formal conc. = [HA] + [A-]
• Ka = =
• [HA]Ka = [H+]F – H+[HA]
[HA] (Ka + H+) = [H+]F
][][
AHA
AA
][
][][
HKa
H
F
HAHA
][
]][[
HA
AH ][
][][
HA
HAFH
][
HKa
KaA