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