Help !I’m

dissolving

Using complex algebra to calculateconcentrations when multiple equilibria are in play

We can look for two more pieces of info:Charge Balance: Electroneutrality of the

solution; the sum of the positive charge in solution equals the sum of negative charges in the solution.

Mass Balance: Conservation of matter; the quantity of all species in a solution containing a particular atom (or group of atoms) must equal to the amount of that atom (or group) delivered to the solution.

We can look for two more pieces of info:Charge Balance: Electroneutrality of the solution;

the sum of the positive charge in solution equals the sum of negative charges in the solution.

Mass Balance: Conservation of matter; the quantity of all species in a solution containing a particular atom (or group of atoms) must equal to the amount of that atom (or group) delivered to the solution.

11

[H+] + 2[Ca2+] = [OH-] + [F-] (4)

2[Ca2+] = [F-] + [HF] (5)

How does the solubility of CaF2 depend on pH?

12

CaF2(s) Ca2+ + 2F- Ksp = [Ca2+][F-]2 = 3.910-11 (1)

F-+H2O HF + OH- 11105.1

][F

][HF][OHKb

(2)

H2O H+ + OH- Kw = [H+][OH-] = 1.010-14 (3)

Five unknowns: [Ca2+], [F-], [HF], [H+], and [OH-]Three equations mean we need more equations…

We can look for two more pieces of info:Charge Balance: Electroneutrality of the solution;

the sum of the positive charge in solution equals the sum of negative charges in the solution.

Mass Balance: Conservation of matter; the quantity of all species in a solution containing a particular atom (or group of atoms) must equal to the amount of that atom (or group) delivered to the solution.

[H+] + 2[Ca2+] = [OH-] + [F-] (4)

2[Ca2+] = [F-] + [HF] (5)

Ksp = [Ca2+][F-]2 = 3.910-11 (1)

11105.1

][F

][HF][OHKb

(2)

Kw = [H+][OH-] = 1.010-14 (3)

(4)

(5)

We need to substitute to get things in terms of Ca2+ or H+

[H+] + 2[Ca2+] = [OH-] + [F-]

2[Ca2+] = [F-] + [HF]

15

11105.1

][F

][HF][OHKb

Ksp = [Ca2+][F-]2 = 3.910-11

Kw = [H+][OH-] = 1.010-14

2[Ca2+] = [F-] + [HF]

Combine eq. 2 and eq. 5, we have…

][2][

][][ 2

Ca

OH

FKF b

][1

][2][

2

OHK

CaF

b

(B)

Combine (B) and eq. 1, we have…

][1

][2][

2

OHK

CaF

b

(B)

2

22

][1

][2][

OHK

CaCaK

bsp

31

2

2

]1

4][

[OH

KKCa bsp

(C)

It’s much simpler if we can consider the pH fixed (how would we do that?)

If we can fix pH…

pH = 3.00 [H+] = 1.0×10-3 M

Kw

[HF-] =1.5[F-]

Kb

[F-] =0.80[Ca2+]Mass

[Ca2+] = 3.9×10-4 M Ksp

[OH-] = 1.0×10-11 M

18

The [Ca] removed from marble stone (largely dissolution of CaCO3) increases as the [H+] of acid rain increases.

Applications of coupled equilibria in the modeling of environmental problems

CaCO3(s) + 2H+(aq)

Ca2+(aq) + CO2(g) + H2O(l)

SO2(g) + H2O(l) H2SO3(aq)

oxidation H2SO4(aq)

www.chem.wits.ac.za/chem212-213-280

Deposits include CaSO4•2H2O (gypsum), which accumulates creating a black residue.

http://pubs.usgs.gov/gip/acidrain/5.html

19

Total [Al] as a function of pH in 1000 Norwegian lakes.

Acid rain also releases Al, Hg, and Pb into the environment.

www.chem.wits.ac.za/chem212-213-280