STPM Chemistry Form 6 Notes – Heterogeneous Ionic Equilibrium (Part 4)

by BerryBerryTeacher

in Berry Reference (Notes)

Heterogeneous equilibrium situation occur when chemical equilibria and chemical reactions involving more than one phase occurs. A good example would be when an ionic compound partially dissolves in water such as calcium carbonate dissolving in water in a process called dissociation. In this Part 4 of Berry Berry Easy notes on Heterogeneous Ionic Equilibrium for STPM Chemistry, Berry Readers will be learning about heterogeneous equilibria, examples of it and be exposed to the list of solubility product constants. Be sure to try out the examples and see how to solve the question.

[Tips: Make sure you know the difference between homogeneous equilibria and heterogeneous equilibria. The former involves chemical species in the same phase, while the later which we are learning now would have species not all in the same phase. So ensure that you know the distinction between the two.]

STPM Chemistry Form 6 Notes – Heterogeneous Ionic Equilibrium (Part 4)

Heterogeneous Equilibria

  • Homogeneous equilibria – all species are in the same phase
  • Heterogeneous equilibria – all the species are not in the same phase. Example: heterogeneous equilibrium is when an ionic compound partially dissolves (in chemical terms, dissociates) in water, the undissolved ionic compound is a solid, and the dissolved portion is aqueous.
Heterogeneous Ionic Equilibrium Precipitation

Heterogeneous Ionic Equilibrium Precipitation

Example: Dissociation of CaCO3 in water
CaCO3(s) <—-> Ca2+(aq) + CO32-(aq)

Equilibrium expression
Ksp = [Ca2+][ CO32-] / [CaCO3]

However because CaCO3 is a pure solid, the concentration relative to itself is one. Therefore, we can ignore it out from the equilibrium expression,
Ksp = [Ca2+][ CO32-]

  • The ‘sp’ after the K term in solubility problems = solubility product, giving Ksp the term solubility product constant.

Solubility Product Constants at 25˚C

AgBr 5.0 x 10-13 CuS 8.5 x 10-36
AgCN 2.2 x 10-16 Fe(OH)2 1.6 x 10-14
Ag2C2O4 6.0 x 10-12 Fe(OH)3 1.1 x 10-36
AgCl 1.0 x 10-10 FeS 3.7 x 10-19
Ag2CrO4 9.0 x 10-12 Hg2Cl2 2.0 x 10-18
AgI 1.5 x 10-16 Hg2I2 1.2 x 10-28
AgOCN 2.3 x 10-7 MgCO3 2.6 x 10-5
AgOH 1.5 x 10-8 MgF2 6.5 x 10-9
Ag2S 1.6 x 10-49 Mg(OH)2 1.2 x 10-11
AgSCN 1.1 x 10-12 Mn(OH)2 4.0 x 10-14
Ag2SO4 1.5 x 10-5 MnS 2.5 x 10-10
Al(OH)3 2.0 x 10-33 Ni(OH)2 2.0 x 10-16
BaCO3 5.0 x 10-9 NiS 2.0 x 10-21
BaCrO4 2.4 x 10-10 PbCO3 3.3 x 10-14
BaF2 1.7 x 10-6 PbCl2 1.7 x 10-4
BaSO4 1.0 x 10-10 PbF2 3.6 x 10-8
CaCO3 1.0 x 10-8 PbI2 1.4 x 10-8
CaF2 3.4 x 10-11 Pb(IO3)2 2.5 x 10-13
CaSO4 2.0 x 10-4 PbSO4 2.0 x 10-8
Cd(OH)2 4.5 x 10-15 RaSO4 4.0 x 10-11
CdS 8.0 x 10-27 SrF2 2.9 x 10-9
CoS 3.0 x 10-26 SrSO4 2.8 x 10-7
CuI 5.0 x 10-12 Zn(OH)2 1.8 x 10-14
Cu(OH)2 1.0 x 10-19 ZnS 1.2 x 10-22

Example:

When a sample of solid AgCl is shaken with water at 25˚C, a solution containing 1.0 x 10-5 mol dm-3 silver ions is produced. Calculate Ksp.

Solution:

AgCl(s) <—-> Ag+ + Cl-
Ksp = [Ag+][Cl-]

[Ag+] = [Cl-] = 1.0 x 10-5
Ksp = (1.0 x 10-5) (1.0 x 10-5) = 1.0 x 10-10

This is the end of Part 4 notes on the topic of Heterogeneous Ionic Equilibrium for STPM Form 6 Chemistry students from Berry Berry Easy. In Part 5, Berry Readers will learn about solubility experiments, their methods and some pointers on predicting precipitation. So do drop by again at the site for more notes. See you again.

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