General Chemistry II 2302102

General Chemistry II2302102 Chemical Equilibrium for Gases and for Sparingly-Soluble Ionic Solids Lecture 2 i.fraser@rmit.edu.au Ian.Fraser@sci.monash.edu.au

Chemical Equilibrium- 2 Lectures Outline - 4 Subtopics • Equilibrium and Le Chatelier’s Principle (Completed) • The Equilibrium Constant (Completed) • Temperature and Pressure Effects (Completed) • Sparingly-Soluble Ionic Compounds in Aqueous Solution

Chemical Equilibrium Objectives - Lecture 2 • By the end of this lecture AND completion of the set problems, you should be able to: • Understand the concepts of: saturated and unsaturated solutions, freely soluble and sparingly soluble ionic compounds. • Understand and know several examples of precipitation reactions. • Understand the definition of the solubility product (Ksp) and its relationship to the solubility of sparingly soluble ionic compounds. • Understand the definition of the Common Ion Effect. • Calculate equilibrium concentrations in precipitation reactions.

Electrical Conductivity Open Circuit Closed circuit

Electrolytes Strong Weak Non Few ionsin solution Many ionsin solution No ionsin solution

Electrolytes An electrolyte is a substance that conducts an electric current when dissolved in water (or in the the molten state)

Electrolytes Examples:Most salts, Acids and bases Strong electrolytes: their water solutions are good conductors (e.g. NaCl solution) Weak electrolytes: their water solutions are poor conductors (e.g. vinegar - acetic acid, CH3COOH) Non-electrolytes: their water solutions are nonconductors (e.g. sugar solution)

e.g. NaCl(s) Na+(aq) + Cl-(aq) AgCl(s) Ag+(aq) + Cl-(aq) Sparingly Soluble Salts • Some salts dissolve readily in water NaCl dissociates in solution to form ions • But other salts are only slightly soluble:

Why are we interested in sparingly soluble salts? • Earth’s crust is dominated by these salts: • feldspars, gypsum, calcite, aluminosilicates, dolomite, and oxides & sulfides of metals • control major geochemical processes • Boiler “scale” - often iron & manganese oxides • “Hard” water - • Waters originating from limestone areas contain Mg2+ and Ca2+ • these ions form soap “scum” precipitates

http://www.jenolancaves.org.au CaCO3(s) + CO2(g) + H2O Ca2+(aq) + 2HCO3-(aq) Caves are formed Stalactites and stalagmites are formed

Terminology • Recall from an earlier lecture on phase equilibria, for solutions in equilibrium with solids: • The equilibrium concentration of a dissolved solid in solution is known as the solubility • Such a solution is called “saturated” • Undersaturated: More solute will dissolve. • Supersaturated: Solution has an excess of solute.

Solubility Terminology Freely soluble - several g (or more) dissolve in 100 g of water • e.g. at 298K, 36 g of NaCl, 122 g of AgNO3 Sparingly soluble - << 1 g dissolves in 100 g of water • e.g. at 298K, 2.4 x 10-4 g of AgCl, 4.4 x 10-13 g of PbS, 9.3 x 10-4 g of CaCO3 Intermediate solubility - ca. 1 g dissolves in 100 g of water (only a few of these) • e.g. 1.02 g of Ag(CH3COO) at 293K, 0.19 g of Ca(OH)2 at 273K

Net Ionic Equation: What is it? A reaction equation between two electrolytes, for example between silver nitrate and sodium chloride, can be written conventionallyas: AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq) precipitate But AgNO3, NaCl, and NaNO3 are ionized in aqueous solution (AgCl is a solid - insoluble)

Net Ionic Equation: What is it? The conventional equation can be rewritten as: Ag+ (aq) + NO3- (aq) + Na+ (aq) + Cl- (aq) AgCl (s) + Na+ (aq) + NO3- (aq) Here, the NO3- and Na+ ions are unchanged. They are simply “spectator” ions - they do not participate in the reaction.

Net Ionic Equation: What is it? It is known (observed) that the actual reaction occurs only between silver ions and chloride ions Ag+ (aq) + Cl- (aq) AgCl(s) The above equation is a net ionic equation.

Spectator Ions Lead nitrate + Potassium chromate Pb2+ + 2 NO3- 2 K+ + CrO42- Lead chromate(s)+ Potassium nitrate PbCrO4(s) 2 K+ + 2 NO3-

Spectator Ions So Net Ionic Equation is: Pb2+ (aq) + CrO42- (aq) PbCrO4(s)

Equilibrium does NOT mean: • Reactions are not occurring Forward and reverse rates of reaction are equal - hence new products are formed at the same rate as they are broken down • The concentrations of all compounds are equal at equilibrium The concentrations of all compounds are determined by the position of the equilibrium - it may favour the products or the reactants

[C]c [D]d K = [A]a [B]b Equilibrium Constant Generally for: aA + bB cC + dD

AgCl(s) Ag+(aq) + Cl-(aq) [Ag+] [Cl-] K = A Saturated Solution is at Equilibrium The equilibrium constant expression is:

AgCl(s) Ag+(aq) + Cl-(aq) AbBa(s) bAa+(aq) + aBb-(aq) [Aa+]b [Bb-]a Kc = [AbBa] The Solubility Product, Ksp The concentration of [AbBa] (a solid) is a constant, and is by convention set  1 in the definition of Kc [Aa+]b [Bb-]a = Ksp  Kc =

The Solubility Product, Ksp Examples: Ksp(AgCl) = [Ag+] [Cl-] Ksp(Fe2S3) = [Fe3+]2 [S2-]3 Ksp(BaSO4) = [Ba2+] [SO42-]

AgCl(s) Ag+(aq) + Cl-(aq) Calculation of Solubility Ksp(AgCl) = [Ag+] [Cl-] = 1.8 x 10-10 mol2 dm-6 Ksp = s x s = 1.8 x 10-10 s = 1.34 x 10-5 mol dm-3

Solubility Products at 25°C

PbF2(s) Pb2+(aq) + 2F-(aq) Calculation of Solubility #2 Ksp(PbF2) = [Pb2+] [F-]2= 3.7 x 10-8 M3 Ksp = s x (2s)2= 3.7 x 10-8 s = 2.1 x 10-3 M [Pb2+] = 2.1 x 10-3 M, [F-] = 4.2 x 10-3 M

The Common Ion Effect The presence of an ion in solution which is common to the electrolyte will decrease the solubility: We’ve just seen that if PbF2 is placed in water, the solubilities of the ions are: [Pb2+] = 2.1 x 10-3 M, [F-] = 4.2 x 10-3 M What happens if PbF2 is placed in a solution of 0.02 M KF? (F- is the “common ion”)

The Common Ion Effect What happens if PbF2 is placed in a solution of 0.02 M KF? (F- is the “common ion”) In this case, [F-] = 2s + 0.02 M Ksp = s x (2s + 0.02)2= 3.7 x 10-8 Ksp = 4s3+ 0.08s2+ 0.0004s = 3.7 x 10-8 Solve as a cubic, or if 0.02 >> s 3.7 x 10-8= s x (0.02)2 s = [Pb2+] = 9.3 x 10-5(c.f. 2.1 x 10-3 M)

PbF2(s) Pb2+(aq) + 2F-(aq) Application of Le Chatelier’s Principle • Recall that: if possible, systems react to an imposed change by reducing the impact of that change • Increase [F-] (by adding KF - soluble) • Force equilibrium to left, and therefore reduce [Pb2+]

bAa+(aq) + aBb-(aq) AbBa(s) Qo = [Aa+]ob [Bb-]oa Precipitation Reactions • A solid will form (precipitate) if the concentrations of the ions exceed the solubility product. First calculate the initial “reaction quotient”, Qo : Then compare Qo with Ksp

Qo = [Aa+]ob [Bb-]oa Precipitation Reactions #2 First calculate the initial “reaction quotient”, Qo : Then compare Qo with Ksp : If Qo> KspPrecipitation Occurs If Qo< KspNo Precipitation

Qo = [Ag+]o [Cl-]o = 2 x 10-4x1 x 10-5 = 2 x 10-9 M2 Precipitation Reactions #3 Example 1. 4 x 10-4 M AgNO3 is mixed with an equal volume of 2 x 10-5 M NaCl. Will a precipitate form? First calculate the initial “reaction quotient”, Qo : Then compare Qo with Ksp (1.8 x 10-10 M2): Is Qo> Ksp Yes! Precipitation Occurs

Cd2+(aq) + 2OH-(aq) Cd(OH)2(s) Precipitation Reactions #4 Example 2.Electroplating with cadmium is a common industrial process. Before waste solutions can be discharged, [Cd2+] must be reduced to < 1 x 10-6 M. This may be achieved by adding sodium hydroxide to precipitate the cadmium as the hydroxide salt: Ksp(Cd(OH)2) = 5.3 x 10-15 M3 What concentration of NaOH is required?

Precipitation Reactions #5 Example 2 cont... Ksp(Cd(OH)2) = 5.3 x 10-15 M3 Cd2+(aq) + 2OH-(aq) Cd(OH)2(s) Ksp = 1 x 10-6x s2= 5.3 x 10-15 s = 7.3 x 10-5 Hence to reduce [Cd2+] < 1 x 10-6 M, [OH-] must be > 7.3 x 10-5 M

Selective Precipitation Problem: Ag+(aq) as silver nitrate is added to a mixture containing 0.001 M Cl- and 0.001 M CrO42-. Ksp (AgCl) = 2.8 x 10-10, Ksp (Ag2CrO4) = 1.9 x 10-12. What will happen as [Ag+] is increased? 1/ Precipitation of AgCl(s) will begin when [Ag+] > 2.8 x 10-7 M. (Qo > Ksp) 2/ Precipitation of Ag2CrO4(s) will begin when [Ag+] > 4.4 x 10-5 M. (Qo > Ksp)

Selective Precipitation 1/ Precipitation of AgCl(s) will begin when [Ag+] > 2.8 x 10-7 M. (Qo > Ksp) 2/ Precipitation of Ag2CrO4(s) will begin when [Ag+] > 4.4 x 10-5 M. (Qo > Ksp) But in order for the value of [Ag+] to exceed 4.4 x 10-5 it is necessary that [Cl-] < 6.4 x 10-6 Thus on addition of silver nitrate, AgCl will precipitate until the concentration of the chloride ion remaining is less than 6.4 x 10-6 M, at which stage precipitation of silver chromate will occur.

Chemical Equilibrium - End of Lecture 2 Objectives Covered in Lecture 2 • After studying this lecture should be able to: • Understand the concepts of: saturated and unsaturated solutions, freely soluble and sparingly soluble ionic compounds. • Understand and know several examples of precipitation reactions. • Understand the definition of the solubility product (Ksp) and its relationship to the solubility of sparingly soluble ionic compounds. • Understand the definition of the Common Ion Effect. • Calculate equilibrium concentrations in precipitation reactions.

General Chemistry II 2302102