Ch 13: Properties of Solutions

15

Chapter 6: Solutions

Objectives

·  units of concentration

·  solution formation

·  reactions in solution – stoichiometry, titration

·  acid – base reactions

·  redox reactions

·  colligative properties

·  mixtures and distillation

·  (colloids)

·  solutions = mixtures of substances which are homogeneous, i.e.- intermingled at molecular level

·  components (may be more than one of each)

·  solvent - major

·  solute - minor

·  say that solute is dissolved in solvent

·  Examples of solutions

·  can be gases, liquids, solids

·  here focus on liquid solutions, especially with water as solvent (aqueous solutions)

6.1 Composition of Solutions - Units of Concentration

·  qualitatively can describe solutions as dilute or concentrated

·  quantitative expressions based on mass or on moles

1.  (a) Mass %

1.  (b) Parts Per Million (ppm)

2.  (a) Mole Fraction (recall gas phase)

·  use symbol Xn where subscript n represents component of interest, eg. XHCl

2.(b) Molarity

units: mol L-1 (SI would prefer mol m-3, but too large)

2.(c) Molality

·  NB: these two easily confused

·  Molarity in terms of volume of solution

·  Molality in terms of mass of solvent

·  do Examples 6.1 & 2, Problems 3-9, odd; read practical details (how to)

6.2 Nature of Dissolved Species

Solution Process (Dissolution) & Solubility

·  an equilibrium:

·  from left, continue adding solute until concentration of solvated (i.e.- solution) particles is high enough that they frequently collide with and attach to the undissolved solute = crystallization

·  this reverse process also encouraged as solvating ability of solvent molecules is used up or occupied

·  eventually reach a point when the rates of dissolution and crystallization are equal, i.e.- at equilibrium (terms in chapter 11)

·  solution in equilibrium with undissolved solute is a saturated solution

·  concentration of solute in solution at that point is its solubility

·  eg. solubility of NaCl = 37.5 g/100mL of water at 0° (less NaCl, undersaturated)

·  possible to form supersaturated solution by preparing at elevated temperature, then cooling carefully (eg. “heat packs”)

How and why do solutions form?

·  recall, intermolecular forces that exist between molecules and ionic particles in pure substances

·  replace these forces with interactions between solvent and solute (the “how”)

·  for liquids dissolving in liquids, terms: miscible, if solution forms; converse: immiscible

·  many examples will be given of solution formation by ionic substances (i.e.- salts), but:

·  solutions also formed by uncharged molecules (eg. fructose, Fig. 6.2)

·  special case, colloids, at end of chapter

·  water is not the only solvent, but it is commonly encountered and is of special relevance to this class

·  terms: solvation and hydration

·  eg. gas solubilities in water, (previous chapter)

·  increase with size due to dispersion forces

·  for solids and liquids - “like dissolves like”

·  polar and ionic solutes in polar solvents

·  non-polar solutes in non-polar solvents

·  eg. acetone, , and smaller alcohols are completely miscible with water due to hydrogen bonding

·  increase the size of hydrocarbon (non-polar) fragments of ketones and alcohols, solubility drops off

·  fat- and water-soluble vitamins

·  eg. salt dissolution, eg. NaCl, K2SO4 (Fig. 6.3) - replace charge-charge interactions with ion - dipole interactions in aqueous solution (strong electrolytes)

·  solvation shell

·  reverse, precipitation reactions

eg. BaCl2(aq) + K2SO4(aq) ® BaSO4(s) + 2 KCl(aq)

K+ and Cl- are spectator ions; could write net ionic reaction

·  solution formation occurs when the forces within pure components and their replacements between solute and solvent are comparable, therefore, energy considerations

Energy Changes & Solution Formation (not in text)

·  i.e.- the “why” solutions form

·  from NaCl example, three energetic steps involved in solution formation

1.  solute - solute interactions (must be overcome)

2.  solvent - solvent interactions (some must be overcome)

3.  solute - solvent interactions (must form)

·  simplest approach (not foolproof), to consider heat (enthalpy) changes

·  net change is algebraic sum of three:

·  calculation by two approaches, depending on data available, eg. for a salt in water:

·  DHsoln = -lattice energy + enthalpy of hydration; (data in Tables)

·  DHosoln = DHo f, soln - DHo f, solid; i.e.- from Hess’s Law (data in Tables)

·  DH2 term not evaluated independently; a part of hydration/solution enthalpy

·  rough correlation of between enthalpy of solution and solubility

·  DHsoln can be exothermic (eg. NaOH forms warm aqueous solution, DHsoln = -45 kJ/mol)

·  DHsoln can be endothermic (eg. NH4NO3 forms chilled aqueous solution, DHsoln = 26 kJ/mol), as long as it is not too endothermic (gain by DH3 comparable to loss by (DH1 + DH2))

·  hence, ionic or polar solutes (eg. NaCl) will not dissolve in non-polar solvents (eg. CCl4), i.e.- DH1 large, DH3 small

·  similarly, non-polar solutes (eg. CCl4) will not be solvated by polar solvents (eg. water), i.e.- DH2 large, DH3 small

Solution Formation, Spontaneity & Disorder (not covered until chapter 8)

·  enthalpy considerations, above, not the only ones

·  why should endothermic process, eg. NH4NO3 dissolution, occur at all?

·  another surprise, two non-polar liquids like CCl4 and octane mix with very little energy change yet they dissolve (in each other) spontaneously

·  two factors must be considered with any chemical change, including dissolution:

1.  “Energy” (“enthalpy” in the context of this chapter). Processes in which the energy content of the system decreases tend to occur spontaneously. eg. falling book

2.  Entropy (disorder, or randomness). Processes in which the disorder of the system increases tend to occur spontaneously.

·  the dissolution of NH4NO3 is endothermic, yet occurs spontaneously because the rigid order of the crystal is totally disrupted as the ions are individually hydrated in solution

·  hence, earlier statement that a substance will not dissolve if DHsoln too endothermic - it is too large to be counteracted by an increase in disorder

·  octane and CCl4 form solutions when mixed in any proportions because each of the molecules has a bigger volume to “roam” around in the solution, i.e.- is more randomized

Factors Affecting Solubility

·  nature of solute & solvent, i.e.- solute - solvent interactions (above)

·  pressure, temp.

Pressure Effects

·  solids and liquids little affected

·  solubility of any gas in any solvent increases as pressure of the gas over the solvent increases

·  quantitated by a form of Henry’s Law:

Sg = kHPg

Sg = solubility of the gas (M) in the solvent

Pg = partial pressure of the gas over the soln

kH = Henry’s Law constant (specific for solute - solvent pair and temp)

( a different form of Henry’s Law in section 6.6, below)

·  eg. solubility of N2(g) in water at 25°C and a partial pressure of 0.78 atm is 5.3 x 10-4 M

·  if partial pressure doubled, what will solubility of N2(g) be?

·  first calculate kH:

·  then, at elevated pressure:

Temperature Effects

·  for most solid and liquid substances, solubility increases with temp.

·  eg. for ionic substances in water

·  solubility of gases in water generally decreases with increasing temp.

explained as increased kinetic energy, better able to escape

·  can be explained by using Le Chatelier’s Principle, i.e.- treat enthalpy of reaction as a component of the dissolution equilibrium:

·  if exothermic, enthalpy a “product”, increased temp., decreases solubility

·  if endothermic, enthalpy a “reactant”, increased temp., increases solubility

·  problem: entropy ignored; wait until chapter 8

·  eg. gas bubbles forming as glass of cold water warms to room temp.

·  eg. lakes become anaerobic if subjected to thermal pollution

6.3 Stoichiometry of Reactions in Solution – Titration

·  some “dissolution” processes involve chemical change, i.e.- “solvent” is also a reactant

·  eg.

·  not reversible; solute cannot be reclaimed by removing solvent

·  or, mix two solutions:

2 Br-(aq) + Cl2(aq) ® 2 Cl-(aq) + Br2(aq)

·  fixed volume of one, eg. 50 mL of 0.060 M NaBr(aq)

·  how much of other to completely react?

·  Concentration? Volume?

·  Do Example 6.5

·  Stages in the above reaction, a titration; at the stoichiometric ratio, the end point – signaled by some indicator

· 2 types here: acid – base, oxidation – reduction (redox)

Acid – Base Reactions

·  water dissociates to a very small extent: H2O(l) ® H+(aq) + OH-(aq)

·  an acid produces a net increase in [H+] over that in water, eg. HCl, HNO3, H2SO4, CH3COOH

·  a base produces a net increase in [OH-] over that in water, eg. NaOH, NH3

·  acid – base titration, Example 6.6 (a H+ transfer)

50.0 mL of acetic acid solution titrated with 1.306 M NaOH; 31.66 mL of latter needed to end point; conc’n of acetic acid?

Oxidation – Reduction (Redox) Reactions

·  electron transfer reactions

· eg. Br- + Cl2 reaction, above

·  Br- oxidized, loses e-

·  Cl2 reduced, gains e-

·  eg. Zn(s) in aqueous acid:

oxidation state: (0) (+1) (+2) (0)

reducing oxidizing oxidized reduced

agent, agent,

reductant oxidant

NB: conjugates (compare, acid - base)

·  not necessary to have ionic components, eg.:

(-3)(+1) (0) (+5)(-2) (+1)(-2)

red. agent ox. agent oxidized reduced

·  another eg.: Zn strip in a CuSO4(aq) solution, over time blue colour of solution disappears, copper “plates out” and Zn strip slowly disappears

overall: Zn(s) + Cu2+(aq) ® Zn2+(aq) + Cu(s)

comprised of two half-reactions:

oxidation: Zn(s) ® Zn2+(aq) + 2 e-

reduction: Cu2+(aq) + 2 e- ® Cu(s)

·  note, overall equation above already balanced with respect to mass and charge

Balancing Redox Equations in aqueous solution

·  2 methods to balance electrons: Oxidation Numbers and Half - Reactions (preferred)

Oxidation Number Method

·  eg., unbalanced (NB: need reactants and products)

(0) (+4) (+3) (0)

·  for Al, change +3; for Mn, change -4

·  to balance Al and Mn, take 4 Al and 3 Mn, then balance O by inspection:

Method of Half - Reactions

·  separate oxidation and reduction processes

·  (could also readily do this one by oxidation numbers)

·  eg. decolorization of permanganate solution by oxalate in aqueous acid solution:

·  unbalanced:

·  incomplete half-reactions:

·  balance first with respect to atom undergoing redox

·  for rest: in aq. acid, plenty of H+ and H2O available (in aq. base, plenty of OH- and H2O avail.)

·  balance O with H2O:

·  then balance H with H+ (from acid):

·  balance charge with e- :

·  similarly for oxalate:

·  no O, H balancing needed

·  charge:

·  multiply half-reactions as appropriate to give same number of electrons in each (gain = loss), then add:

·  overall (may need to simplify in some cases):

·  likewise, Examples 6.8 (in acid) and 6.9 (in base)

·  if in aqueous base, treat initially as if in acid, then add equal amounts of OH- to each side to neutralize acid (your text suggests a slightly different method; you choose one method)

·  Example in base:

Al(s) + S(s) ® Al(OH)3(s) + HS-(aq)

·  half reactions:

oxidation: Al(s) ® Al(OH)3(s)

reduction: S(s) ® HS-(aq)

·  for mass, redox-active atoms already balanced

·  balance O with H2O:

Al(s) + 3 H2O(l) ® Al(OH)3(s)

S(s) ® HS-(aq)

·  balance H, first with H+:

Al(s) + 3 H2O(l) ® Al(OH)3(s) + 3 H+(aq)

S(s) + H+(aq) ® HS-(aq)

·  then, since in base, add hydroxides to both sides to neutralize acid on one side:

Al(s) + 3 H2O(l) + 3 OH-(aq) ® Al(OH)3(s) + 3 H2O(aq)

simplified: Al(s) + 3 OH-(aq) ® Al(OH)3(s)

S(s) + H2O(aq) ® HS-(aq) + OH-(aq)

·  balance charge with e-:

Al(s) + 3 OH-(aq) ® Al(OH)3(s) + 3 e-

S(s) + H2O(aq) + 2 e- ® HS-(aq) + OH-(aq)

·  multiply half-reactions to balance numbers of electrons on opposite sides:

by 2x: 2 Al(s) + 6 OH-(aq) ® 2 Al(OH)3(s) + 6 e-

by 3x: 3 S(s) + 3 H2O(aq) + 6 e- ® 3 HS-(aq) + 3 OH-(aq)

·  add half-reactions and simplify, if necessary:

2 Al(s) + 3 S(s) + 3 H2O(aq) + 6 OH-(aq) ® 2 Al(OH)3(s) + 3 HS-(aq) + 3 OH-(aq)

simplify: 2 Al(s) + 3 S(s) + 3 H2O(aq) + 3 OH-(aq) ® 2 Al(OH)3(s) + 3 HS-(aq)

Redox Titration

·  KMnO4 example (dark purple)

MnO4-(aq) + 5 Fe2+(aq) + 8 H+(aq) ® Mn2+(aq) + 5 Fe3+(aq) + 4 H2O(l)

Add KMnO4 solution slowly to Fe2+ solution with stirring, slight persistence of purple colour is the end point (Fig. 6.9)

Disproportionation

·  one type of molecule goes in both redox directions in same reaction

eg. 2 H2O2 (l) ® 2 H2O (l) + O2 (g)

Ox. # (-1) (-2) (0)

Example 6.10

6.5 Colligative Properties of Solutions

·  solution properties that depend upon number of particles dissolved (i.e.- concentration) in a given solvent but not on the kind of solute

·  four colligative properties:

1.  vapour pressure reduction

2.  boiling point elevation

3.  freezing point depression

4.  osmotic pressure (related to 1.)

Changes in Vapour Pressure: Raoult’s Law

·  non-volatile solute decreases vapour pressure of volatile solvent above the solution

·  illustrated by experiment: 2 beakers with same volume of liquid, one pure solvent, one with non-volatile solute in same solvent

·  in closed chamber, volume decreases in pure solvent, increases in solution beaker

·  result of dynamic equilibria:

·  in pure solvent, evaporation and condensation rates equal

·  in solution, evaporation rate decreased due to interaction with solute, but condensation rate same, \decreased vapour pressure

·  hence net transfer of liquid to solution beaker

·  extent of vapour pressure lowering proportional to concentration, independent of which solute

·  quantitatively described by Raoult’s Law

PA = vapour pressure of soln

XA = mol fraction of the solvent

PAo = vapour pressure of pure solvent

·  example:

·  PH2Oo = 17.5 mm of Hg at 20°C. If glucose added so that XH2O = 0.8, then:

·  PH2O = 0.8 x 17.5 = 14 mm of Hg

·  similarly in Example 6.11 calculating vapour pressure of benzene with and without naphthalene in solution