Chapter 13

13.1 Ionic Compounds in Aqueous Solution Aqueous solution = A solution in which water is the solvent (aq). Example: A solution of water (the solvent) and NaCl (the solute) Aqueous solutions can be electrolytes or non electrolytes. Electrolytes and non-electrolytes are solutes of ionic salt solutions. Do electrolytes conduct electricity? Yes Are non-electrolytes conductors? No Are all electrolytes conductors? Yes Are all conductors electrolytes? No

Conductors vs. Nonconductors Pure Substance Mixtures Elements Compound Alloys Electrolytic Sol’n All metals: Cu, Ag, Fe All ionic in liquid state: NaBr (I), KNO3 (I) Stainless steel, Sterling silver Water solutions of NaBr, HCl, NH3 (aqueous solution) Pure Substance Mixtures Elements Compounds Non-electrolytic Sol’n All nonmetals: I2, P4 All covalent compounds in liquid state: HBr (I), Al2Cl6 (I), All solid compounds: Sucrose, NaBr (s), AlBr (s) Water solutions of sucrose, isopropyl alcohol, ethyl alcohol, glycerin (aqueous solutions)

Theory of Ionization Theory of Ionization - Some water solutions conduct electricity. These solutions are called charges called electrolytes. Strong electrolytes: NaCl(s) + H2O(l)  Na+1(aq) + Cl-1(aq) HCl(g) + H2O(l)  H+1(aq) + Cl-1(aq) Weak electrolyte: HC2H3O2(l) + H2O(l)  H+1(aq) + C2H3O2-1(aq)

Theory of Ionization In 1887, Svante Arrhenius (Sweden) proposed the theory of ionization. Some substances break into smaller substances with charges. He based his ideas on observations of changes in freezing and boiling points with different molal concentrations Note: Some reactions get a single arrow () and some get a double arrow () - equilibrium. Arrhenius proposed that when some chemicals are dissolved in water, they produce particles with charges.

Ionization ionic compounds dissociate NaCl(s) + H2O -----> Na+ (aq) + Cl- (aq) MgCl2 (s) + H2O -----> Mg+2(aq)+ 2C l- (aq) acids ionize (covalent dissociation) HCl(g) + H2O -----> H+(aq) + Cl-(aq) H2SO4(g) + H2O -----> 2H+(aq) + SO42-(aq) Substances that are not ionic, electrolytes, or acids do not dissociate/ionize.

Dissolving Ionic Compounds Hydration: Water molecules surround each ion in solution; the entire ions from the crystal dissolves and hydrated ions become uniformly dissolved.

Heat of Hydration energy released when ions are surrounded by water molecules. The # of water molecules used depends on the size and charge of the ion. ↑ Heat released (more negative) as the size of the ion ↓ Li+1 -523 kJ/mole vs Na+1 -418 kJ/mole ↑ Heat released (more negative) as the charge of the ion ↑ Na+1 -418 kJ/mole vs Mg+2 -1949 kJ/mole Li and Mg are close to the same size, so charge is more important

Heat of Hydration Three interactions contribute to the heats of solution (by forming a solution) Step 1 dissociation - separation of ions (they already exist, H2O separates them) — solute-solute connections break apart = energy absorbed Step 2 solvent — solvent-solvent connections break apart = energy absorbed Step 3 hydration – Solute-solvent - particles are surrounded by water = energy released

Heat of Hydration Endothermic Energy Level Diagram Hydrate E Solution Time Hydrate Solution Solute

Heat of Hydration Exothermic Energy Level Diagram Hydrate 2 E 1 Solute 3 1 Time Hydrate Solute Solution If step #1 + step #2 are less than step #3, than the overall reaction is exothermic

Dissociation The separation of ions that occurs when an ionic compound dissolves. Ex. A 1.0 M solution of sodium chloride contains: 1 mol of Na+ ions and 1 mol of Cl- ions. NaCl(s) ---------> Na+(aq) + Cl-(aq) 1 mol 1 mol 1 mol Total of 2 moles of ions

Dissociation A 1.0 M solution of calcium chloride contains: CaCl2(s) ----> Ca2+(aq) + 2Cl-(aq) 1 mol of Ca2+ ions and 2 mol of Cl- ions a total of 3 mole of ions.

Dissociation (a) Dissolve Al2(SO4)3 in water. (b) How many moles of aluminum ions and sulfate ions are produced by dissolving 1 mol of Al2(SO4)3. (c) What is the total number of moles of ions produced by dissolving 1 mol of Al2(SO4)3? (a) Al2(SO4)3 -----> 2Al3+(aq) + 3SO4-2(aq). (b) 1 mol ----> 2 mol + 3 mol. (c) 2 mol Al3+ + 3mol SO42- = 5 mol of solute ions

Solubility Equilibria No ionic compound has infinite solubility. No ionic compound has zero solubililty. Rough rules of solubililty (using the solubility tables): If more than 1 gram per 100g H20 before saturation = soluble If .1 gram to 1 gram per 100g H20 = slightly soluble If less than .1 gram per 100g H20 = insoluble

Solubility Equilibria Very slightly soluble ionic compounds – when placed in water, an equilibrium is established between the solid compound and its ions in solution: Example: AgCl(s) Ag+(aq) + Cl-(aq) Fe(OH)3(s) Fe3+(aq) + 3OH-(aq) Ag2S(s) 2Ag+(aq + S2-(aq)

Solubility Equilibria Precipitation Reactions = Soluble compounds form insoluble products. Type of rxn: double replacement - remember – reactants are soluble in water Ex1: Silver nitrate + magnesium chloride KCl(aq) + AgNO3(aq) ---> KNO3(aq) + AgCl(s)

Solubility Equilibria Net ionic equations – double replacement reactions and other reactions of ions in aqueous solutions are represented as ‘net ionic equations.’ Steps: 1. write an equation & show soluble compounds as dissociated ions 2. write a net ionic equation – only those compounds and ions that undergo a chemical change in a reaction in an aqueous sol’n and does not include spectator ions (ions found on the reactants and products side).

Net ionic equations Ex1: Molecular Total Ionic net ionic equation KCl(aq) + AgNO3(aq) ---> KNO3(aq) + AgCl(s) Total Ionic net ionic equation Spectator Ioncs

Net Ionic Equations Calcium chloride + aluminum carbonate Molecular Total Ionic Net Ionic Spectator Ions

13.2 Molecular Electrolytes Molecular solutes can form electrolytic solutions if they are highly polar. Ionization versus dissociation Dissociation = The separation of ions that occurs when the ionic compound dissolves Ionization = The formation of ions that occurs when a polar covalent compound dissolves in water (water rips apart molecules and turns them into ions

Molecular Electrolytes Ionization example = HCl in water H2O + HCl -----> H3O+ + Cl- When a hydrogen chloride molecule ionizes in water, its hydrogen ion bonds covalently to a water molecule. A hydronium ion and a chloride ion are formed

Hydronium Ion: H3O+ The H+ ion attracts other molecules or ions so strongly that it does not normally exist, so the H+ ion becomes covalently bonded to oxygen.

Substances which form electrolytic solutions 1. Acids HX ex: HCl, HNO3 polar 2. Bases MOH ex: NaOH ionic 3. Salts MX ex: NaCl, KBr, CaCO3 ionic H = Hydrogen M = Metal OH = Hydroxide X = NM or P-ion Why? The solute pulls ions into solutions

Which of the following form electrolytic solutions? MgBr2 C8H18 KOH C12H22O11 HNO3

Strong vs. weak electrolytes Some compounds ionize / dissociate completely, while others don’t. Strong electrolyte – a compound that when dissolved/ionized, yields 100% ions. Distinguishing factor of strong electrolytes – to whatever extent they dissolve in water, they yield only ions: HCl, HBr and HI are 100% ionized in dilute aqueous solutions.

Strong vs. weak electrolytes Weak electrolyte – a solute that yields a relatively low concentration of ions in an aqueous solution. HF(aq) + H2O(l)  H3O+ (aq) + F-(aq) In an aqueous solution, the majority of HF molecules are present as dissolved HF molecules.

Strong vs. weak electrolytes In general, the extent to which a solute ionizes in solution depends on the bonds within the molecules of the solute and the strength of attraction to solvent molecules. Note: If the strength of bonds in solute molecules < the attractive forces of the water dipoles, then the covalent bonds break and the molecule separates into ions.

13.3 Properties of Electrolyte Solutions Conductivity of Solutions To compare the conductivities of strong and weak electrolytes, the conductivities of solutions of equal concentration must be compared. Ionization of pure water H2O(l) + H2O(l)  H3O+ (aq) + OH-(aq) So why does water that comes out of the tap conduct electricity? It contains a high enough concentration of dissolved ions to make it a better conductor than pure water.

Colligative Properties of Electrolytic Solutions Properties that depend on the concentration of the solute particles. Freezing point and boiling point are colligative properties. Freezing point depression – the difference between the freezing points of a pure solvent and a nonelectrolyte solution in it. Solutions that conduct electricity contain electrolytes.

Colligative Properties of Electrolytic Solutions Ionic compounds dissociate: NaCl(s) + H2O(l) yields Na+(aq) + Cl-(aq) MgCl2(s) + H2O(l) yields Mg+2(aq) + 2Cl-(aq) Acids ionize (dissociation of a covalent compound): HCl(g) + H2O(l) yields H+(aq) + Cl-(aq) H2SO4(l) + H2O(l) yields 2H+(aq) + SO4-2(aq) Substances that are not acids, bases, and salts do not dissociate/ionize. When solutes dissolve in liquids, they lower the freezing point.

Freezing Point Depression Two factors affect the degree of change in the temperature: the amount of the solute and the nature of the solvent. ∆tf = kf (m)(x) x = # of ions produced when the solute dissolves kf = constant (kf water = -1.86 oC/m m = molality (moles solute) Kg solvent

Why does freezing point depression occur? O Na+ O H H ……… O …. . H H   H H Cl- The solute (NaCl) interferes with crystal formation. (ex: antifreeze)   As the number of solute particles increase, the freezing point decreases.

Freezing Point Depression Ex1: Calculate the freezing point of 10.00 grams of NaCl in 200.0 grams of water. ∆tf = kf (m)(x) Grams  moles 10.00 g NaCl x 1 mole NaCl = .1709 moles NaCl 58.5 g NaCl Molality m = .1709 moles = m = .8547 m .2000 kg Change in temperature ∆tf = (-1.86 oC/m)(.8547 m)(2) = -3.179 oC New Freezing point ∆tf = tf - ti  -3.179 oC = x – 0 oC  x = -3.179 oC

Boiling Point Elevation Boiling point elevation – when solutes dissolve in liquids, they raise the boiling points. Same concept as freezing point depression except boiling point increases. kb water = 0.512 oC/m Why does boiling point elevation occur? The solute takes up space on the surface of a liquid. This decreases the ability of the liquid to evaporate. Thus, the vapor pressure decreases. Boiling occurs when the atmospheric pressure equals the vapor pressure. So, an increase in energy is needed to increase the vapor pressure to reach the atmospheric pressure.

Boiling Point Elevation = solvent = solute “A” “B” Which would produce more vapor? A Which would have a higher vapor pressure? A Which would take less energy to raise the vapor pressure to atmospheric pressure? A Which would have a higher boiling point? B

Boiling Point Elevation Ex1: Calculate the boiling point of a solution of 10.00 grams of NaCl in 200.0 grams of water. ∆tb = kb(m)(x) Grams  moles 10.00 g NaCl x 1 mole NaCl = .1709 moles NaCl 58.5 g NaCl Molality m = .1709 moles = .8547 m .2000 kg Change in temperature  ∆tb = (.512 oC/m)(.8547 m)(2) = .8752 oC New Temperature ∆tb = tf - ti  .8752 oC = x – 100 oC = 100.8752 oC

Ion Pairing When experiments are done regarding freezing point depression and boiling point elevation, the actual answers are different than the theoretical answers Example: a solution of NaCl in water:Sodium Chloride can dissociate at a rate of 100% if the concentration of the solution is very low. With increased concentration, ions may come in contact with each other and rejoin resulting in less than 100% dissociation. Only at low, low concentrations do solutions have their “x” factor approach the theoretical value. Concentration (molality) Actual change in the freezing point Theoretical change in the freezing point % dissociation .1 - 0.346 - 0.372 93 % .01 - 0.0361 - 0.0372 97 % .001 - 0.00366 - 0.00372 98 % .0001 - 0.000372 100 %