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More on solutions…

More on solutions…. Dilution. Used for calculating how much concentrated acid should be used to make a more dilute acid solution M 1 V 1 = M 2 V 2 concentrated diluted mol (L) = mol (L) L L mol = mol . Dissociation.

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More on solutions…

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  1. More on solutions…

  2. Dilution • Used for calculating how much concentrated acid should be used to make a more dilute acid solution • M1V1 = M2V2 concentrated diluted • mol (L) = mol (L) L L mol = mol

  3. Dissociation • When ionic compounds dissolve in water, the ions are removed from the crystal lattice and dispersed throughout the solution. • This breaking up of the ions is known as “dissociation.” Cl- CO32- Ca2+ Na+ Ca2+ Na+ CO32- NaCl (aq) Cl- CaCO3 (aq) CO32- Na+ Cl- Ca2+

  4. Molarity of ions in solution • If a solute totally dissolves in water when forming a solution, the molarity of the ions in solution can be determined. • 1 M NaCl consists of 1 M Na+ ions and 1 M Cl- ions • 1 M CaCO3 consists of 1 M Ca2+ ions and 1 M CO3 2- ions • 1 M AlCl3 consists of 1 M Al3+ ions and 3 M Cl- ions • What would happen with 1 M Ca3(PO4)2 ?

  5. Dissociation Equations • also called Ionic Equations • shows how an ionic compound dissociates (breaks up) while in solution. • NaCl (s) Na+ (aq) + Cl- (aq) • CaCO3(s)  Ca2+(aq) + CO3 2-(aq) • K2SO4(s)  2 K+(aq) + SO4 2-(aq)

  6. Dissociation Equation Homework • Write the Dissociation Equations for the following. • Potassium fluoride • Magnesium iodide • Aluminum sulfate • Calcium hydroxide • Aluminum hydroxide • Hydrobromic acid • Sulfuric acid • Phosphoric acid

  7. Homework Answers H2O H2O H2O H2O H2O H2O H2O H2O • KF (s) K+(aq) + F-(aq) • MgI2(s) Mg+2(aq) + 2 I-(aq) • Al2(SO4)3(s) 2 Al+3(aq) + 3 SO4-2(aq) • Ca(OH)2(s) Ca+2(aq) + 2 OH-(aq) • Al(OH)3(s) Al+3(aq) + 3 OH-(aq) • HBr (s or aq) H+(aq) + Br-(aq) • H2SO4(s or aq) 2 H+(aq) + SO4-2(aq) • H3PO4(s or aq) 3 H+(aq) + PO4-3(aq)

  8. Electrolytes vs Nonelectrolytes • Electrolyte – compound that conducts electricity in aqueous solutions or in its molten state due to the formation of ions • Examples on next two slides • Nonelectrolyte - Does not separate into ions in its molten state or aqueous state; does not conduct • Ex. Large Organic Molecules and other covalent compounds • C6H12O6, C12H22O11, CCl4

  9. Strong Electrolytes • 100% dissociation in water • Separates into ions; 100% ionized • Types of strong electrolytes • Strong, non-organic acids (start with H) • Ex. HI, HBr, HCl, H2SO4, HNO3, HClO4 • Strong Group 1 and 2 bases (bases are ionic compounds with a metal and the hydroxide ion (OH-)) • Ex. NaOH, Ca(OH)2, KOH, Ba(OH)2 • Soluble salts (Ionic compounds that dissolve in water); Remember, ionic compounds are metal and non-metal or compounds with polyatomic ions • Ex: NaF, KBr, FeCl3, NH4NO3

  10. Weak Electrolytes • Do not dissociate completely • Do not separate into ions 100%; not fully ionized • Types of weak electrolytes • Organic acids (start with H and contain carbon) • Ex. HC2H3O2 (acetic acid) • Bases with metals not in group 1 or 2 • Ex. Fe(OH)2, Al(OH)3

  11. Colligative Properties • Properties which depend upon the number of solute particles in solution, but are independent of the nature of the solute particles • 1) Vapor Pressure Lowering • 2) Freezing Point Depression • 3) Boiling Point Elevation

  12. Vapor Pressure Lowering • Vapor pressure is a measure of the tendency of molecules to escape from the surface of a liquid or solid • Vapor pressure is decreased proportionally to the number of solute particles in a solution

  13. Freezing Point Depression • When a substance freezes, the particles of the solid take on an orderly pattern; the presence of the solute particles disrupts this pattern. • The magnitude of freezing point depression is proportional to the number of solute particles in solution

  14. Boiling Point Elevation • When a substance boils, the particles of the liquid must attain enough energy to overcome intermolecular forces and escape into the gas form; the presence of solute ions makes it harder for these molecules to escape. • The magnitude of boiling point elevation is proportional to the number of solute particles in solution.

  15. Calculations for FPD and BPE • Formula for Freezing Point Depression ΔTf = Kfim • Formula for Boiling Point Elevation ΔTb = Kbim ΔT is change in (boiling point or freezing point) temperature Kfand Kb are constants for the solvent iis the dissociation factor for the solute mis the molality for the solution

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