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20 B Week IV Chapters 11

• Chapter 10, 11( except 11.4 and and 11.6 -7) • Intermolecular potentials. ion-dipole, ion-ion • Solutions. Interactions in solution, Boiling Point Elevation, Freezing Point Depression and Osmotic Pressure, Electrolyte solutions

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20 B Week IV Chapters 11

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  1. • Chapter 10, 11( except 11.4 and and 11.6 -7) • Intermolecular potentials. ion-dipole, ion-ion • Solutions. Interactions in solution, Boiling Point Elevation, Freezing Point Depression and Osmotic Pressure, Electrolyte solutions • Dissolution reactions(rxns) and Arrhenius type Acid/Base rxns Midterm Friday: Chaps 9(no 9.7), 10, 11.-11.3 One side of 1 page notes(must be hand written), closed book Review Session Today @ 2-3 pm, in FRANZ 1178 20 B Week IV Chapters 11

  2. e In Solutions, for example when NaCl(s) is dissolved in H2O(l). + H2O Dissolution of a polar solid by a polar solid by a polar liquid A non-polar liquid e.g., benzene, would not dissolve NaCl? NaCl(s) + H2O(l) Na+(aq) +Cl-(aq) (aq) means an aqueous solution, where water is the solvent, major component. The solute is NaCl, which is dissolved, minor component Water molecules solvates the ions the Cation (Na+) and the Anion (Cl-). The forces at play here are Ion dipole forces Fig. 10-6, p. 450

  3. -2∂ First Solvation Shell of the Solvent -2∂ +∂ -2∂ + -2∂ +∂ +∂ Solvated Na+ Fig. 10-6a, p. 450

  4. Dissolution of K2SO4 in water 2nd solvation shell Fig. 11-3, p. 480

  5. In a solution of solvent A and solute B, the important thermodynamic variables are V, T, molar Concentration of [A] and [B] and the relative composition X The Molarity, moles of A or B/ Liters of Solution [A]=nA/Vsol and [B]=nB/Vsol in units of mols L-1 The composition in Mole Fraction: XA= nA/(nA + nB) and XB= nB/(nA + nB) XA + XB =1 and XA=1 -XB Because Volume depends on the Temperature, the Molarity is not always the Best variable, the Molality {B} = Moles of Solute/ kg of Solvent mol kg-1 {B}= nB//kg of solvent, since the density of water is defined as kg/L, Therefore, for ideal aqueous solutions the Molality is, in effect, the moles of solute/Liters of solvent.

  6. The dissolution reaction with water as a solvent: A(s)A(aq) A is a molecular solid and the molecular units, or monomers, do not dissociate in solution. Like fructose with lots of OH ( hydroxyl groups) for forming H-bonds and which makes the monomer more stable inaqsoln Electro Static potential

  7. The Hydrogen bonds between the water molecules of the solvents makes the molecule more stable in solution than in the solid H-bonds Fructose C6H12O6 Hydrated Fructose C6H12O6

  8. The dissolution of salts such as NaCl(s), NaOH(s)(basic in soln, sodium hydroxide) and NH4Cl(acidic in soln, Ammonium Chloride ) AB(s)  A+(aq) + B+(aq) In some case like NaCl no reaction occurs with the solvent just hydration + H2O

  9. Lone pair + NH4Cl(s)  NH+4(aq) + Cl-(aq) NH+4(aq) + H2O(l)H3O+((aq) + NH3((aq) Proton Transfer Notice that the Number of Elements and the Amount of Charge are balanced On both sides reaction the rxn Such a reaction has the correct Stoichiometry and is said to be balanced

  10. NH+4(aq) + H2O(l)  H3O+(aq) + NH3(aq) By definition, the Ammonium ion NH+4 is an Arrhenius acid (Svante Arrhenius) since it increased the concentration(molarity) of the H3O+(aq), the hydronium ion in solution by reacting with the solvent, H2O in the case +

  11. In the general an Acid AH, the solute, in the Arrhenius sense will react with the solvent H2O, in this case, it could have been AH(aq) + H2O(l)  H3O+(aq) + A-(aq) A strong acid such as HCl will completely dissociate in solution HCl(g) + H2O(l)  H3O+(aq) + Cl-(aq) The reaction goes to completion: all the HCl molecules produce H3O+(aq)

  12. For a relatively weak Acids such as HF, the reaction does not go to completion HF(g) + H2O(l)  H3O+(aq) + F-(aq) And there is still lots of HF in solution solvated by H2O molecules, the rxn therefore goes to equilibrium HF forms a strong H-bonded network

  13. Types of Acids Table 11-1, p. 483

  14. The Neutralization reaction H2O(l) + H2O(l)  H3O+(aq) + OH-(aq) where H2O(l) amphoteric acts both as an acid as well as a base ths reaction goes to equilibrium  H+ +

  15. NaOH(s)  Na+(aq) + OH-(aq) Where OH-, the hydroxide ion, is a strong base In acid base reactions for example: HCl(g) + H2O(l)  H3O+(aq) + Cl-(aq) HCl is the strong Acid and H2O is the Base HCl + NaOH  H2O + NaCl  Na+(aq) + Cl-(aq) Which is just salt water

  16. Electrolyte Solutions can Carry current Dissolution of K2SO4(s)  2K+(aq) + SO-4(aq) Fig. 11-3, p. 480

  17. The Solubility limit of Potassium Sulfate in aqueous soln is 120 gL-1 at 25 °C Molecules that dissolve in solution to produce ions are called Electrolytes. Some compounds have rather limited solubility in water e.g., example of BaSO4(s) can dissolve to 2.5 mgL-1 BaSO4(s)  Ba++(aq) + SO2-4(aq) The equilibrium of this reaction is therefore on the side of Ba++(aq) + SO2-4(aq)  BaSO4(s) And the solid would precipitate out of soln

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