Recall:Solutions

1. Recall:Solutions • A solution is a homogeneous mixture. • A solution is composed of a solute dissolved in a solvent. • Solutions exist in all three physical states: Chapter 14

2. Add:Gases in Solution • Temperature effects the solubility of gases. • The higher the temperature, the lower the solubility of a gas in solution. • An example is carbon dioxide in soda: • Less CO2 escapes when you open a cold soda than when you open the soda warm Chapter 14

3. 4.1The Dissolution Process • When a soluble crystal is placed into a solvent, it begins to dissolve. • When a sugar crystal is placed in water, the water molecules attack the crystal and begin pulling part of it away and into solution. • The sugar molecules are held within a cluster of water molecules called a solvent cage. Chapter 14

4. 4.1Dissolving of Ionic Compounds • When a sodium chloride crystal is place in water, the water molecules attack the edge of the crystal. • In an ionic compound, the water molecules pull individual ions off of the crystal. • The anions are surrounded by the positively charged hydrogens on water. • The cations are surrounded by the negatively charged oxygen on water. Chapter 14

5. 4.5 Calculations with Reactions in Solution 1. How to describe the solution quantitatively 2. How to prepare the solution from a pure substance 3. How to dilute a solution from a more concentrated one 4. How to use stoichiometry with solutions

6. moles of solute M = molarity = liters of solution volume of KI solution moles KI grams KI 1 L 2.80 mol KI 166 g KI x x x 1000 mL 1 L soln 1 mol KI 4.5 Solution Stoichiometry The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. What mass of KI is required to make 500. mL of a 2.80 M KI solution? M KI M KI 500. mL = 232 g KI

7. Preparing a Solution of Known Concentration

8. Dilution Add Solvent Moles of solute before dilution (i) Moles of solute after dilution (f) = = MfVf MiVi Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution.

9. 0.200 M x 0.0600 L = MfVf 4.00 M Vi = Mi How would you prepare 60.0 mL of 0.200 M HNO3 from a stock solution of 4.00 M HNO3? MiVi = MfVf Mi = 4.00 M Vi = ? L Mf = 0.200 M Vf = 0.0600 L = 0.00300 L = 3.00 mL Dilute 3.00 mL of acid with water to a total volume of 60.0 mL.

10. A 25.0ml sample of 0.0040 M solution of NaBr in water is treated with a 0.0025M aqueous solution of AgNO3. What volume of the AgNO3 solution is required to react completely with the NaBr solution?

11. 4.6 Techniques: Gravimetric Analysis • Dissolve unknown substance in water • React unknown with known substance to form a precipitate • Filter and dry precipitate • Weigh precipitate • Use chemical formula and mass of precipitate to determine amount of unknown ion

12. 187.77 g AgBr 0.100 mol AlBr3 3 mol AgBr 37.5 mL soln × × × 1 mol AgBr 1000 mL soln 1 mol AlBr3 Gravimetric Analysis Problem • What mass of silver bromide is produced from the reaction of 37.5 mL of 0.100 M aluminum bromide with excess silver nitrate solution? AlBr3(aq) + 3 AgNO3(aq) → 3 AgBr(s) + Al(NO3)3(aq) • We want g AgBr, we have volume of AlBr3 = 2.11 g AgBr Chapter 14

13. 4.7-4.8 Techniques: Titrations In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point – the point at which the reaction is complete Slowly add base to unknown acid UNTIL the indicator changes color Indicator – substance that changes color at (or near) the equivalence point

14. H2SO4 + 2NaOH 2H2O + Na2SO4 5Fe2+ + MnO4- + 8H+ Mn2+ + 5Fe3+ + 4H2O Titrations can be used in the analysis of Acid-base reactions Redox reactions

15. H2SO4 + 2NaOH 2H2O + Na2SO4 M M rxn volume acid moles red moles base volume base base acid coef. 4.50 mol H2SO4 2 mol NaOH 1000 ml soln x x x 1000 mL soln 1 mol H2SO4 1.420 mol NaOH What volume of a 1.420 M NaOH solution is required to titrate 25.00 mL of a 4.50 M H2SO4 solution? WRITE THE CHEMICAL EQUATION! 25.00 mL = 158 mL

16. M V rxn volume red moles red moles oxid M oxid oxid red coef. 5 mol Fe2+ 1 0.1327 mol KMnO4 x x 5Fe2+ + MnO4- + 8H+ Mn2+ + 5Fe3+ + 4H2O x 1 mol KMnO4 0.02500 L Fe2+ 1 L 16.42 mL of 0.1327 M KMnO4 solution is needed to oxidize 25.00 mL of an acidic FeSO4 solution. What is the molarity of the iron solution? WRITE THE CHEMICAL EQUATION! 16.42 mL = 0.01642 L 25.00 mL = 0.02500 L 0.01642 L = 0.4358 M End Solution Stoichiometry

17. What you need to know: • Definitions of solution, solute, solvent, electrolyte, non-electrolyte, hydration. • Types of aqueous reactions • Precipitation (selected solubility rules/ exceptions) • Neutralization rxns • Reduction – oxidation reactions • Combination, decomposition, combustion, displacement

18. What you need to know: • How to find oxidation numbers • How to identify substance oxidized, substance reduced, reducing agent, oxidizing agent • How to write half reactions • Calculate molarity and dilutions • Understand theoretical use of gravimetric analysis and titrations and perform accompanying calculations

19. What you need to know: • How to find oxidation numbers • Interpret activity series • Calculate molarity and dilutions • Understand theoretical use of gravimetric analysis and titrations and perform accompanying calculations