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1. SolutionsChapter 14 Hein and Arena Eugene PasserChemistry DepartmentBronx Community College © John Wiley and Sons, Inc Version 1.1

2. General Properties of Solutions

3. A solution is a system in which one or more solutes are homogeneously dissolved in a solvent. • A solute is a compound that is dissolved in a solvent. • The solvent is the dissolving agent or the most abundant component in the solution.

4. Solubility(…and the dissolution process)

5. Solubility describes the amount of a substance that will dissolve in a specified amount of solvent.

6. Dissolution of sodium chloride in water. Na+ and Cl- ions hydrated by H2O molecules. 15.2

7. Terms that describe the extent of solubility of a solute in a solvent: • very soluble ( greatest degree of dissolution) • soluble • moderately soluble • slightly soluble • Insoluble (least degree of dissolution)

8. miscible: liquids that are soluble in each other. • Terms that describe the solubility of liquids: methyl alcohol and water • immiscible: liquids that are insoluble in each other. oil and water

9. Factors Related to Solubility

10. The Nature of theSolute and Solvent

11. The general rule for predicting solubility is “like dissolves like”.

12. Polar compounds tend to be more soluble in polar solvents than nonpolar solvents. • Nonpolar compounds tend to be more soluble in nonpolar solvents than in polar solvents.

13. Rate ofDissolving Solids

14. Factors that affect solubility are: • Particle size; smaller particles dissolve faster due to greater surface area. • Temperature; higher temperatures generally increase the dissolution rate. • Interactions between solute and solvent; mixing the solution increases contact between solvent molecules and the solute surface.

15. Solutions:A Reaction Medium

16. sodium chloride reacts with silver nitrate when dissolved in water NaCl(aq) + AgNO3(aq) → AgCl(s) +NaNO3(aq)

17. Concentrationof Solutions

18. A dilute solution contains a relatively small amount of dissolved solute. • A concentrated solution contains a relatively large amount of dissolved solute.

19. Mass Percent Solution (wt/wt)

20. Mass percent expresses the concentration of a solution as the percent of solute in a given mass of solution.

21. What is the mass percent of sodium hydroxide in a solution that is made by dissolving 8.00 g NaOH in 50.0 g H2O? grams of solute (NaOH) = 8.0 g grams of solvent (H2O) = 50.0 g

22. What masses of potassium chloride and water are needed to make 250. g of 5.00% (by wt/wt) solution? The percent expresses the mass of the solute. Dissolving 12.5 g KCl in 238 g H2O gives a 5.00% KCl solution. 250.g = total mass of solution 250.g – 12.5 g = 238 g H2O

23. A 34.0% (by wt/wt) sulfuric-acid solution has a density of 1.25 g/mL. How many grams of H2SO4 are contained in 1.00 L of this solution? Grams of solution are determined from the solution density. Step 1. Determine grams of solution. 1.00 L = 1.00 x 103 mL

24. A 34.0% (by wt/wt) sulfuric-acid solution has a density of 1.25 g/mL. How many grams of H2SO4 are contained in 1.00 L of this solution? Solve the mass percent equation for grams of solute. [(g H2SO4)/(g solution)] x 100 = mass % g H2SO4 = [34.0 % (g solution)]/100 g H2SO4 = [34.0%(1250g solution)]/100 = 425gH2SO4

25. Mass/Volume Percent(wt/v)

26. Mass /volume percent expresses the concentration of solute as g solute per mlsolution.

27. A 3.0% (by wt/vol) H2O2 solution is commonly used as a topical antiseptic to prevent infection. What volume of this solution will contain 10. g of H2O2? Solve the mass/volume equation for mL of solution.

28. Volume Percent (vol./vol.)

29. The volume percent is the volume of a liquid per total volume of solution.

30. Molarity

31. Molarity of a solution is the number of moles of solute per liter of solution.

32. What is the molarity of a solution containing 1.4 mol of acetic acid (HC2H3O2) in 250. ml of solution? It is necessary to convert 250. mL to L since molarity = mol/L.

33. How many grams of potassium hydroxide are required to prepare 600. mL of 0.450 M KOH solution? The data are: The calculation is:

34. Calculate the number of moles of nitric acid in 325 mL of 16 MHNO3? Use the equation: moles = liters x M Substitute the data given in the problem and solve:

35. What volume of 0.250 M solution can be prepared from 16.0 g of potassium carbonate? The data are: The conversion is:

36. How many mL of 2.00 M HCL will react with 28.0 g NaOH? Step 1 Write and balance the equation for the reaction. HCl(aq) + NaOH(aq)→ NaCl(aq) + H2O(aq) Step 2 Find the number of moles of NaOH in 28.0 g NaOH.

37. How many mL of 2.00 M HCL will react with 28.0 g NaOH? Step 3 Solve for moles and volume of HCl.

38. Note: volumes are not always additive.

39. Dilution and Concentration Problems

40. If a solution is diluted by adding pure solvent: • the volume of the solution increases. • the number of moles of solute remain the same. • the ratio of mol./L gets smaller. • the molarity decreases.

41. If a solution is concentrated by removing the solvent via evaporation: • the volume of the solution decreases. • the number of moles of solute remain the same. • the ratio of mol./L gets larger. • the molarity increases.

42. Calculate the molarity of a sodium hydroxide solution that is prepared by mixing 100. mL of 0.20 M NaOH with 150. mL of water. Assume volumes are additive. Step 1 Calculate the moles of NaOH in the original solution. 0.020 mol is the number of moles of NaOH in the original 100. mL of the 0.20 M NaOH solution.

43. Calculate the molarity of a sodium hydroxide solution that is prepared by mixing 100. mL of 0.20 M NaOH with 150. mL of water. Assume volumes are additive. Step 2 Solve for the new molarity. solution volume = 100. mL + 150. mL = 250. mL 0.020 mol NaOH = 0.080 M NaOH 0.250 L

44. The End