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Solutions

Solutions. Chapter 14. Common Solutions. Chemical solutions encountered in everyday life: air coffee tap water gasoline shampoo cough syrup orange soda Gatorade. Solutions. . . . the components of a mixture are uniformly intermingled (the mixture is homogeneous ). A Solute.

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Solutions

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  1. Solutions • Chapter 14

  2. Common Solutions • Chemical solutions encountered in everyday life: • air coffee • tap water gasoline • shampoo cough syrup • orange soda Gatorade

  3. Solutions • . . . the components of a mixture are uniformly intermingled (the mixture is homogeneous).

  4. A Solute • dissolves in water (or other “solvent”) • changes phase (if different from the solvent) • is present in lesser amount (if the same phase as the solvent)

  5. A Solvent • retains its phase(if different from the solute) • is present in greater amount (if the same phase as the solute)

  6. Figure 14.1: When solid sodium chloride dissolves, the ions are dispersed randomly throughout the solution

  7. Aqueous Solutions • Aqueous solutions are solutions in which water is the solvent. • Aqueous solutions are the most common type of solution.

  8. Some Properties of Water • Water is able to dissolve so many substances because: • Water is “bent” or V-shaped. • The O-H bonds are covalent. • Water is a polar molecule. • Hydration occurs when salts dissolve in water.

  9. Water is a polar molecule because it is a bent molecule. The hydrogen end is + while the oxygen end is -, Delta () is a partial charge--less than 1.

  10. Polar water molecules interact with the positive and negative ions of a salt, assisting in the dissolving process. This process is called hydration.

  11. Solubility • The general rule for solubility is: • “Like dissolves like.” • Polar water molecules can dissolve other polar molecules such as alcohol and, also, ionic substances such as NaCl. • Nonpolar molecules can dissolve other nonpolar molecules but not polar or ionic substances. Gasoline can dissolve grease.

  12. Miscibility • Miscible -- two substances that will mix together in any proportion to make a solution. Alcohol and water are miscible because they are both polar and form hydrogen bonds. • Immiscible -- two substances that will not dissolve in each other. Oil and vinegar are immiscible because oil is nonpolar and vinegar is polar.

  13. Figure 14.6: An oil layer floating on water

  14. Structure & Solubility • Like dissolves like. • Hydrophobic --water-fearing. Fat soluble vitamins such as A, D, E, & K. • Hydrophilic --water-loving. Water soluble vitamins such as B & C. • Hypervitaminosis--excessive buildup of vitamins A, D, E, & K in the body.

  15. Solubility • How does the rule “Like dissolves like.” apply to cleaning paint brushes used for latex paint as opposed to those used with oil-based paint?

  16. Common Terms of Solution Concentration • Saturated - when a solution contains as much solute as will dissolve at that temperature. • Unsaturated - when a solution has not reached its limit of solubility. • Supersaturated - when a solution has more of the solute dissolved than it normally would --very unstable.

  17. Common Terms of Solution Concentration • Stock - routinely used solutions prepared in concentrated form. • Concentrated - relatively large ratio of solute to solvent. (5.0 M NaCl) • Dilute - relatively small ratio of solute to solvent. (0.01 M NaCl)

  18. Qualitative -- general and relative. dilute concentrated Quantitative -- mathematically defined. Mass % -- biology Molarity -- most solutions in chemistry Normality -- chemical titrations Molality -- molar mass, freezing point depressions, & boiling point elevations Solution Composition

  19. Mass % • Mass (weight) percent =

  20. Mass % Calculations • If 1.00 g of ethanol is added to 100.0 g of water, what is the mass % of the ethanol?

  21. Mass % Calculations • Cow’s milk typically contains 4.5 % by mass of the sugar lactose, C12H22O11. Calculate the mass of lactose present in 175 g of milk. • Mass % = • mass of solute = (mass %)(mass of solution)/(100%) • mass of solute = (4.5 % lactose)(175 g)/(100%) • mass of solute = 7.9 g lactose

  22. Molarity • Molarity (M) = moles of solute per volume of solution in liters:

  23. Molarity Calculations • If 1.00 g of ethanol is dissolved in enough water to make 101 mL of solution, what is the molarity of the solution?

  24. Molarity Calculations • Calculate the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution. • (11.5g NaOH/1.50L)(1 mol NaOH/40.00g NaOH) • = 0.192 M NaOH

  25. Molarity Calculations • Calculate the molarity of a solution prepared by dissolving 1.56 g of gaseous HCl into enough water to make 26.8 mL of solution. • (1.56g HCl/26.8mL)(1mol/36.5g)(1000mL/1L) = 1.59 M HCl solution

  26. Molarity Calculations • How many moles of nitrate ions are present in 25.00 mL of a 0.75 M Co(NO3)2 solution? • (25.00mL)(1L/1000mL)(0.75mol Co(NO3)2/1L) • (2 mol NO3-/1 mol Co(NO3)2) = 3.8 x 10-2 mol NO3-

  27. Molarity Calculations • Typical blood serum is about 0.14M NaCl. What volume of blood contains 1.0 mg of NaCl? • (1.0mg NaCl)(1g/1000mg)(1mol/58.45g)(1L/0.14mol) = 1.2 x 10-4 L blood serum

  28. Molarity Calculations • A chemist needs 1.00 L of an aqueous 0.200 M K2Cr2O7 solution. How much solid potassium dichromate must be massed to make this solution? • (1.00L)(0.200mol K2Cr2O7/L)(294.2g/1mol) = 58.8 g K2Cr2O7

  29. Standard Solution • A standard solution is a solution whose concentration is accurately known. • A volumetric flask is used to make a standard solution.

  30. Steps involved in making a standard solution.

  31. Dilution of Stock Solutions • When diluting stock solutions, the moles of solute after dilution must equal the moles of solute before dilution. • moles of solute before dilution = moles of solute after dilution • Stock solutions are diluted using either a measuring or a delivery pipet and a volumetric flask.

  32. Steps to dilute a stock solution.

  33. Dilution Calculations • What volume of 16 M sulfuric acid must be used to prepare 1.5 L of a 0.10 M H2SO4 solution? • (0.10mol H2SO4/1L)(1.5L)(1L/16mol)(1000mL/1L) • = 9.4 mL conc H2SO4

  34. Dilution Calculations • What volume of 12 M HCl must be taken to prepare 0.75 L of 0.25 M HCl? • (0.75L)(0.25mol HCl/L)(1L/12mol)(1000mL/1L) = 16 mL conc. HCl

  35. Steps For Solving Solution Stoichiometry Problems • 1. Write the balanced reaction. • 2. Calculate moles of reactants. • 3. Determine limiting reactant. • 4. Calculate moles of required reactant/product. • 5. Convert to grams or volume, as required.

  36. Precipitation Calculations • When aqueous solutions of Na2SO4 & Pb(NO3)2 are mixed, PbSO4 precipitates. Calculate the mass of PbSO4 formed when 1.25 L of 0.0500 M Pb(NO3)2 & 2.00 L of 0.0250 M Na2SO4 are mixed. • 1. Pb(NO3)2(aq) + Na2SO4(aq) ----> PbSO4(s) + 2NaNO3(aq) • 2. (1.25L)(0.0500mol Pb(NO3)2/L) = 0.0625 mol Pb(NO3)2 • (2.00L)(0.0250mol Na2SO4/L) = 0.0500 mol Na2SO4

  37. Precipitation CalculationsContinued • 3. (0.0625mol Pb(NO3)2)(1mol Na2SO4/1mol Pb(NO3)2 = • 0.0625 mol Na2SO4 • Na2SO4is the limiting reactant. • 4. (0.0500mol Na2SO4)(1mol PbSO4/1mol Na2SO4) • (303.3g/1mol) = 15.2 g PbSO4

  38. Neutralization Reactions • Acids and bases react to neutralize each other and form a salt and water. This type of reaction is called a neutralization reaction. • HCl(aq) +NaOH(aq) ----> NaCl(aq) + HOH(l)

  39. Acid-Base Calculations • What volume of a 0.100 M HCl solution is needed to neutralize 25.0 mL of 0.350 M NaOH? • HCl(aq) + NaOH(aq) ----> HOH(l) + NaCl(aq) • (25.0mL)(0.350mol NaOH/1L)(1mol HCl/1mol NaOH)(1L/0.100mol) = 87.5 mL HCl solution

  40. Normality • Acid-Base Equivalents = (moles) (total (+) charge) • N = M ( total (+) charge)

  41. Normality Calculations • .250 M H3PO4 =______N • N = M(total(+) charge) • N = (0.250)(3) • N = 0.750 N H3PO4

  42. Normality Calculations • A solution of sulfuric acid contains 86 g of H2SO4 per liter of solution. What is its normality? • (86g H2SO4/1L)(1mol/98.0g) = 0.88 M H2SO4 • N = M(total (+) charge) • N = (0.88M)(2) • N = 1.8 N H2SO4

  43. Normality Stoichiometry Calculations • What volume of a 0.075 N NaOH solution is required to react exactly with 0.135 L of 0.45 N H3PO4? • Na = 0.45 N • Va = 0.135 L • Nb = 0.075 N • Vb = ? NaVa = NbVb Vb = (NaVa)/Nb Vb = (0.45N)(0.135L)/(0.075N) Vb = 0.81 L NaOH

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