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Chapter 15:

Chapter 15:. Solutions. Students will learn. Solution composition – solvent, solute Solubility Types of solutions – saturated, unsaturated, supersaturated Concentration units - % by mass, molarity , normality Solution stoichiometry. Solution. Two substances well, evenly mixed

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Chapter 15:

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  1. Chapter 15: Solutions

  2. Students will learn • Solution composition – solvent, solute • Solubility • Types of solutions – saturated, unsaturated, supersaturated • Concentration units - % by mass, molarity, normality • Solution stoichiometry

  3. Solution • Two substances well, evenly mixed • Solute – the substance that dissolves (usually less in amount) • Solvent – the substance that does dissolving (more in amount) (Ex) Salt solution: salt = solute; water = solvent (Ex) 10% alcohol solution: alcohol = solute; water = solvent • Also called homogeneous mixture • Can be liquid, solid (=alloy), gas • The solute particle size < 10-7 cm

  4. Solute – Solvent Interactions • Use the rule: “Like dissolves like” (nonpolar solutes dissolve in nonpolar solvents and polar solutes dissolve in polar solvents Ex) oil (nonpolar) and water (polar): dissolves? Ex) salt (ionic) and water (polar): dissolves? • immiscible – one liquid doesn’t dissolve in another liquid (Ex) Oil (nonpolar) and water (polar) • miscible – one liquid dissolves in another liquid (Ex) ethanol (polar) in water (polar)

  5. Types of Solution based on Solubility • Saturated solution: contains the maximum amount of dissolved solute • Unsaturated solution: contains less solute than a saturated solution • Supersaturated solution: contains more dissolved solute than a saturated solution * Solubility of sugar at 25 ˚C = 91 g /100 mL water

  6. Ways to Express Concentrations • Qualitative = in words • weak, strong, concentrated, or diluted • Quantitative • mass percent • molarity • normality • molality* • ppm/ppb*

  7. Mass Percent

  8. Examples (Pg 482) What is the percent-by-mass concentration of glucose in a solution made my dissolving 5.5 g of glucose in 78.2 g of water?

  9. (Pg 483) What mass of water must be added to 435 g of formaldehyde to prepare a 40.0% (by mass) solution of formaldehyde? This solution, called formalin, is used to preserve biological specimens.

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

  11. Examples 1. You have 1.00 mol of sugar in 125.0 mL of solution. Calculate the concentration in units of molarity.

  12. 2) Pg 485: Calculate the molarity of a solution prepared by dissolving 1.00 g of ethanol, C2H5OH, in enough water to give a final volume of 101 mL. Molar mass of ethanol = 46.1 g/mol

  13. 3) Pg 486: Give the concentrations of the ions in each of the following solutions. a) 0.10 M Na2CO3 b) 0.010 M Al2(SO4)3

  14. Standard Solution Solution whose concentration is accurately known Usually made from a pure solid (Ex) How much solid K2Cr2O7 (molar mass = 294.2 g) must be weighed out to make 1.00 L of 0.200 M solution?

  15. Making a Standard Solution Weigh out a sample of solute. Transfer to a volumetric flask. Add enough solvent to mark on flask.

  16. Dilution By adding more solvent The amount of solute remains unchanged The concentration decreases

  17. 100. mL of 1.0 M NaOH solution • 1.0 M x 0.100 L = 0.100 mol NaOH • Mbefore x Vbefore • Water is added to the final volume of 500. mL • 0.100 mol ÷ 0.500 L = 0.200 M • Mafter x Vafter • Mbefore x Vbefore = Mafter x Vafter

  18. Example What is the volume of a 2.00 MNaOH solution needed to make 150.0 mL of a 0.800 MNaOH solution?

  19. Stoichiometry # of particles # of particles mass moles moles mass volume (gas @STP) volume (gas @STP) solution solution (Wanted Quantity) (Given Quantity)

  20. Example 10.0 mL of a 0.30 M sodium phosphate solution reacts with 20.0 mL of a 0.20 M lead(II) nitrate solution (assume no volume change). • What precipitate will form? • What mass of precipitate will form? (Molar mass of Pb = 811.5 g/mol) • What is the molarityof nitrate ions left in solution after the reaction is complete? Assume the volume is additive.

  21. Neutralization Reaction • An acid-base reaction is called a neutralization reaction. • For a strong acid and base reaction: • H+(aq) + OH–(aq)  H2O(l) • # equivalences of acid = # of equivalences of base

  22. Equivalence, Equiv • One equivalent of acid – amount of acid that furnishes 1 mol of H+ ions. • (Ex) 1 mol HCl = 1 equivalence • (Ex) 1 mol H2SO4 = 2 equivalences • One equivalent of base – amount of base that furnishes 1 mol of OH ions • (Ex) 1 mol NaOH = 1 equivalence • 1 mol Ca(OH)2 = 2 equivalences • Equivalent weight – mass in grams of 1 equivalent of acid or base. • (Ex) Equivalent weight of H2SO4 = 98 g /2 equiv = 49 g/equiv

  23. Normality, N (Ex) If Ba(OH)2 is used as a base, how many equivalents of Ba(OH)2 are there in 4 mol Ba(OH)2?

  24. Molarity vs. Normality N = (# of H+/OH‒)·M (Ex) What is the normality of 0.1 M H2SO4 solution?

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