1 / 26

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

vonda
Télécharger la présentation

Chapter 15:

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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?

More Related