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II III I C. Johannesson I. The Nature of Solutions Solutions.

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Presentation on theme: "II III I C. Johannesson I. The Nature of Solutions Solutions."— Presentation transcript:

1 II III I C. Johannesson I. The Nature of Solutions Solutions

2 Solution  a mixture of two or more substances that is identical throughout  can be physically separated  composed of solutes and solvents the substance in the smallest amount and the one that dissolves in the solvent the substance in the larger amount that dissolves the solute Colloids (milk, fog, jello) are considered solutions Iced Tea Mix (solute) Water (solvent) Iced Tea (solution) Salt water is considered a solution. How can it be physically separated?

3 C. Johannesson A. Definitions  Solution -  Solution - homogeneous mixture Solvent Solvent - present in greater amount Solute Solute - substance being dissolved

4 C. Johannesson A. Definitions Solute Solute - KMnO 4 Solvent Solvent - H 2 O

5 C. Johannesson B. Solvation  Solvation –  Solvation – the process of dissolving solute particles are separated and pulled into solution solute particles are surrounded by solvent particles

6 C. Johannesson B. Solvation Strong Electrolyte Non- Electrolyte solute exists as ions only - + salt - + sugar solute exists as molecules only - + acetic acid Weak Electrolyte solute exists as ions and molecules DISSOCIATIONIONIZATION View animation online.animation

7 C. Johannesson B. Solvation  Dissociation separation of an ionic solid into aqueous ions NaCl(s)  Na + (aq) + Cl – (aq)

8 C. Johannesson B. Solvation  Ionization breaking apart of some polar molecules into aqueous ions HNO 3 (aq) + H 2 O(l)  H 3 O + (aq) + NO 3 – (aq)

9 C. Johannesson B. Solvation  Molecular Solvation molecules stay intact C 6 H 12 O 6 (s)  C 6 H 12 O 6 (aq)

10 Factors in Explaining Solubility  In most cases, “like dissolves like.” –This means that polar solvents dissolve polar (or ionic) solutes and nonpolar solvents dissolve nonpolar solutes. –The relative force of attraction of the solute for the solvent is a major factor in their solubility.

11 C. Johannesson B. Solvation NONPOLAR POLAR “Like Dissolves Like”

12 C. Johannesson B. Solvation  Soap/Detergent polar “head” with long nonpolar “tail” dissolves nonpolar grease in polar water

13 Figure 12.1: Immiscible and miscible liquids. Photo courtesy of American Color.

14 Solutes Change Solvents  The amount of solute in a solution determines how much the physical properties of the solvent are changed  Examples: Lowering the Freezing Point The freezing point of a liquid solvent decreases when a solute is dissolved in it. Ex. Pure water freezes at 32 0 F (0 0 C), but when salt is dissolved in it, the freezing point is lowered. This is why people use salt to melt ice. Raising the Boiling Point The boiling point of a solution is higher than the boiling point of the solvent. Therefore, a solution can remain a liquid at a higher temperature than its pure solvent. Ex. The boiling point of pure water is 212 0 F (100 0 C), but when salt is dissolved in it, the boiling point is higher. This is why it takes salt water longer to boil than fresh water.

15 Concentration  the amount of solute dissolved in a solvent at a given temperature described as dilute if it has a low concentration of solute described as saturated if it has a high concentration of solute described as supersaturated ifsupersaturated contains more dissolved solute than normally possible

16 C. Johannesson C. Solubility SATURATED SOLUTION no more solute dissolves UNSATURATED SOLUTION more solute dissolves SUPERSATURATED SOLUTION becomes unstable, crystals form concentration

17 Comparison of unsaturated and saturated solutions.

18 Crystallization from a supersaturated solution of sodium acetate.

19 Solubility  the amount of solute that dissolves in a certain amount of a solvent at a given temperature and pressure to produce a saturated solution  influenced by: Temperature Pressure Solids  increased temperature causes them to be more soluble and vice versa Gases  increased temperature causes them to be less soluble and vice versa Ex. Iced Coffee Solids  increased pressure has no effect on solubility Gases  increased pressure causes them to be more soluble and vice versa Ex. Soda, “The Bends” What do we call things that are not soluble?

20 C. Johannesson C. Solubility  Solubility maximum grams of solute that will dissolve in 100 g of solvent at a given temperature varies with temp based on a saturated soln

21 C. Johannesson C. Solubility  Solubility Curve shows the dependence of solubility on temperature

22 C. Johannesson C. Solubility  Solids are more soluble at... high temperatures.  Gases are more soluble at... low temperatures & high pressures (Henry’s Law). EX: nitrogen narcosis, the “bends,” soda

23 Solid Solutions  Solid solutions of metals are referred to as alloys. –Brass is an alloy composed of copper and zinc. –Bronze is an alloy of copper and tin. –Pewter is an alloy of zinc and tin.

24 II III I C. Johannesson II. Concentration Solutions

25 C. Johannesson A. Concentration  The amount of solute in a solution.  Describing Concentration % by mass - medicated creams % by volume - rubbing alcohol ppm, ppb - water contaminants molarity - used by chemists molality - used by chemists

26 C. Johannesson A. Concentration SAWS Water Quality Report - June 2000

27 C. Johannesson B. Molality mass of solvent only 1 kg water = 1 L water

28 C. Johannesson B. Molality  Find the molality of a solution containing 75 g of MgCl 2 in 250 mL of water. 75 g MgCl 2 1 mol MgCl 2 95.21 g MgCl 2 = 3.2 m MgCl 2 0.25 kg water

29 C. Johannesson B. Molality  How many grams of NaCl are req’d to make a 1.54m solution using 0.500 kg of water? 0.500 kg water1.54 mol NaCl 1 kg water = 45.0 g NaCl 58.44 g NaCl 1 mol NaCl

30 C. Johannesson C. Dilution  Preparation of a desired solution by adding water to a concentrate.  Moles of solute remain the same.

31 C. Johannesson C. Dilution  What volume of 15.8M HNO 3 is required to make 250 mL of a 6.0M solution? GIVEN: M 1 = 15.8M V 1 = ? M 2 = 6.0M V 2 = 250 mL WORK: M 1 V 1 = M 2 V 2 (15.8M) V 1 = (6.0M)(250mL) V 1 = 95 mL of 15.8M HNO 3

32 C. Johannesson D. Preparing Solutions  500 mL of 1.54M NaCl 500 mL water 45.0 g NaCl mass 45.0 g of NaCl add water until total volume is 500 mL mass 45.0 g of NaCl add 0.500 kg of water 500 mL mark 500 mL volumetric flask  1.54m NaCl in 0.500 kg of water

33 C. Johannesson D. Preparing Solutions Copyright © 1995-1996 NT Curriculum Project, UW-Madison (above: “Filling the volumetric flask”)

34 C. Johannesson D. Preparing Solutions  250 mL of 6.0M HNO 3 by dilution measure 95 mL of 15.8M HNO 3 95 mL of 15.8M HNO 3 water for safety 250 mL mark combine with water until total volume is 250 mL Safety: “Do as you oughtta, add the acid to the watta!”


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