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Unit 14 – Solutions 14.1 Solubility 14.2 Solution Composition 14.3 Mass Percent 14.4 Molarity 14.5 Dilution 14.6 Stoichiometry of Solution Reactions 14.7.

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Presentation on theme: "Unit 14 – Solutions 14.1 Solubility 14.2 Solution Composition 14.3 Mass Percent 14.4 Molarity 14.5 Dilution 14.6 Stoichiometry of Solution Reactions 14.7."— Presentation transcript:

1 Unit 14 – Solutions 14.1 Solubility 14.2 Solution Composition 14.3 Mass Percent 14.4 Molarity 14.5 Dilution 14.6 Stoichiometry of Solution Reactions 14.7 Neutralization Reactions 14.8 Normality Text Pages

2 Unit 14 – Solutions Upon completion of this unit, you should be able to do the following: Calculate the molarity of a solution given its composition. Determine the amount of substance needed to prepare a solution with a definite molar concentration. Understand the process of dissolving. Learn why certain components dissolve in water Understand mass percent and learn how to calculate it Understand molarity. Learn to use molarity to calculate the number of moles of solute present. Learn to calculate the concentration of a solution made by diluting a stock solution. Understand the strategy for solving stoichiometric problems for solution reactions. Learn how to do calculations involved in acid-base reactions. Learn about normality and equivalent weight. Learn to use these concepts in stoichiometric calculations.

3 Unit 14 – Solutions A solution is a homogeneous mixture in which the components are uniformly distributed The substance present in the largest amount is called the solvent The other substances are called solutes. Aqueous solutions have water as the solvent.

4 Solubility When sodium chloride is dissolved in water, the resulting solution conducts electricity. This convinces us that the solution contains ions that can move. When the solid dissolves, the ions are separated and dispersed throughout the solution. The strong ionic forces that hold the sodium chloride crystal together are overcome by the strong attractions between the ions and the polar water molecules. The strong forces holding the positive and negative ions in the solid are replaced by strong water-ion interactions and the solid dissolves.

5 Solubility Water also dissolves many nonionic substances.
Ethanol (C2H5OH) is very soluble in water. Ethanol contains a polar O-H bond similar to water, which makes it very compatible with water. Ethanol forms hydrogen bonds with water.

6 Solubility Many substances do not dissolve in water. Petroleum, for example, does not mix with water. Petroleum is nonpolar, with an almost equal sharing of the bonding electrons. The resulting molecule cannot form attractions with the polar water molecules and this prevents it from being soluble.

7 Solubility

8 Solubility “Like dissolves like” Water dissolves most polar solutes.
A nonpolar solvent is required to dissolve a nonpolar solute. “Grease” is composed of nonpolar molecules and requires a nonpolar solvent to dissolve it, or to remove a grease stain from clothing.

9 Solution Composition Even for soluble substances, there is a limit as to how much solute can be dissolved in a given amount of solvent. When a solution contains as much solvent as can be dissolved at that temperature, it is saturated. A solution that has not reached the limit of solute that will dissolve in it is said to be unsaturated. A relatively large amount of solute is dissolved in a concentrated solution. A relatively small amount of solute is dissolved is a dilute solution.

10 Mass Percent Describing the composition of a solution means specifying the amount of solute present in a given quantity of solution. The amount of solute is typically given in mass or moles. The amount of solution is typically given in mass or volume. Mass percent (sometimes called weight percent) expresses the mass of solute present in a given mass of solution.

11 Mass Percent Mass Percent = x 100 = x 100

12 Mass Percent What is the mass percent of NaCl in a solution prepared from 1.0 g of NaCl in 48 g of water? = x 100 = x 100 = 2.0 %

13 Mass Percent Example 14.1 page 428 A solution is prepared by mixing 1.00 g of ethanol, C2H5OH, with 100 g of water. Calculate the mass percent of ethanol in this solution.

14 Mass Percent Example 14.2, page 429 Although milk is not a true solution (it really is a suspension of tiny globules of fat, protein, and other substrates in milk), it does contain a dissolved sugar called lactose. Cow’s milk typically contains 4.5% by mass of lactose, C12H22O11. Calculate the mass of lactose present in 175 g of milk.

15 Molarity The concentration of a solution is the amount of solute in a given volume of solution. Molarity is the number of moles of solute in one liter of solution. M = molarity = =

16 Molarity Example 14.3, page 430 Calculate the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.5 L of solution.

17 Molarity Example 14.6, page 432 How many moles of Ag+ ions are present in 25 mL of a 0.75 M AgNO3 solution?

18 Standard Solution - a solution whose molarity is accurately known.
Figure 14.7 – Steps involved in the preparation of a standard aqueous solution. (a) Put a weighed amount of solute into a volumetric flask and add a small amount of water. (b) Dissolve the solid in the water by gently swirling the flask with the stopper in place. (c) Add more water until the level of the solution just reaches the mark etched on the neck of the flask. Then mix the solution thoroughly by inverting the flask several times.

19 Standard Solution Example 14.7, page 433 To analyze the alcohol content of a certain wine, a chemist needs 1.00 L of an aqueous M potassium dichromate (K2Cr2O7) solution. How much solid potassium dichromate (MW = g/mol) must be weighed out to make this solution?

20 Dilution The process of adding more solvent to a concentrated solution is called dilution. The concentrated solutions are called stock solutions. A dilution calculation involves determining how much solvent must be added to a stock solution to achieve a solution of the desired concentration. Only solvent (typically water) is added in the dilution. The amount of solute in the final, more dilute solution, is the same as amount of solute in the original, more concentrated solution. Moles of solute present = volume of dilute solution x molarity of dilute solution

21 Dilution Example 14.8, page 436 What amount of 16 M sulfuric acid must be used to prepare 1.5L of a 0.10 M H2SO4 solution?

22 Stoichiometry of Solution Reactions

23 Stoichiometry of Solution Reactions
Example 14.9, page 437 Calculate the mass of solid NaCl that must be added to 1.5L of a M AgNO3 solution to precipitate all of the Ag+ ions in the form of AgCl. Calculate the mass of AgCl formed.

24 Neutralization Reactions
Remember that an acid is a substance that furnishes an H+ ion. A strong acid completely dissociates in water. HCl (aq) → H+ (aq) + Cl- (aq) A base is a substance that furnishes an OH- ion. A strong base completely dissociates in water. NaOH (aq) → Na + (aq) + OH- (aq) When a strong acid and strong base react, the net ionic equation is H+ (aq) + OH- (aq) → H2O (l) An acid-base reaction is often called a neutralization reaction.

25 Neutralization Reactions
Example 14.11, page 440 What volume of a M HCl solution is needed to neutralize 25.0 mL of a M NaOH solution?

26 Normality Normality is another unit of concentration used, especially with acids and bases. One equivalent of an acid is equal to the amount of that acid that can furnish 1 mol of H+ ions. One equivalent of a base is the amount of that base that can furnish 1 mol of OH- ions. The equivalent weight of an acid or a base is the mass in grams of 1 equivalent of that acid or base. Normality is defined as the number of equivalents of solute per liter of solution

27 Normality

28 Normality Example 14.12, page 442 Phosphoric acid can furnish three H+ ions per molecule. Calculate the equivalent weight of H3PO4. Example 14.13, page 443 A solution of sulfuric acid contains 86 g of H2SO4 per liter of solution. Calculate the normality of this solution.

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