Chapter 15: Solutions 15.1 Solubility 15.2 Solution Composition 15.3 Mass Percent 15.4 Molarity 15.7 Neutralization Reactions

Solutions Solutions are homogeneous mixtures. Mixtures in which the components are uniformly intermingled Solvent: the substance present in the highest percentage Solute: the dissolved substance, which is present in lesser amount Aqueous solutions: solutions with water as the solvent 2

Solutions (cont.)

The Solution Process: Ionic Compounds When ionic compounds dissolve in water they dissociate into ions and become hydrated. When solute particles are surrounded by solvent molecules we say they are solvated. Figure 15.1: When solid sodium chloride dissolves, the ions are dispersed randomly throughout the solution. 3

Solution: Solid in Liquid Salt in water - separate into ions Figure 15.2: Polar water molecules interact with the positive and negative ions of a salt. These interactions replace the strong ionic forces holding the ions together in the undissolved solid, thus assisting in the dissolving process.

The Solution Process: Covalent Molecules Covalent molecules that are small and have “polar” groups tend to be soluble in water. The ability to H-bond with water enhances solubility. O H C 4

Solubility When one substance (solute) dissolves in another (solvent), it is said to be soluble. Salt is soluble in water Bromine is soluble in methylene chloride When one substance does not dissolve in another, it is said to be insoluble. Oil is insoluble in water 5

Solubility (cont.) There is usually a limit to the solubility of one substance in another. Gases are always soluble in each other Some liquids are always mutually soluble

Solutions & Solubility Molecules that are similar in structure tend to form solutions: “like dissolves like” 6

Solutions & Solubility (cont.) The solubility of the solute in the solvent depends on the temperature. Higher temp = greater solubility of solid in liquid Lower temp = greater solubility of gas in liquid The solubility of gases depends on the pressure. Higher pressure = greater solubility

Figure 15.6: An oil layer floating on water.

Describing Solutions Qualitatively A concentrated solution has a high proportion of solute to solution. A dilute solution has a low proportion of solute to solution. 7

Describing Solutions Qualitatively (cont.) A saturated solution has the maximum amount of solute that will dissolve in the solvent. Depends on temp An unsaturated solution has less than the saturation limit. A supersaturated solution has more than the saturation limit. Unstable

Describing Solutions Quantitatively (cont.) Solutions have variable composition. To describe a solution accurately, you need to describe the components and their relative amounts. Concentration: the amount of solute in a given amount of solution Occasionally amount of solvent 8

Solution Concentration mass percent = mass of solute X 100% mass of solution Molarity = moles of solute liters of solution (mass of solute + mass of solvent) 9

Solution Concentration Percentage Mass percent = grams of solute per 100 g of solution 5.0% NaCl has 5.0 g of NaCl in every 100 g of solution Mass of solution = mass of Solute + mass of solvent Divide the mass of solute by the mass of solution and multiply by 100% to get mass percent. 9

Mass Percent #15.1 A solution is prepared by mixing 2.50 g if calcium chloride with 50.0g of water. Calculate the mass percent of calcium chloride in this solution mass percent = mass of solute X 100% mass of solution 4.76% CaCl2 mass percent = 2.50g X 100 = 2.50 + 50.0g 9

Mass Percent #15.2 Concentrated hydrochloric acid solution contains 37.2% by mass HCl. What mass of HCl is contained in 35.5g of concentrated HCl? mass percent = mass of solute X 100% mass of solution 13.2 g HCl 37.2 = mass of solute X 100 = 35.5g 9

Solution Concentration Molarity Moles of solute per 1 liter of solution Used because it describes how many moles of solute in each liter of solution If a sugar solution concentration is 2.0 M, 1 liter of solution contains 2.0 moles of sugar, 2 liters = 4.0 moles sugar, 0.5 liters = 1.0 mole sugar, etc. molarity = moles of solute liters of solution 10

Molarity #15.3 Calculate the molarity of a solution prepared by dissolving 15.6 g of solid KBr in enough water to make 1.25 L of solution Molarity = moles of solute liters of solution 15.6 g KBr (1 mol /119.0g) = 0.131 mol KBr Molarity = 0.131 mol = 1.25 L 0.105 M 9

Molarity #15.4 Calculate the molarity of a solution prepared by dissolving 2.80 g of solid NaCl in enough water to make 135 mL of solution Molarity = moles of solute liter of solution 2.80 g NaCl (1 mol /58.44g) = 0.0479 mol KBr 135 mL (1 L /1000 mL) = 0.135 L Molarity = 0.0479 mol = 0.135 L 0.355 M 9

Molarity & Dissociation The molarity of the ionic compound allows you to determine the molarity of the dissolved ions. 11

Molarity & Dissociation (cont.) CaCl2(aq) = Ca+2(aq) + 2 Cl-1(aq) A 1.0 M CaCl2(aq) solution contains 1.0 moles of CaCl2 in each liter of solution Because each CaCl2 dissociates to one Ca+2, 1.0 M CaCl2 = 1.0 M Ca+2 Because each CaCl2 dissociates to 2 Cl-1, 1.0 M CaCl2 = 2.0 M Cl-1

Calculating Ion Concentration from Molarity #15.5 Give the concentrations of all ions in each of the following solutions: 1.20 M Na2SO4 0.750 M K2CrO4 2.40M Na+, 1.20M SO42- 1.50M K+, 0.750M CrO42- Na2SO4  Na+(aq) + SO42-(aq) 1 2 1 1.20 Na2SO4  2.40 Na+(aq) + 1.20 SO42-(aq) K2CrO4  K+(aq) + CrO42-(aq) 1 2 1 0.750 K2CrO4  1.50 K+(aq) + 0.750 CrO42-(aq) 9

Dilution Dilution: adding solvent to decrease the concentration of a solution The amount of solute stays the same, but the concentration decreases. 12 25

Dilution (cont.) Dilution Formula M1 x V1 = M2 x V2 # Moles/L · # L = # moles In dilution we take a certain number of moles of solute and dilute to a bigger volume. Concentrations and volumes can be most units as long as they are consistent.

M1 x V1 = moles of solute = M2 x V2 Dilution (cont.) remains constant M = moles of solute volume (L) M1 x V1 = moles of solute = M2 x V2 decreases Add water, therefore increases Initial conditions Final conditions Molarity before Dilution Volume before Dilution Molarity after Dilution Volume after Dilution

Dilution #15.6 A What volume of 19M NaOH must be used to prepare 1.0 L of a 0.15M NaOH solution? M1 x V1 = M2 x V2 19 M x V1 = 0.15M x 1.0 L V1 = 0.15M x 1.0 L = 0.0079 L = 7.9 mL 19 M 9 28

Dilution #15.6 B What volume of 19M NaOH must be used to prepare 1.0 L of a 0.15M NaOH solution? M1 x V1 = M2 x V2 19 M x V1 = 0.15M x 1.0 L V1 = 0.15M x 1.0 L = 0.0079 L = 7.9 mL 19 M 9 29

15.7 Neutralization Reactions 18 30

Neutralization Reactions Acid-Base reactions are also called neutralization reactions. Often we use neutralization reactions to determine the concentration of an unknown acid or base. The procedure is called a titration. With this procedure we can add just enough acid solution to neutralize a known volume of a base solution. Or vice-versa 18 31

Neutralization Reactions When a strong acid and a strong base react, the net ionic reaction is H+(aq) + OH-(aq)  H2O(l) 18 32

Neutralization Reactions # 15.7a What volume of 0.150 M HNO3 solution is needed to neutralize 45.0 mL of a 0.550 M KOH solution? Answer = 165 mL 18 33

Neutralization Reactions # 15.7b What volume of 1.00 X 10-2 M HCL solution is needed to neutralize 35.0 mL of a 5.00 X 10-3 M Ba(OH)2 solution? Answer = 35.0 mL 18 34