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

Chapter 15: Solutions

15.1 Solubility15.2 Solution Composition15.3 Mass Percent15.4 Molarity15.7 Neutralization Reactions

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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


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 15.2: 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.

Solution: Solid in Liquid

• Salt in water - separate into ions


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.OH

H

C O

H

HH H

O

H

H


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


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”


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.


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


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)


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.


Mass Percent #15.1


mass of solution

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

4.76% CaCl2mass percent = 2.50g X 100 =

2.50 + 50.0g


Mass Percent #15.2


mass of solution

Concentrated hydrochloric acid solution contains 37.2% by mass HCl. What mass of HCl is contained in 35.5g of concentrated HCl?

13.2 g HCl37.2 = mass of solute X 100 =

35.5g


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 soluteliters of solution


Molarity #15.3

Molarity = moles of solute liters of solution

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

0.105 MMolarity = 0.131 mol =

1.25 L

15.6 g KBr (1 mol /119.0g) = 0.131 mol KBr


Molarity #15.4

Molarity = moles of solute liter of solution

Calculate the molarity of a solution prepared by dissolving 2.80 g of solid NaCl in enough water to make 135 mL of solution

0.355 MMolarity = 0.0479 mol =

0.135 L

2.80 g NaCl (1 mol /58.44g) = 0.0479 mol KBr

135 mL (1 L /1000 mL) = 0.135 L


Molarity & Dissociation

• The molarity of the ionic compound allows you to determine the molarity of the dissolved ions.


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. 1.20 M Na2SO4

2. 0.750 M K2CrO4

Na2SO4 Na+(aq) + SO42-(aq)

1.20 Na2SO4 2.40 Na+(aq) + 1.20 SO42-(aq)

1 12

2.40M Na+, 1.20M SO42-

K2CrO4 K+(aq) + CrO42-(aq)

0.750 K2CrO4 1.50 K+(aq) + 0.750 CrO42-(aq)

1 12

1.50M K+, 0.750M CrO42-


Dilution

• Dilution: adding solvent to decrease the concentration of a solution

• The amount of solute stays the same, but the concentration decreases.


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.


Dilution (cont.)

M = moles of solute volume (L)

M1 x V1 = moles of solute = M2 x V2

remains constant

decreasesAdd water, therefore increases

Initial conditions Final conditions

Molarity before Dilution

Volume before Dilution

Molarity after Dilution

Volume after Dilution


Dilution #15.6 A

M1 x V1 = M2 x V2

What volume of 19M NaOH must be used to prepare 1.0 L of a 0.15M NaOH solution?

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


Dilution #15.6 B

M1 x V1 = M2 x V2

What volume of 19M NaOH must be used to prepare 1.0 L of a 0.15M NaOH solution?

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


15.7 Neutralization Reactions


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


Neutralization Reactions

• When a strong acid and a strong base react, the net ionic reaction is

H+(aq) + OH-(aq) H2O(l)


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


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

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

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