1 Solutions Chapter 14. 2 Solutions Solutions are homogeneous mixtures Solute is the dissolved substance –Seems to “disappear” or “Takes on the state”

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Solutions

Chapter 14

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Solutions• Solutions are homogeneous mixtures• Solute is the dissolved substance

– Seems to “disappear” or “Takes on the state” of the solvent

• Solvent is the substance the solute dissolves in– Does not appear to change state– When both solute and solvent have the same state, the solvent is the

component present in the highest percentage

• Solutions in which the solvent is water are called aqueous solutions– Water is often called the universal solvent

• Solutions that contain metal solutes and solvent are called alloys

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The Solution Process - Ionic Compounds

• When ionic compounds dissolve in water they dissociate into ions– ions become surrounded by water molecules -

hydrated

• When solute particles are surrounded by solvent molecules we say they are solvated

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Figure 14.1: When solid sodium chloride dissolves, the ions are dispersed randomly throughout the solution

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Figure 14.2: Polar water molecules interact with the positive and negative ions of a salt

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The Solution ProcessCovalent Molecules

• Covalent molecules that are small and have “polar” groups tend to be soluble in water

• The ability to H-bond with water enhances solubilityOH

H

C O

H

HH H

O

H

H

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Figure 14.3: Polar O—H bond similar to those in the water molecule

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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 they are said to be insoluble– Oil is insoluble in Water

• 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

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Solutions & Solubility

• Molecules that are similar in structure tend to form solutions– Like dissolves like

• The solubility of the solute in the solvent depends on the temperature– Higher Temp = Larger solubility of solid in liquid– Lower Temp =Larger solubility of gas in liquid

• The solubility of gases depends on the pressure– Higher pressure = Larger solubility

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Figure 14.4: The structure of common table sugar (called sucrose)

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Figure 14.5: A molecule typical of those found in petroleum

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Figure 14.6: An oil layer floating on water

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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

• 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

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Describing Solutions Quantitatively

• Solutions have variable composition

• To describe a solution accurately, you need to describe the components and their relative amounts

• Concentration = amount of solute in a given amount of solution– Occasionally amount of solvent

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Figure 14.7: Preparation of a standard aqueous solution

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Solution ConcentrationPercentage

• Parts Per Hundred

• % = grams of solute per 100 g of solution– Mass Percent or Percent by Mass– 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%

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Solution ConcentrationMolarity

• moles of solute per 1 liter of solution

• used because it describes how many molecules 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

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Molarity & Dissociation• the molarity of the ionic compound allows you to determine the

molarity of the dissolved ions• 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– 1 L = 1.0 moles CaCl2, 2 L = 2.0 moles CaCl2, 0.5 L = 0.5 moles

CaCl2

• Because each CaCl2 dissociates to one Ca+2, 1.0 M CaCl2 = 1.0 M Ca+2

– 1 L = 1.0 moles Ca+2, 2 L = 2.0 moles Ca+2, 0.5 L = 0.5 moles Ca+2

• Because each CaCl2 dissociates to 2 Cl-1, 1.0 M CaCl2 = 2.0 M Cl-1

– 1 L = 2.0 moles Cl-1, 2 L = 4.0 moles Cl-1, 0.5 L = 1.0 moles Cl-1

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Dilution

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

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

• Dilution Formula

M1 x V1 = M2 x V2

• Concentrations and Volumes can be most units as long as consistent

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Figure 14.8: Acetic acid dilution

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Solution Stoichiometry• Many reactions occur in solution, therefore you

need to be able to predict amounts of reactants and products in terms of concentrations and volumes as well as masses

• Basic strategy is the same1. Balance the Equation

2. Change Given Amounts to Moles

3. Determine Limiting Reactant

4. Calculate Moles of Required Substance

5. Convert Moles of the Required Substance into the Desired Unit

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Example

1. Write and Balance the Reaction

The reaction is a Precipitation Reaction. The reaction involves Cl-1 ions from NaCl reacting with Ag+1 ions

from AgNO3 to form AgCl(s). Therefore we get

Ag+1(aq) + Cl-1(aq) AgCl(s)

After Balancing we get

Ag+1(aq) + Cl-1(aq) AgCl(s)

Calculate the Mass of Solid NaCl required to Precipitate all the Ag+1 ions from 1.50 L of a

0.100 M AgNO3 Solution

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2. Change the Given Amounts to MolesWe are given 1.50 L of 0.100 M AgNO3

Since 1 AgNO3 dissociates into 1 Ag+1

The concentration of Ag+1 = 0.100 M

1 L Solution = 0.100 mol Ag+1

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Ag mol 0.150 Solution L1

Ag mol 0.100x L1.50

ExampleCalculate the Mass of Solid NaCl required to Precipitate all the Ag+1 ions from 1.50 L of a

0.100 M AgNO3 Solution

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3. Determine the Limiting Reactant

Since we are going to precipitate all the Ag+1

by adding Cl-1 , the Ag+1 is the Limiting Reactant

4. Determine the Number of Moles of the Required Substance

We need to calculate the Moles of Cl-1 Required to precipitate 0.150 moles of Ag+1

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11 Cl mol 0.150

Ag mol 1

Cl mol 1x Ag mol 0.150

ExampleCalculate the Mass of Solid NaCl required to Precipitate all the Ag+1 ions from 1.50 L of a

0.100 M AgNO3 Solution

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5. Convert Moles of the Required Substance into the Desired Unit

We need 0.150 moles of Cl-1

Since 1 NaCl dissociates into 1 Cl-1

The moles of NaCl needed = 0.150 moles

1 mol NaCl = 58.44 g NaCl

NaCl g 8.76 NaCl mol 1NaCl g 58.44

x NaCl mol 0.150

ExampleCalculate the Mass of Solid NaCl required to Precipitate all the Ag+1 ions from 1.50 L of a

0.100 M AgNO3 Solution

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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 visa versa

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Normality

• Normality is a concentration unit used mainly for acids and bases

• One equivalent of an acid is the amount of acid that can furnish 1 mol of H+1

• One equivalent of a base is the amount of base that can furnish 1 mol of OH-1

• The equivalent weight is the mass of 1 equivalent of an acid or base

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Equivalents• 1 mol HCl = 1 mol H+1 = 1 equivalent HCl

– Therefore the equivalent weight of HCl = the Molar Mass of HCl = 36.45 g

• 1 mol H2SO4 = 2 mol H+1 = 2 equiv H2SO4

– Therefore the equivalent weight of H2SO4 = one-half the Molar Mass of H2SO4 = ½(98.07 g) = 49.04 g

• 1 mol NaOH = 1 mol OH-1 = 1 equivalent NaOH– Therefore the equivalent weight of NaOH = the Molar Mass of NaOH =

40.00 g

• 1 mol KOH = 1 mol OH-1 = 1 equiv KOH– Therefore the equivalent weight of KOH = the Molar Mass of KOH = 56.11

g

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Solution ConcentrationNormality

• equivalents of solute per 1 liter of solution• used because it describes how many H+1 or OH-1

in each liter of solution• If an acid solution concentration is 2.0 N , 1 liter

of solution contains 2.0 equiv of acid which means 2 mol H+1

– 2 liters = 4.0 equiv acid = 4.0 mol H+1 – 0.5 liters = 1.0 equiv acid = 1.0 mol H+1

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Normality

Normality = equivalents of solute

liters of solution

Liters x Normality = Equivalents of Solute

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Normality and Neutralization

• One Equivalent of Acid exactly neutralizes One Equivalent of Base

• Can be used to simplify neutralization stoichiometry problems to the equation

Nacid x Vacid = Nbase x Vbase

• The volumes can be most any unit, as long as they are consistent

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Example

Determine the quantities and units in the problemAcid Solution Base

Solution

Normality 0.45 N 0.075 N

Volume 0.135 L ? L

Solve the Formula for the Unknown Quantity

What volume of 0.075 N KOH is required to Neutralize 0.135 L of 0.45 N H3PO4

base

acidx acid base

basebaseacidacid

NVNV

Vx N Vx N

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Plug the Values into the Equation and SolveAcid Solution Base

Solution

Normality 0.45 N 0.075 N

Volume 0.135 L ? L

L0.81 N 0.075

L0.135N 0.45V

NVNV

x

base

base

acidx acid base

ExampleWhat volume of 0.075 N KOH is required to

Neutralize 0.135 L of 0.45 N H3PO4