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Introductory Chemistry: A Foundation, 6 th Ed. Introductory Chemistry, 6 th Ed. Basic Chemistry, 6 th Ed. by Steven S. Zumdahl & Donald J. DeCoste University of Illinois. Chapter 15 Solutions. Solutions. Solutions are homogeneous mixtures.

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Introductory chemistry a foundation 6 th ed introductory chemistry 6 th ed basic chemistry 6 th ed

Introductory Chemistry: A Foundation, 6th Ed. Introductory Chemistry, 6th Ed. Basic Chemistry, 6th Ed.

by Steven S. Zumdahl& Donald J. DeCoste

University of Illinois


Chapter 15 solutions
Chapter 15Solutions


Solutions
Solutions

  • Solutions are homogeneous mixtures.

  • Solvent: the substance present in the highest percentage

  • Solute: the dissolved substance, which is present in lesser amount


Solutions cont
Solutions (cont.)


The solution process ionic compounds
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.


The solution process covalent molecules

H

O

H

H

C

O

H

H

H

H

O

H

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.


Solubility
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
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
Solutions & Solubility

  • Molecules that are similar in structure tend to form solutions: “like dissolves like”


Solutions solubility cont
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


Describing solutions qualitatively
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
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
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 percentage
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.


Solution concentration molarity

moles of solute

liters of solution

molarity =

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 dissociation
Molarity & Dissociation

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


Molarity dissociation cont
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


Dilution
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 (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.


Solution stoichiometry
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.


Solution stoichiometry cont
Solution Stoichiometry (cont.)

  • Basic strategy is the same:

    • Balance the equation

    • Change given amounts to moles (M x V = #moles)

    • Determine limiting reactant

    • Calculate moles of required substance

    • Convert moles of the required substance into the desired unit


Example 1
Example #1:

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.


Example 1 cont
Example #1 (cont.)

  • Write and balance the reaction:

    • The reaction is a precipitation reaction. It 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) (balanced)


Example 1 cont1
Example #1 (cont.)

  • Change the given amounts to moles:

    • We are given 1.50 L of 0.100 M AgNO3 . Since 1 AgNO3 dissociates into 1 Ag+1


Example 2
Example #2:

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.


Example 2 cont
Example #2 (cont.)

  • 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

  • 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


Example 2 cont1
Example #2 (cont.)

  • 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


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


Normality
Normality

  • Normality: 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

  • Equivalent weight: the mass of 1 equivalent of an acid or base



Solution concentration normality
Solution Concentration Normality

  • Equivalents of solute per 1 liter of solution

  • Used because it describes how many H+ or OH- in each liter of solution


Solution concentration normality cont
Solution Concentration Normality (cont.)

  • 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


Normality1

equivalents of solute

liters of solution

Normality =

Liters x normality =

equivalents of solute

Normality


Normality and neutralization
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

    # equiv/L x # L = # equiv

    # eqivacid = # equivbase

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


Example 3
Example #3:

What volume of 0.075 N KOH is required to

neutralize 0.135 L of 0.45 N H3PO4?


Example 3 cont
Example #3 (cont.)

  • Determine the quantities and units in the problem

    Acid Solution Base Solution

    Normality 0.45 N 0.075 N

    Volume 0.135 L ? L

  • Solve the formula for the unknown quantity


Example 3 cont1
Example #3 (cont.)

  • Plug the values into the equation and solve

    Acid Solution Base Solution

    Normality 0.45 N 0.075 N

    Volume 0.135 L ? L


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