Recall:Solutions

Chapter 14 1

Recall:Solutions• A solution is a homogeneous mixture.• A solution is composed of a solute dissolved in a solvent.

• Solutions exist in all three physical states:

Chapter 14 2

Add:Gases in Solution• Temperature effects the solubility of gases.

• The higher the temperature, the lower the solubility of a gas in solution.

• An example is carbon dioxide in soda:– Less CO2 escapes when you open a cold soda than

when you open the soda warm

Chapter 14 3

4.1The Dissolution Process• When a soluble crystal is placed into a solvent, it

begins to dissolve.

• When a sugar crystal is placed in water, the water

molecules attack the crystal and begin pulling part of it

away and into solution.

• The sugar molecules are held within a cluster of water molecules called a solvent

cage.

Chapter 14 4

4.1Dissolving of Ionic Compounds• When a sodium chloride crystal is place in water,

the water molecules attack the edge of the crystal.• In an ionic compound, the

water molecules pull individual ions off of the

crystal.

• The anions are surrounded by the positively charged hydrogens on water.

• The cations are surrounded by the negatively charged

oxygen on water.

4.5 Calculations with Reactions in Solution

1. How to describe the solution quantitatively

2. How to prepare the solution from a pure substance

3. How to dilute a solution from a more concentrated one

4. How to use stoichiometry with solutions

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4.5 Solution Stoichiometry

The concentration of a solution is the amount of solute present in a given quantity of solvent or solution.

M = molarity =moles of solute

liters of solution

What mass of KI is required to make 500. mL of a 2.80 M KI solution?

volume of KI solution moles KI grams KIM KI M KI

500. mL = 232 g KI166 g KI

1 mol KIx

2.80 mol KI

1 L solnx

1 L

1000 mLx

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Preparing a Solution of Known Concentration

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Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution.

DilutionAdd Solvent

Moles of solutebefore dilution (i)

Moles of soluteafter dilution (f)=

MiVi MfVf=

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How would you prepare 60.0 mL of 0.200 M HNO3 from a stock solution of 4.00 M HNO3?

MiVi = MfVf

Mi = 4.00 M Mf = 0.200 M Vf = 0.0600 L Vi = ? L

Vi =MfVf

Mi

= 0.200 M x 0.0600 L4.00 M

= 0.00300 L = 3.00 mL

Dilute 3.00 mL of acid with water to a total volume of 60.0 mL.

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• A 25.0ml sample of 0.0040 M solution of NaBr in water is treated with a 0.0025M aqueous solution of AgNO3. What volume of the AgNO3 solution is required to react completely with the NaBr solution?

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4.6 Techniques: Gravimetric Analysis1. Dissolve unknown substance in water

2. React unknown with known substance to form a precipitate

3. Filter and dry precipitate

4. Weigh precipitate

5. Use chemical formula and mass of precipitate to determine amount of unknown ion

Chapter 14 12

Gravimetric Analysis Problem• What mass of silver bromide is produced from the

reaction of 37.5 mL of 0.100 M aluminum bromide with excess silver nitrate solution?

AlBr3(aq) + 3 AgNO3(aq) → 3 AgBr(s) + Al(NO3)3(aq)

• We want g AgBr, we have volume of AlBr3

= 2.11 g AgBr

37.5 mL soln ×3 mol AgBr

1 mol AlBr3

0.100 mol AlBr3

1000 mL soln×

1 mol AgBr

187.77 g AgBr×

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4.7-4.8 Techniques: TitrationsIn a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete.

Equivalence point – the point at which the reaction is complete

Slowly add baseto unknown acid

UNTIL

the indicatorchanges color

Indicator – substance that changes color at (or near) the equivalence point

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Titrations can be used in the analysis of

Acid-base reactions

Redox reactions

H2SO4 + 2NaOH 2H2O + Na2SO4

5Fe2+ + MnO4- + 8H+ Mn2+ + 5Fe3+ + 4H2O

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What volume of a 1.420 M NaOH solution is required to titrate 25.00 mL of a 4.50 M H2SO4 solution?

WRITE THE CHEMICAL EQUATION!

volume acid moles red moles base volume base

H2SO4 + 2NaOH 2H2O + Na2SO4

4.50 mol H2SO4

1000 mL solnx

2 mol NaOH

1 mol H2SO4

x1000 ml soln

1.420 mol NaOHx25.00 mL = 158 mL

M

acid

rxn

coef.

M

base

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WRITE THE CHEMICAL EQUATION!

volume red moles red moles oxid M oxid

0.1327 mol KMnO4

1 Lx

5 mol Fe2+

1 mol KMnO4

x1

0.02500 L Fe2+x0.01642 L = 0.4358 M

M

red

rxn

coef.

V

oxid

5Fe2+ + MnO4- + 8H+ Mn2+ + 5Fe3+ + 4H2O

16.42 mL of 0.1327 M KMnO4 solution is needed to oxidize 25.00 mL of an acidic FeSO4 solution. What is the molarity of the iron solution?

16.42 mL = 0.01642 L 25.00 mL = 0.02500 L

End Solution Stoichiometry

What you need to know:

• Definitions of solution, solute, solvent, electrolyte, non-electrolyte, hydration.

• Types of aqueous reactions– Precipitation (selected solubility rules/

exceptions)– Neutralization rxns– Reduction – oxidation reactions

• Combination, decomposition, combustion, displacement

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What you need to know:• How to find oxidation numbers

• How to identify substance oxidized, substance reduced, reducing agent, oxidizing agent

• How to write half reactions

• Calculate molarity and dilutions

• Understand theoretical use of gravimetric analysis and titrations and perform accompanying calculations

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What you need to know:

• How to find oxidation numbers

• Interpret activity series

• Calculate molarity and dilutions

• Understand theoretical use of gravimetric analysis and titrations and perform accompanying calculations