Chapter 4

Aqueous solutions

Types of reactions

Parts of Solutions

Solution- homogeneous mixture. Solute- what gets dissolved. Solvent- what does the dissolving.

Soluble- can be dissolved. Miscible- liquids dissolve in each other.

Aqueous solutions

Dissolved in water. Water is a good solvent because

the molecules are polar. The oxygen atoms have a partial

negative charge. The hydrogen atoms have a

partial positive charge. The angle is 105º.

Hydration

The process of breaking the ions of salts apart.

Ions have charges and attract the opposite charges on the water molecules.

Hydration

H HO

HH O

HH

O

H HO

HHO

HH

O

HH

OH

H

O

HH

O

Solubility

How much of a substance will dissolve in a given amount of water.• Usually listed in g/100 mL

Solubilities vary greatly, but if they do dissolve the ions are separated, and they can move around.

Water can also dissolve non-ionic compounds if they have polar bonds.

Electrolytes

Electricity is moving charges. The ions that are dissolved can move. Solutions of ionic compounds can

conduct electricity and are therefore called “electrolytes.”

Types of solutions

Solutions are classified three ways: • Strong electrolytes- completely dissociate (fall

apart into ions).• Many ions- Conduct well.

• Weak electrolytes- Partially fall apart into ions.• Few ions -Conduct electricity slightly.

• Non-electrolytes- Don’t fall apart.• No ions- these sol’ns don’t conduct.

Types of solutions

Acids- form H+ ions when dissolved.• Strong acids fall apart completely.

• many ions

• There are just 6 strong acids – MEMORIZE THEM!

• Sulfuric acid: H2SO4

• Nitric acid: HNO3

• Hydrochloric acid: HCl

• Hydrobromic acid: HBr

• Hydroiodic acid: HI

• Perchloric acid: HClO4

• Weak acids- don’t dissociate completely.

Types of solutions

Bases - form OH- ions when dissolved.• Strong bases- many ions. Most common:

• Lithium hydroxide:  LiOH 

• Sodium hydroxide:  NaOH 

• Potassium hydroxide:  KOH 

• Rubidium hydroxide:  RbOH 

• Calcium hydroxide:  Ca(OH)2 

• Strontium hydroxide:  Sr(OH)2 

• Barium hydroxide:  Ba(OH)2

MOM to the rescue!Milk of magnesia = aqueous suspension of magnesium hydroxide

Is Mg(OH)2 a strong base?

Yes, technically… since all of the dissolved magnesium hydroxide dissociates.

However, it has a very low solubility in water.

Measuring Solutions

Concentration- how much is dissolved.

1 M = 1 mol solute / 1 liter solution

nsol' Lsolute mol

MMolarity

What is this???

A one molar solution!!!

Ha ha ha ha ha ha!

1 L

Molarity Practice Problems

Calculate the molarity of a solution with 34.6 g of NaCl dissolved in 125 mL of solution.

How many grams of HCl would be required to make 50.0 mL of a 2.7 M solution?

What would the concentration be if you used 27g of CaCl2 to make 500. mL of solution?• What is the concentration of each ion?

Calculate the molarity of a solution with 34.6 g of NaCl dissolved in 125 mL of solution.

nsol' Lsolute mol

Molarity

L 0.125

g 44.58mol 1NaCl g 34.6

Molarity

M 74.4Molarity

How many grams of HCl would be required to make 50.0 mL of a 2.7 M solution?

HCl mol 1

HCl g 36.46

nsol' Lsolute mol

Molarity

L 0.0500HCl mol

M 7.2

mol 513.0HCl mol HCl g 92.4

What would the concentration be if you used 27g of CaCl2 to make 500. mL of solution? What is the concentration of each ion?

L 0.500

g 10.981mol 1CaCl g27

Molarity

2

2CaClMolarity 0.49 M

nsol' Lsolute mol

Molarity

[Ca2+]=0.49M

[Cl-]=0.98M

Why?

Because for every one CaCl2 there are two Cl- and one Ca2+… thus for every 1 mol CaCl2, 1 mole of Ca2+ and 2 moles of Cl- form.

Molarity Practice Problems

Calculate the concentration of a solution made by dissolving 45.6 g of Fe2(SO4)3 to 475 mL.

• What is the concentration of each ion?

L 0.475

g 399.88mol 1)(SOFe g 45.6

)(SOFe of Molarity

342

342

240M.0

[Fe2+]=

[SO42-]=

2(.240 M)=.480 M3(.240 M)=.720 M

This is because for every 1 Fe2(SO4)3 particle there are 2 iron ions

and 3 sulfate ions

Practice Problems – Making Solutions

Describe how to make 100.0 mL of a 1.0 M K2Cr2O4 solution.

mol 1

g 246.2

L 0.1000

OCrK molM .01 422

mol 10.0OCrK mol 422 g 24

Answer: Measure out 24 g of K2Cr2O4 in a volumetric flask. Add enough distilled water to dissolve the solid. Dilute to a total solution volume of 100.0 mL.

Practice Problems – Making Solutions

Describe how to make 250. mL of an 2.0 M copper (II) sulfate dihydrate solution.

mol 1

g 195.61

L 0.250

O2HCuSO molM .02 24

mol 50.0O2HCuSO mol 24 g 98

Answer: Measure out 98 g of CuSO4•2H2O in a volumetric flask. Add enough distilled water to dissolve the solid. Dilute to a total solution volume of 250. mL.

Dilution

Adding more solvent to a known solution. The moles of solute stay the same.

•M1 V1 = M2 V2

Stock solution is a solution of known concentration used to make more dilute solutions

Dilution Practice Problems

What volume of a 1.7 M solutions is needed to make 250 mL of a 0.50 M solution?

M1V1 = M2V2

(1.7 M)(V1) = (0.50 M)(250 mL)

V1 = 74 mL

Dilution Practice Problems

18.5 mL of 2.3 M HCl is added to 250 mL of water. What is the concentration of the resulting solution?

M1V1 = M2V2

(2.3 M)(18.5 mL) = (M2)(250 mL+18.5 mL)

M2 = 0.16 M

Dilution Practice Problems

You have a 4.0 M stock solution. Describe how to make 1.0L of a .75 M solution.

M1V1 = M2V2

(4.0 M)(V1) = (0.75 M)(1.0 L)

V1 = 0.19 L = 190 mL

Measure out 190 mL of 4.0 M stock solution. Dilute with enough water to make 1.0 L solution

Dilution Practice Problems

25 mL 0.67 M of H2SO4 is added to 35

mL of 0.40 M CaCl2 . What mass CaSO4

is formed? H2SO4 + CaCl2 CaSO4 + 2HCl

L 0.025L

mol0.67 SOH mol 42 mol 017.0

L 0.035L

mol 0.40CaCl mol 2 mol 014.0

Dilution Practice Problems

25 mL 0.67 M of H2SO4 is added to 35

mL of 0.40 M CaCl2 . What mass CaSO4

is formed? H2SO4 + CaCl2 CaSO4 + 2HCl

2 4 44

2 4

0.014 mol CaCl 1 mol CaSO 136.14 g CaSO1.9 g CaSO

1 mol CaCl 1 mol CaSO

0.017mol 0.014mol

LR

Types of Reactions

Precipitation reactions• When aqueous solutions of ionic compounds

are poured together a solid forms.

• A solid that forms from mixed solutions is a precipitate

• If you’re not a part of the solution, your part of the precipitate! LOL! :-D

Precipitation reactions

Sample reaction… Overall (molecular) eq’n:

3NaOH(aq) + FeCl3(aq) NaCl(aq) + Fe(OH)3(s)

The above rxn is really… Total ionic eq’n:

3Na+(aq)+3OH-(aq)+Fe+3(aq) + 3Cl-(aq) Na+(aq) + 3Cl-(aq) +Fe(OH)3(s)

So all that really happens is… Net ionic eq’n:

3OH-(aq) + Fe+3(aq) Fe(OH)3(s)

(Double replacement reaction)

Precipitation reaction

We can predict the products Can only be certain by experimenting The anion and cation switch partners

Practice Problems – Predicting Products of Precipitation Rxns

AgNO3(aq) + KCl(aq)

Zn(NO3)2(aq) + BaCr2O7(aq)

CdCl2(aq) + Na2S(aq)

KNO3( ) + AgCl( ) Ag+ NO3

- K+ Cl-

ZnCr2O7( ) + Ba(NO3)2( ) Zn2+ NO3

-Ba2+ Cr2O7

2-

Cd2+ Cl- Na+ S2-

2NaCl( ) + CdS( )

Precipitations Reactions

Only happen if one of the products is insoluble

Otherwise all the ions stay in solution- nothing has happened.

Need to memorize the rules for solubility! • See Solubility Rules Handout!

Solubility Rules

1) All nitrates are soluble

2) Alkali metals ions and NH4+ ions are soluble

3) Halides are soluble except Ag+, Pb+2, and Hg2+2

4) Most sulfates are soluble, except Pb+2, Ba+2, Hg+2,and Ca+2

5) Most hydroxides are slightly soluble (insoluble) except NaOH and KOH

6) Sulfides, carbonates, chromates, and phosphates are insoluble

Lower number rules supersede so Na2S is soluble

Practice Problems – Predicting Products of Precipitation Rxns

AgNO3(aq) + KCl(aq)

Zn(NO3)2(aq) + BaCr2O7(aq)

CdCl2(aq) + Na2S(aq)

KNO3( ) + AgCl( ) Ag+ NO3

- K+ Cl-

ZnCr2O7( ) + Ba(NO3)2( )

2NaCl( ) + CdS( )

Zn2+ NO3-

Ba2+ Cr2O72-

Cd2+ Cl- Na+ S2-

aq s

s aq

aq s

Three Types of Equations

Molecular Equation- written as whole formulas, not the ions.• K2CrO4(aq) + Ba(NO3)2(aq) BaCrO4(s) + 2KNO3(aq)

Complete Ionic Equation shows dissolved electrolytes as the ions.• 2K+ + CrO4

-2 + Ba+2 + 2 NO3- BaCrO4(s) + 2K+ + 2 NO3

-

• Spectator ions are those that don’t react.

Net Ionic equation shows only those ions that react, not the spectator ions• Ba+2 + CrO4

-2 BaCrO4(s)

Practice Problem

Write the three types of equations for the reactions when these solutions are mixed.• iron (III) sulfate and potassium sulfide

• lead (II) nitrate and sulfuric acid

iron (III) sulfate and potassium sulfide

Fe2(SO4)3(aq) + 3K2S(aq) Fe2S3(s) + 3K2SO4(aq)

2Fe3+(aq) + 3SO42-(aq) + 6K+(aq) + 3S2-(aq)

Fe2S3(s) + 6K+(aq) + 3SO42-(aq)

2Fe3+(aq) + 3S2-(aq) Fe2S3(s)

lead (II) nitrate and sulfuric acid

Pb(NO3)2(aq) + H2SO4(aq) PbSO4(s) + 2HNO3(aq)

Pb2+(aq) + 2NO3-(aq) + 2H+ + SO4

2-(aq)

PbSO4(s) + 2H+(aq) + 2NO3-

(aq)

Pb2+(aq) + SO42-(aq) PbSO4(s)

Stoichiometry of Precipitation-Practice Problems

Exactly the same, except you may have to figure out what the pieces are.

• What mass of solid is formed when 100.00 mL of 0.100 M barium chloride is mixed with 100.00 mL of 0.100 M sodium hydroxide?

• What volume of 0.204 M HCl is needed to precipitate the silver from 50.ml of 0.0500 M silver nitrate solution ?

Stoichiometry of Precipitation-Practice Problems

What mass of solid is formed when 100.00 mL of 0.100 M barium chloride is mixed with 100.00 mL of 0.100 M sodium hydroxide?

BaCl2(aq) + 2NaOH(aq) Ba(OH)2(s) + 2NaCl(aq)

0.01 mol 0.01 mol LR

2

22

Ba(OH) mol 1

Ba(OH) g 171.35

NaOH mol 2

Ba(OH) mol 1NaOH mol 0.01 2Ba(OH) g0.857

Stoichiometry of Precipitation-Practice Problems

• What volume of 0.204 M HCl is needed to precipitate the silver from 50.ml of 0.0500 M silver nitrate solution ?

HCl(aq) + AgNO3(aq) HNO3(aq) + AgCl(s)

? mL 0.0025 mol

3

3

AgNO mol 1HCl mol 1AgNO mol 0.0025

HCl mol 0025.0

L x0.0025mol

M 420.0 12mL.012Lvolume

Types of Reactions

Acid-Base For our purposes an acid is a proton

donor. a base is a proton acceptor usually OH-

What is the net ionic equation for the reaction of HCl(aq) and KOH(aq)?

Acid + Base salt + water H+ + OH- H2O

Acid - Base Reactions

Often called a neutralization reaction Because the acid neutralizes the base.

Often titrate to determine concentrations. Solution of known concentration (titrant) is

added to the unknown (analyte) until the equivalence point is reached where enough titrant has been added to neutralize it.

Titration

Where the indicator changes color is the endpoint.

Not always exactly at the equivalence point, but should pick an indicator that changes color very close to the endpoint.

# of H+ x MA x VA = # of OH- x MB x VB

or

NA x VA = NB x VB

Titration Practice Problems

A 50.00 mL sample of aqueous Ca(OH)2 requires

34.66 mL of 0.0980 M Nitric acid for neutralization.

What is [Ca(OH)2 ]?

NaVa = NbVb

(0.0980N)(34.66mL)=(Nb)(50.00mL)

Nb=0.06793

NCa(OH)2=(M)(2)

[Ca(OH)2]=0.03397M

Titration Practice Problems 75 mL of 0.25M HCl is mixed with 225 mL of 0.055 M

Ba(OH)2. What is the concentration of the excess H+ or OH-?

H mol 0.01875L 0.075L

mol 0.25

-OH mol 0.02475L 0.225L

mol 0.110

[H+] = 0.25 M [OH-] = 0.110 M

Excess [OH-] = 0.02475 mol – 0.01875 mol = 0.00600 mol

L) .225L (.075mol 0.00600

]OH[ excess M 0.000200

Types of Reactions

Oxidation-Reduction (“Redox”) Ionic compounds are formed through the

transfer of electrons An oxidation-reduction rxn involves the

transfer of electrons We need a way of keeping track –

oxidation states

Oxidation States

A way of keeping track of the electrons. Not necessarily true of what is in nature,

but it works. need the rules for assigning

• memorize these!

Rules for assigning oxidation states

The oxidation state of an atom in an element is zero (i.e. Na(s), O2(g), O3(g), Hg(l))

Oxidation state for monoatomic ions are the same as their charge. (i.e. Na+, Cl-)

Oxygen is assigned an oxidation state of -2 in its covalent compounds except as a peroxide (such as H2O2).

In compounds with nonmetals hydrogen is assigned the oxidation state +1.

In its compounds fluorine is always –1. The sum of the oxidation states must be zero in

compounds or equal the charge of the ion.

Oxidation States PracticeAssign the oxidation states to each element in the following.

•CO2

•NO3-

•H2SO4

•Fe2O3

•Fe3O4

-2

-2

-2

-2

-2

+4

+5

+6

-2

+1

+3

+8/3

Oxidation-Reduction

Electrons are transferred, so the oxidation states change.• 2Na + Cl2 2NaCl

• CH4 + 2O2 CO2 + 2H2O

OIL RIG• Oxidation is the loss of electrons.

• Reduction is the gain of electrons. LEO GER

• Losing electrons - oxidation

• Gaining electrons - reduction

0 0 +1 -1

-4 +1 0 +4 -2 +1 -2

Oxidation-Reduction

Oxidation means an increase in oxidation state - lose electrons.

Reduction means a decrease in oxidation state - gain electrons.

The substance that is oxidized is called the reducing agent.

The substance that is reduced is called the oxidizing agent.