Zumdahl 6th Ed, Chapter 4 Sections 1-6.4.1 Water, the Common Solvent4.2 The Nature of Aqueous Solutions: Strong and Weak Electrolytes4.3 The Composition of Solutions4.4 Types of Chemical Reactions

precipitation, acid-base, redox reactions4.5 Precipitation Reactions4.6 Describing Reactions in SolutionUse Molarity to keep track of number of moles; serial dilution.

Problems: 4.10-11, 4.13-18, 4.20-21, 4.23-25, 4.27-28, 4.30-34

Discussion: 4.1, 4.2, 4.4, 4.5, 4.6

Lecture 6: Lec4a Chemical Reactions in solutions

Water: The best known solvent

Because of hydrogen bond formation, water boils at a much higher temperature than CH4 (90 K), which has the same molecular mass (to three sig figs).

Polar and Nonpolar Molecules

Nonpolar: Electrons Equally Shared

Polar: Electrons Unequally Shared

Which of the following molecules is non-polar?

1. HCl2. CHCl33. CO4. BrCl5. NH3

6. None of the above

Dissolve a salt in water

Polar water molecules interact with the positive and negative ions of a salt,

making the dissolving process possible.

Electrical conductivity - The flow of electrical current in a solution is ameasure of the solubility of ionic compounds or a measurement of the presence of ions in solution.

Electrolyte - A substance that conducts a current when dissolved in water. Soluble ionic compound dissociate completely andmay conduct a large current, and are called StrongElectrolytes.

(When Sodium Chloride dissolves into water the ions become solvated,and are surrounded by water molecules. These ions are called “aqueous”and are free to move through out the solution, and are conductingelectricity, or helping electrons to move through out the solution

The Role of Water as a Solvent: The solubility of Ionic Compounds…

) ( ) ( ) ( )2NaCl s H O Na aq Cl aq+ −+ → +

Electrical Conductivity of Ionic Solutions

Completely IonizeStrong Acid (HCl) and Strong Base (NaOH)

( ) ( ) ( )( ) ( ) ( )3 2 n

HCl aq H aq Cl aq

H aq H O aq H O H

+ −

+ + +

→ +

≡ ≡( ) ( ) ( )NaOH aq Na aq OH aq+ −→ +

Partially Ionize in Solution Weak Acid (HAc) and a Weak Base (NH3)

( ) ( ) ( )( ) ( )3

HAc aq H aq Ac aq

Ac aq CH COO aq

+ −

− −

+

≡( ) ( ) ( )3 4NH aq NH aq OH aq+ −→ +

Strong Electrolytes• Produce ions in aqueous solution and conduct

electricity well. • Strong electrolytes are soluble salts, strong acids and

strong bases.• Strong acids produce H+ ions when they dissolve in

water.• HCl, HNO3 and H2SO4 are strong acids

HNO3(aq) → H+(aq) + NO3-(aq)

• NaOH and KOH are strong bases:NaOH(s) → Na+(aq) + OH-(aq) (Produce OH- ions)

All of the above species are ionized nearly 100% in water.

Weak Electrolytes

• Produce relatively few ions in aqueous solution• The most common weak electrolytes are weak acids and

weak bases.• Acetic acid is a typical weak acid:

• Ammonia is a common weak base:

• Weak bases like NH3 and weak acids like acetic acid don’t conduct electricity well; they “partially”ionize.

• These two (acetic acid and ammonia) ionize only ~1%

( ) ( ) ( )3 2 3 2HCH COO aq H aq C H O aq+ −+

( ) ( ) ( ) ( )3 2 4NH aq H O NH aq OH aq+ −+ → +

NonelectrolytesDissolve in water but produce no ions in solution.Nonelectrolytes do not conduct electricity because they

dissolve as whole molecules, and produce no ions. Common nonelectrolytes include ethanol and table

sugar (sucrose, C12H22O11)

Non electrolytes contain polar parts; can hydrgeonbond with water.

Nonpolar covalent compounds can’t form hydrogen bonds and have little or no interactions with water molecules. Examples are the hydrocarbons in gasoline and oil, which don’t mix with water.

Interaction of Water and EthanolA water soluble non-electrolyte, has polar components (COH) that hydrogen-bonds; fits in with water network.

Molecules that contain carbon and water! CxH2yOy or Cx(H2O)y

CH2OHCC

C

CC C

C C

C

CH2OH

CH2OH

O

O

O

H

HH

H

H

HH

HOH

OH

OH

OHHO

Sucrose C12H22O11 , C12(H2O)11 a disaccharideGlucose C6H12O6 , (CH2O)6 a monosaccharide

http://chemistry.about.com/od/chemistrydemonstrations/a/acidsugardemo.htm

Carbohydrates

Sucrose

Problem: How many moles of each ion type are in each of the following:a) 4.0 moles of sodium carbonate dissolved in water

Other examples later:•Both mass and moles are conserved in solutions, even if chemicalforms change.

a) Na2CO3 (s) 2 Na+(aq) + CO3

-2(aq)

moles of Na+ = 4.0 moles Na2CO3 x

= 8.0 moles Na+ and 4.0 moles of CO3-2 are present

H2O

2 mol Na+

1 mol Na2CO3

Determining Moles of Ions in Aqueous Solutions of Ionic Compounds

Dissolve a salt in water• How do you describe it?• Problem: Take 28 grams of NaCl and put

it in 250 mls of solution. • What do you write on the bottle?

• Possibilities?

Figure 4.9: Steps involved in the preparation of a standard solution.

M = =Moles of Solute molLiters of Solution L

The solute is material dissolved into the solvent

In sea water, water is the solvent, and NaCl, MgCl2, etc are the solutes.In brass, copper is the solvent (90%), and zinc is the solute (10%).

Compare with density which is the mass of the solute in the solution.

Molarity (Concentration of Solutions) = M

WW W W

moles Mmass molesd M Molarity M M MVolume Volume Volume

⋅= = = ⋅ = ⋅ = ⋅

Does it matter whether the salt dissociates or not?

Problem: A solution of sodium phosphate is prepared by dissolving 3.95 g of sodium phosphate in water and diluting it to 300.0 mL. What is the molarity, M, of the salt? What about (the M-ity of) each of the ion types?

Strategy?

Prepare a solution

Problem: A solution of sodium phosphate is prepared by dissolving 3.95 g of sodium phosphate in water and diluting it to 300.0 mL. What is the molarity, M, of the salt? What about each of the ions? Density?

Strategy: Think about what you would write on the bottle to describe the contents. The concentration is the quantity that will not change when you pour some out. Concentration can be expressed in moles per liter (Molarity) or grams per liter (Density).

Prepare a solution

( )3 4

3.95 13.20.30

163.94Na POW

mass g gd lVolume LgM mol

= = =

=

Problem: A solution of sodium phosphate is prepared by dissolving 3.95 g of sodium phosphate in water and diluting it to 300.0 mL. What is the molarity, M, of the salt? What about each of the ions? Density?

Strategy: Calculate the Concentration (as Molarity) of the salt.

Prepare a solution

( )

( )

( )

3 4

3 4

3 4

13.20.0803

163.94

3.95 0.0241

0.0241 0.08030.30

Na PO

W

Na PO

W

Na PO

gd lC MgM

molgn mol

Mn molC MV L

= = =

= =

= = =

Strategy: Start with the Molarity of the salt, and the chemical reaction.

Reaction shows: 3 moles Na+ ions per mole of loss of salt.

Assume the rxn goes to completion (no reactant left).

Concentration of Ions

( )

3 41 1 ( )

1 33 3. .

1 1

mole Reaction mole loss Na PO

mole Reaction moles Na ionsmoles Na Nas c Na

mole Reaction Reaction

+

+ ++

=

=

= =

( ) ( )( ) ( )

( ) ( )( ) ( )

3 43 4

343

4 3 43 4

. .3 0.0803 0.241

. .

. .0.0803

. .

s c NaM Na M Na PO M M

s c Na PO

s c POM PO M Na PO M

s c Na PO

++

−−

= ⋅ = ⋅ =

= ⋅ =

( ) ( ) ( )33 4 43Na PO s Na aq PO aq+ −→ +

Dilution: Convert a Stock solution to a required concentration

Dilution Problem• A medical team needs from pharmacy 500mls

isotonic saline (0.14M) for a patient. You have 1 liter of 1.50 M saline (NaCl) solution. What do you (in pharmacy) do?

Dilute using Volumetric Pipets

Dilution Problem• A medical team needs from pharmacy 500mls isotonic saline

(0.14M) for a patient. You have 1 liter of 1.50M saline (NaCl) solution. What do you (in pharmacy) do?

• Need to find the initial volume of the 1.50M NaClneeded to make 500 mls of 0.14M NaCl. Use the fact that the number of moles Is what counts and what is the same (i.e. conserved).

( )( ) 0

0

0

0.14 0.5 0.070

1.501.50 0.14 0.5

0.14 0.5 0.0467 46.71.50

moles NaCl need M L moles

moles NaCl available M VM V M L

M LV L mlsM

= ⋅ =

= ⋅

⋅ = ⋅⋅

= = =

Take 46.7mls of stock solution and dilute to 500 mls and voila!

General Dilution Problem• Need volume • Have molarity . What is ?• General Answer: Moles are conserved so:

1 1@V molarity M

oM oV

1

1 1

o

o o

M Mmoles V M V M

>= ⋅ = ⋅

Rearrange the general expression for Vo.

Can rearrange the general expression for any of the 4 quantities.

Two step serial Dilution• Begin with a 3.5M (stock) solution of NaCl. Take 10 mls

of this solution and dilute to 100 mls with water. Then tak 5 mls of the new solution and dilute to 300 mls with water (as a second serial dilution). What is the concentration of the salt after the second dilution?

Two step serial Dilution• Begin with a 3.5M (stock) solution of NaCl. Take 10 mls

of this solution and dilute to 100 mls with water. Then tak 5 mls of the new solution and dilute to 300 mls with water (as a second serial dilution). What is the concentration of the salt after the second dilution?

Strategy: Note: final concentration means molarity, M

1

2 1

22

2

* 3.5 10 leave volume in mls, n is in millimoles

3.5 10 10: 3.5 First Dilution factor100 100

10* : 5 3.5 5 moles taken from first dilution100

3:

moln M V n mlsl

n molM M lV

n M V n M mls mol

nnM MV V

= = ⋅

⋅ ⎛ ⎞= = = ⋅ ⇐⎜ ⎟⎝ ⎠

⎛ ⎞= = ⋅ = ⋅ ⋅⎜ ⎟⎝ ⎠

= = =

10.5 510 5100 3.5 0.00583

300 100 300mol

l

⎛ ⎞⋅ ⋅⎜ ⎟ ⎛ ⎞ ⎛ ⎞⎝ ⎠ = ⋅ ⋅ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

First Dilution Second Dilution

Serial Dilution as a product of dilution factors

Precipitation ReactionPb(NO3)2 is in beaker (colorless and soluble in water) and NaI is in graduate (also colorless and soluble).

The mixture produces a yellow ppt. which is PbI2.

All 4 ions are in solution. What pairings are possible? Hint: 6 is max. Many can be ruled out.

23Pb NO

Na I

+ −

+ −

Table in Text

Rules 1 and 2, for our purposes all are soluble (not most).

Think Ammonium Nitrate the soluble salt.

More info in chapter 8.

Molecular equation,

Total ionic equation,

and net ionic equation

Predicting if a precipitate forms, and which?

Pb(NO3)2(aq) + NaCl(aq) →Pb+2(aq) + 2 NO3

- (aq) + Na+ (aq) + Cl-(aq)

If any of the possible new species formed by combining anions with cations is insoluble, then that precipitate will form. Salts of sodium and nitrates do not ppt.

USE TABLE 4.1

In this case, PbCl2 is the only salt possibly insoluble, and a precipitate forms.

Note: Like the demo with PbI2(s) above.

End of Lecture

• Extra Infromation available in the EXTRA file also on line.