the nature of aqueous solutions and molarity and solution stoichiometry n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
The Nature of Aqueous Solutions and Molarity and Solution Stoichiometry PowerPoint Presentation
Download Presentation
The Nature of Aqueous Solutions and Molarity and Solution Stoichiometry

Loading in 2 Seconds...

play fullscreen
1 / 25

The Nature of Aqueous Solutions and Molarity and Solution Stoichiometry - PowerPoint PPT Presentation


  • 112 Views
  • Uploaded on

Lecture #10. The Nature of Aqueous Solutions and Molarity and Solution Stoichiometry. Chemistry 142 B James B. Callis, Instructor Autumn Quarter, 2004. The Role of Water as a Solvent: The solubility of Ionic Compounds.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'The Nature of Aqueous Solutions and Molarity and Solution Stoichiometry' - evangelia


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
the nature of aqueous solutions and molarity and solution stoichiometry

Lecture #10

The Nature of Aqueous Solutionsand Molarity and Solution Stoichiometry

Chemistry 142 B

James B. Callis, Instructor

Autumn Quarter, 2004

slide3

The Role of Water as a Solvent:

The solubility of Ionic Compounds

Electrical conductivity - The flow of electrical current in a solution is a

measure 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 and

may conduct a large current, and are called Strong

Electrolytes.

NaCl(s)+ H2O(l) Na+(aq) + Cl -(aq)

When sodium chloride dissolves in water the ions become solvated,

and are surrounded by water molecules. These ions are called “aqueous”

ions and are free to move throughout the solution, and can conduct

electricity.

strong electrolytes
Strong Electrolytes

Produce many ions in aqueous solution and conduct electricity well.

The most common strong electrolytes are soluble salts, strong acids and strong bases.

Acids are substances that produce H+ ion when they dissolve in water.

HCl, HNO3 and H2SO4 are common strong acids HNO3(aq) -> H+(aq) + NO3-(aq)

NaOH and KOH are a common stong bases:

NaOH(s) -> Na+(aq) + OH-(aq)

All of the above species are ionized nearly 100%

weak electrolytes
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:

HC2H3O2(aq) -> H+(aq) + C2H3O2-(aq)

Ammonia is a common weak base:

NH3(aq) + H2O(l) -> NH4+(aq) + OH-(aq)

Both of these species are ionized only 1%

nonelectrolytes
Nonelectrolytes

Dissolve in water but produce no ions in solution.

Nonelectrolytes do not conduct electricity because they dissolve as whole molecules, not ions.

Common nonelectrolytes include ethanol and table sugar (sucrose, C12H22O11)

slide11

Molarity (Concentration of Solutions) = M

Moles of Solute mol

Liters of Solution L

M = =

solute = material dissolved into the solvent

In air, nitrogen is the solvent and oxygen, carbon dioxide, etc.

are the solutes.

In sea water, water is the solvent, and salt, magnesium chloride, etc.

are the solutes.

In brass , copper is the solvent (90%), and Zinc is the solute (10%).

slide13

Determining Moles of Ions in Aqueous

Solutions of Ionic Compounds - I

Problem 10-1: How many moles of each ion are in each of the following:

a) 4.0 moles of sodium carbonate dissolved in water

b) 46.5 g of rubidium fluoride dissolved in water

c) 5.14 x 1021 formula units of iron (III) chloride dissolved in water

d) 75.0 mL of 0.56 M scandium bromide dissolved in water

e) 7.8 moles of ammonium sulfate dissolved in water

a) Na2CO3(s) 2 Na+(aq) + CO32-(aq)

H2O

2 mol Na+

1 mol Na2CO3

slide14

Determining Moles of Ions in Aqueous

Solutions of Ionic Compounds - II

H2O

b) RbF(s) Rb+ (aq)+ F - (aq)

moles of RbF =

H2O

c) FeCl3(s) Fe3+ (aq) + 3 Cl - (aq)

moles of FeCl3 = 9.32 x 1021 formula units x

moles of Cl - =

moles Fe3+ =

slide15

Determining Moles of Ions in Aqueous

Solutions of Ionic Compounds - III

H2O

d) ScBr3(s) Sc3+(aq) + 3 Br -(aq)

Converting from volume to moles:

1 L

103 mL

Moles of ScBr3 = 75.0 mL x x

Moles of Br - =

H2O

e) (NH4)2SO4 (s) 2 NH4+ (aq) + SO42- (aq)

2 mol NH4+

1 mol(NH4)2SO4

Moles of NH4+ = 7.8 moles (NH4)2SO4 x = 15.6 mol NH4+

and 7.8 mol SO42-are also present.

problem 10 2 preparing a solution 1
Problem 10-2: Preparing a Solution - 1

Problem 10-2: Prepare a solution of Sodium Phosphate by dissolving 3.95g of Sodium Phosphate into water and diluting it to 300.0 ml or 0.300 L.

What is the Molarity of the salt and each of the ions?

Na3PO4 (s) + H2O(l) = 3 Na+(aq) + PO43-(aq)

problem 10 2 preparing a solution 2
Problem 10-2: Preparing a Solution - 2

Molar mass of Na3PO4 = g / mol

mol Na3PO4 =

dissolve and dilute to 300.0 mL = volume of solution

M in Na3PO4(aq) =

M in PO43- ions =

M in Na+ ions =

problem 10 3 dilution of solutions
Problem 10-3: Dilution of Solutions

Take 25.00 mL of the 0.0400 M KMnO4

Dilute the 25.00 mL to 1.000 L. - What is the resulting molarity (M) of the diluted solution?

# moles KMnO4 = Vol1 x C1 =

C2 = final M KMnO4 = # moles / Vol2 =

Note: V1 x C1 = moles solute = V2 x C2

slide19

Problem 10-4: Calculating Amounts of Reactants and

Products for a Reaction in Solution

Al(OH)3 (s) + 3 HCl(aq) 3 H2O (l) + AlCl3 (aq)

Problem: Given 10.0 g Al(OH)3(s), what

volume of 1.50 M HCl(aq) is required to

neutralize the base?

Mass (g) of Al(OH)3

÷M (g/mol)

10.0 g Al(OH)3

Moles of Al(OH)3

x molar ratio

Moles of HCl

÷ M (mol/L)

Volume (L) of HCl

slide20

Problem 10-5: Solving Limiting Reactant

Problems in Solution - Precipitation

Problem: Lead has been used as a glaze for pottery for years, and can be

a problem if not fired properly in an oven, and is leachable from the

pottery. Vinegar is used in leaching tests, followed by lead precipitated

as a sulfide. If 257.8 ml of a 0.0468 M solution of lead nitrate is added

to 156.00 ml of a 0.095 M solution of sodium sulfide, what mass of

solid lead sulfide will be formed?

Plan: This is a limiting-reactant problem because the amounts of two

reactants are given. After writing the balanced equation, determine the

limiting reactant, then calculate the moles of product. Convert moles of

product to mass of the product using the molar mass.

Solution: Write the formulas for each ionic compound using the names

of ions and their charges in Tables 2.3-2.5. Write the balanced equation:

Pb(NO3)2 (aq) + Na2S(aq) PbS(s) + 2 NaNO3 (aq)

slide21

Volume

and Conc.

Of Pb(NO3)2

solution

Volume

and Conc.

of Na2S

solution

RR of Na2S

RR of Pb(NO3)2

RRmin x stoich coeff for PbS

= Amount (mol)of PbS

Mass (g) of PbS

slide22

Problem 10-5: Solving Limiting Reactant

Problems in Solution – Precipitation, cont.

RRPb(NO3)2 = (V x C)/stoich coeff. =

RRNa2S = (V x C) /stoich coeff. =

(stoich. coeff. = 1 for both.)

Therefore is the Limiting Reactant!

Calculation of product yield:

Moles PbS =

Mass of PbS =

slide23

Problem 10-6: Another Limiting Reactant

Problem in Solution - Precipitation

Problem: When aqueous silver nitrate and sodium chromate solutions

are mixed, a reaction occurs that forms solid silver chromate and a

solution of sodium nitrate. If 257.8 ml of a 0.0468 M solution of silver

nitrate is added to 156.00 ml of a 0.095 M solution of sodium chromate,

what mass of solid silver chromate (M = 331.8 g/mol) will be formed?

Plan: It is a limiting-reactant problem because the amounts of two

reactants are given. After writing the balanced equation, determine the

limiting reactant, then calculate the moles of product. Convert moles of

product to mass of the product using the molar mass.

Solution: Write the formulas for each ionic compound using the names

of ions and their charges in Tables 2.3-2.5. Write the balanced equation:

Tables=> Ag ion = , nitrate = , Na ion = , chromate =

Therefore balanced reaction is:

AgNO3(aq) + Na2CrO4(aq) Ag2CrO4(s) + NaNO3(aq)

slide24

Problem 10-6: Another Limiting Reactant

Problem in Solution – Precipitation, cont.

• RRAgNO3= V x C/reac. coeff. =

• RRNa2CrO4= V x C / reac. Coeff. =

__________________ is the Limiting Reactant

Calculation of product yield:

Mass Ag2CrO4 = RRmin x reac. coeff. x MM

answers to problems in lecture 10
Answers to Problems in Lecture #10
  • (a) 8.0 moles Na+ and 4.0 moles of CO32- , (b) 0.445 mol Rb+ and 0.445 mol F -, (c) 0.0465 mol Cl – and 0.0155 mol Fe3+, (d) 0.126 mol Br –, (e) 15.6 mol NH4+ and 7.8 mol SO42-
  • 0.0803 M in PO43- ions, 0.241 M in Na+ ions
  • 0.00100 M
  • 256 mL
  • 2.89 g PbS
  • 2.00 g Ag2CrO4