Transcript The Nature of Aqueous Solutions and Molarity and Solution Stoichiometry Chemistry 142 B

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
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.
How Water Dissolves an Ionic
Substance
Interaction of Water and Ethanol
Electrical Conductivity of Ionic Solutions
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
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
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)
Molarity (Concentration of Solutions) = M
M=
Moles of Solute =
Liters of Solution
mol
L
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%).
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)
b)
c)
d)
e)
4.0 moles of sodium carbonate dissolved in water
46.5 g of rubidium fluoride dissolved in water
5.14 x 1021 formula units of iron (III) chloride dissolved in water
75.0 mL of 0.56 M scandium bromide dissolved in water
7.8 moles of ammonium sulfate dissolved in water
a) Na2CO3(s)
H2O
2 Na+(aq) + CO32-(aq)
2 mol Na+
1 mol Na2CO3
Determining Moles of Ions in Aqueous
Solutions of Ionic Compounds - II
b)
RbF(s)
H2O
Rb+ (aq) + F - (aq)
moles of RbF =
c)
FeCl3(s)
H2 O
Fe3+ (aq) + 3 Cl - (aq)
moles of FeCl3 = 9.32 x 1021 formula units x
moles of Cl - =
moles Fe3+ =
Determining Moles of Ions in Aqueous
Solutions of Ionic Compounds - III
d) ScBr3(s)
H2O
Sc3+(aq) + 3 Br -(aq)
Converting from volume to moles:
Moles of ScBr3 = 75.0 mL x 13L x
10 mL
Moles of Br - =
e) (NH4)2SO4 (s)
H2O
2 NH4+ (aq) + SO42- (aq)
+
2
mol
NH
4
Moles of NH4 = 7.8 moles (NH4)2SO4 x
= 15.6 mol NH4+
1 mol(NH4)2SO4
+
and 7.8 mol SO42- are also present.
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
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
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
Problem 10-4: Calculating Amounts of Reactants and
Products for a Reaction in Solution
Al(OH)3 (s) + 3 HCl (aq)
Mass (g) of Al(OH)3
÷M (g/mol)
Moles of Al(OH)3
x molar ratio
Moles of HCl
÷ M (mol/L)
Volume (L) of HCl
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?
10.0 g Al(OH)3
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)
Volume
and Conc.
Of Pb(NO3)2
solution
Volume
and Conc.
of Na2S
solution
RR of Pb(NO3)2
RR of Na2S
RRmin x stoich coeff for PbS
= Amount (mol)of PbS
Mass (g) of PbS
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
Calculation of product yield:
Moles PbS =
Mass of PbS =
is the Limiting Reactant!
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)
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
1. (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 SO422. 0.0803 M in PO43- ions, 0.241 M in Na+ ions
3. 0.00100 M
4. 256 mL
5. 2.89 g PbS
6. 2.00 g Ag2CrO4