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Chapter 4
Aqueous solutions
Types of reactions
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4.3 Composition of Solutions
 Calculate the molarity of a solution with 34.6 g
of NaCl dissolved in 125 mL of solution.
 Convert grams to moles. How? Convert mL
to L. How? Divide moles by liters
 (34.6 g)(1 mol)
(58.5 g)(0.125 L)
 = 4.77 M
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Dilution
 What volume of a 1.7 M solution is needed to
make 250. mL of a 0.500 M solution? Steps.
 V1M1 = V2M2
 (1.7)(x) = (250.)(0.500)
 So, x = 73.5 ml
 Add this to enough distilled H20 to make a
total volume 250. ml
 Be careful! If you just add it to 250. mL then
you get a volume of 323 mL and this throws
off your molarity.
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4.7 Stoichiometry of Precipitation
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?
 Strategy: need to convert both reactants to
moles, then determine LR then use stoich.
 Have 0.01 moles of both
 BaCl2 + 2NaOH  Ba(OH)2 + 2NaCl

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BaCl2 continued
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?
 Have 0.010 moles of both
 BaCl2 + 2NaOH  Ba(OH)2 + 2NaCl
 NaOH is limiting & makes 0.005 mol BaCl2
 0.005 mol BaCl2 ÷ 208 g/mol = 0.85 g formed

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Titration
 Where indicator changes color is endpoint.
 Not always at the equivalence point.
 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 ]?

# of H+ x MA x VA = # of OH- x MB x VB
(unbalanced reaction)
(1)(0.0980)(34.65) = (2)(50)(x) = 0.0340 M
 but look at Zumdahl method page 158!

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Acid-Base Reaction
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75 mL of 0.25 M HCl is mixed with 225 mL of 0.055
M Ba(OH)2. What is the concentration of the excess
H+ or OH- ? We can do this in mmoles since the
units will cancel. Use mL and mmoles is same as L
and moles. Steps. . .
Mmole HCl = 75 mL • 0.25 M = 18.75 mmole;
Since 1:1 mole ratio of H+ : HCl; H1+ = 18.75 mmole
Mmole Ba(OH)2 = 225 mL • 0.055 M = 12.37 mmole
2:1 mole ratio OH- : Ba(OH)2 ; OH- = 24.75 mmole
H : OH mole ratio = 1:1; so 6 mmoles xs OH in (75 +
225) ml = 0.02 M OH1-
Balancing Redox Example
 A breathalyzer uses potassium
dichromate to test for ethanol because
the orange potassium dichromate
changes to the green chromium 3+ ion in
the presence of alcohol.
 Write and balance the Breathalyzer
equation given the following:
 The reactants are K2Cr2O7, HCl and
C2H5OH.
 The products are CrCl3, CO2, KCl & H2O.
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Example
Write and balance the Breathalyzer equation
given the following:
Reactants: K2Cr2O7, HCl and C2H5OH.
Products: CrCl3, CO2, KCl & H2O.
 Write the formula equation:


K2Cr2O7 + HCl + C2H5OH CrCl3 + CO2 + KCl + H2O

Write the ionic equation (solubility poem):

2K1+ Cr2O72- + H1+ + Cl1- + C2H5OH Cr3+ 3Cl1- + CO2 + K1+ + Cl1- + H2O

Assign oxidation numbers

2K1+ Cr2O72- + H1+ + Cl1- + C2H5OH Cr3+ 3Cl1- + CO2 + K1+ + Cl1- + H2O
+1 +6 -2 +1
-1 -2+1 -2+1 +3 -1
+4 -2 +1
-1 +1 -2
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Example

Write the formula equation:
K2Cr2O7 + HCl + C2H5OH CrCl3 + CO2 + KCl + H2O
Write the ionic equation (solubility poem):
2K1+ Cr2O72- + H1+ + Cl1- + C2H5OH Cr3+ 3Cl1- + CO2 + K1+ + Cl1- + H2O
Assign oxidation numbers
2K1+ Cr2O72- + H1+ + Cl1- + C2H5OH Cr3+ + 3Cl1- + CO2 + K1+ + Cl1- + H2O
+1 +6 +2 +1 -1 -2 +1 -2+1 +3 -1
+4 -2 +1
-1 +1 -2

Delete substances where no element changes its
oxidation number:

2K1+ Cr2O72- + H1+ Cl1- + C2H5OH Cr3+ + 3Cl1- + CO2 + K1+ + Cl1- + H2O
+1 +6 -2 +1 -1 -2 +1 -2+1 +3
-1
+4 -2 +1
-1 +1 -2

Only those where oxidation number changes :
Cr2O72- + C2H5OH Cr3+ + CO2
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+6
-2
+3
+4
Balancing Redox Equations

K2Cr2O7 + HCl + C2H5OH CrCl3 + CO2 + KCl + H2O

Only those where oxidation number changes :
Cr2O72- + C2H5OH Cr3+ - + CO2
+6
-2
+3
 For
+4
reactions in acidic solution: 8-step
procedure.
 Write separate half-reactions
Cr2O72- Cr3+
C2H5OH CO2
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Balancing Redox Equations
Cr2O72- Cr3+ C2H5OH CO2
 For each half-reaction balance all
reactants except H and O
Cr2O72- Cr3+
C2H5OH CO2
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Balancing Redox Equations
Cr2O72- Cr3+ C2H5OH CO2
 Balance O using H2O
Cr2O72- Cr3+ + 7H2O
3H2O + C2H5OH CO2
 Balance H using H1+
14H1+ + Cr2O72- Cr3+ + 7H2O
3H2O + C2H5OH CO2 + 12H1+
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Example
14H1+ + Cr2O72- Cr3+ + 7H2O
C2H5OH CO2 + 12H1+
 Balance the charge using e-
3H2O +
6e- + 14H1+ + Cr2O72- Cr3+ + 7H2O
3H2O + C2H5OH CO2 + 12H1+ + 12e Multiply equations to equalize the charges
2(6e- + 14H1+ + Cr2O72- Cr3+ + 7H2O)
3H2O + C2H5OH CO2 + 12H1+ + 12eGet:
12e- + 28H1+ + 2Cr2O72-  Cr3+ + 14H2O
3H2O + C2H5OH CO2 + 12H1+ + 12e14
Example
12e- + 28H1+ + 2Cr2O72-  Cr3+ + 14H2O
3H2O + C2H5OH CO2 + 12H1+ + 12e-
 Add equations and cancel identical
species (cancel coefficients if needed)
12e- + 1628H1+ + 2Cr2O72- Cr3+ + 1114H2O
3H2O + C2H5OH CO2 + 12H1+ + 12e-
• 16H
1+
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+ 2Cr2O72- + C2H5OH CO2 + Cr3+ + 11H2O
Acidic Solution
16H1+ + 2Cr2O72- + C2H5OH CO2 + Cr3+ + 11H2O
 Combine the net ions to form the original
compounds and check that everything is
balanced.
The original compounds were:
K2Cr2O7 + HCl + C2H5OH CrCl3 + CO2 + KCl +
H2O
So the balanced equation is . . .(write it out)
2K2Cr2O7 + 16HCl + C2H5OH CrCl3 + 2CO2 + 4KCl + 11H2O
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Basic Solution
 Do everything you would with acid, but
we have to add a step because . . .
 Because there is no H1+ in basic solution.
 So, add OH1- sufficient to convert the H1+
to water (OH1- + H1+ H2O)
 Be sure to add the OH1- to both sides of
the reaction (conservation of mass law)
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Cr(OH)3 + OCl- + OH- CrO4-2 + Cl- + H2O
we’ll do this on the board to get . . .
2Cr(OH)3 + 3OCl- + 4OH- CrO4-2 + 3Cl- + 5H2O
Redox Titration Example
 The iron content of iron ore can be
determined by titration with standard
KMnO4 solution. The iron ore is dissolved
in excess HCl, and the iron reduced to
Fe+2 ions. This solution is then titrated
with KMnO4 solution, producing Fe+3 and
Mn+2 ions in acidic solution.
 If it requires 41.95 mL of 0.205 M KMnO4
to titrate a solution made with 6.128 g of
iron ore, what percent of the ore was iron?
Steps. . .
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Example
Iron ore + xs HCl  iron reduced to Fe+2. Then titrated w/ KMnO4
solution  Fe+3 + Mn+2 ions in acidic solution.
41.95 mL of 0.205 M KMnO4 titrates 6.128 g of iron ore, what percent is
iron?
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Write NR Fe2+ reacting with MnO41(8H + 5Fe2+ + MnO41- 5Fe3+ + Mn2+ + 4H20)
Do stoich (calc moles MnO41- 1st) to get
0.0086 moles MnO41- + 0.043 moles Fe2+
Find g of Fe2+ = 0.043 mol x 55.847 g/mol = 2.40 g
(2.40 g Fe/6.128 g Fe ore) x 100% = 39.2%
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