Transcript No Slide Title

Reactions in Aqueous Solutions
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
A solution is a homogenous mixture of 2
or more substances.
Solvent: the substance present in the larger amount.
Solute: the substance present in the smaller amount.
Solution
Solvent
Solute
Soft drink (l)
H2O
Sugar, CO2
Milk (l)
H2O
fat, protein, sugar
Air (g)
N2
O2, Ar, CH4
Steel (s)
Fe
C
aqueous solutions of
KMnO4
An electrolyte is a substance that, when dissolved in
water, results in a solution that can conduct electricity.
A nonelectrolyte is a substance that, when dissolved,
results in a solution that does not conduct electricity.
Strong electrolyte
Weak electrolyte
Nonelectrolyte
Why do they conduct electricity in solution?
Ionic Compounds dissociate (break apart) to
form ions when dissolved in water
NaCl (s)
H2O
Na+ (aq) + Cl- (aq)
Crystal Lattice
Ions can pass along (conduct) free electrons (electricity)
Hydration is the process in which an ion is surrounded
by water molecules arranged in a specific manner.
+
www.youtube.com/watch?v=AN4KifV12DA
Water is polar
High e- density near oxygen (-)
d-
d+
Low e- density near hydrogen (+)
-
Strong Electrolyte – 100% dissociation
CaCl2 (s)
(NH4)2S (s)
H2O
H2O
Ca+2 (aq) + 2Cl- (aq)
2NH4+ (aq) + S-2 (aq)
Weak Electrolyte – do not completely dissociate
CH3COOH
~95%
CH3COO- (aq) + H+ (aq)
~5%
A reversible reaction. The reaction occurs in both directions.
Acetic acid is a weak electrolyte because its ionization
in water is incomplete.
6
Nonelectrolytes do not conduct electricity
No cations (+) and anions (-) form in solution
C6H12O6 (s)
H2O
C6H12O6 (aq)
(Molecules)
Precipitation Reactions
Precipitate – insoluble solid that separates from solution
Aqueous: soluble
Pb(NO3)2 (aq) + 2NaI (aq)
Precipitate: insoluble solid
PbI2 (s) + 2NaNO3 (aq)
Chemical equation
Pb2+ + 2NO3- + 2Na+ + 2I-
PbI2 (s) + 2Na+ + 2NO3-
ionic equation
Pb2+ + 2IPbI2
PbI2 (s)
net ionic equation
Na+ and NO3- are spectator ions
8
Precipitation of Lead Iodide
2NaI (aq) + Pb(NO3)2 (aq)
Pb2+ + 2I-
PbI2 (s)
Na+ + 2I- + Pb2+ + 2NO3-
Na+
Na+
PbI2
Ions precipitate due to a stronger electrostatic
attraction than forces from H2O
Large charge & Small ion size lead to greater
attraction and make them less soluble.
Examples of Insoluble Compounds
CdS
PbS
Ni(OH)2
Al(OH)3
10
Solubility is the maximum amount of solute that will dissolve in a
given quantity of solvent at a specific temperature.
Most are single charged
Smaller charges form weaker ionic
bonds and are easier to separate
and NH4+ ions
11
Example Soluble (aq) or Insoluble (s)
Classify the following ionic compounds as soluble or insoluble using
the Solubility chart
(a) silver sulfate (Ag2SO4)
(a) Ag2SO4 (s) is insoluble.
(b) calcium carbonate (CaCO3)
(b) This is a carbonate and Ca is a Group 2A
metal. Therefore, CaCO3 (s) is insoluble.
(c) sodium phosphate (Na3PO4).
(c) Sodium is an alkali metal (Group 1A) so
Na3PO4 (aq) is soluble.
(d) Barium Hydroxide (Ba(OH)2)
(e) Lead (II) Iodide (PbI2)
(d) Hydroxides are normally insoluble, but
Barium is one of the exceptions along with all
alkali metals making it soluble (aq).
(e) Iodides are normally soluble, but it is
insoluble (s) because bound to lead
Writing Net Ionic Equations
1. Write the balanced molecular equation with phase states.
Swap the anions of reactants to get products. (re-criss-cross)
Al2(SO4)3(aq) + 6NaOH(aq) → 2Al(OH)3(s) + 3Na2SO4(aq)
*Do not carry the subscripts over (except for polyatomic ions)
2. Write the ionic equation showing the soluble product dissociated
into ions, but the insoluble product left together.
2Al+3 + 3SO4-2 + 6Na+ + 6OH- → 2Al(OH)3(s) + 6Na+ + 3SO4-2
3. Cancel the spectator ions on both sides of the ionic equation.
6Na+ + 3SO4-2
4. Check that charges and number of atoms are balanced
2Al+3 (aq) + 6OH-(aq) → 2Al(OH)3(s)
Al+3 (aq) + 3OH-(aq) → Al(OH)3(s)
Example Net Ionic Equation
Predict what happens when potassium phosphate (K3PO4)
solution is mixed with calcium nitrate [Ca(NO3)2] solution.
Write the net ionic equation.
*Swap anions
K+ + NO3-1
Ca2+ +
Balance the reactants and products to get the chemical equation
Use Solubility chart to determine (s) or (aq) for each
Example Solution
Step 2: Separate the soluble (aq) compounds to their ions for the ionic
equation. Leave the insoluble (s) precipitate together
*Remember: Ions have charges, Show them (K ≠ K+)
Step 3: Cancel the spectator ions (K+ and NO3-1) on each side of the
equation, we obtain the net ionic equation:
Writing Net Ionic Equations Practice
Pb(NO3)2 and LiCl
Al2(SO4)3 and NaBr
Na2S and Ba(OH)2
Li2CO3 and Ca(NO3)2
Al2(SO4)3 and NH4OH
H3PO4 and Ba(OH)2
Precipitation Reactions: Crash Course
Chemistry #9
www.youtube.com/watch?v=IIu16dy3ThI
Argyria:
Silver
ingestion
and NH4+ ions
and NH4+ ions
Chemistry In Action:
An Undesirable Precipitation Reaction
Ca2+ (aq) + 2HCO3-1 (aq)
CaCO3 (s) + CO2 (aq) + H2O (l)
Calcium Carbonate
build up
The fix?
HCl reacts to form
soluble CaCl2
CaCO3(s) + 2HCl →
CaCl2(aq) + H2O + CO2
18
Properties of Acids
Have a sour taste. Vinegar owes its taste to acetic acid. Citrus
fruits contain citric acid.
Cause color changes in plant dyes.
React with certain metals to produce hydrogen
gas.
2HCl (aq) + Mg (s)
MgCl2 (aq) + H2 (g)
React with carbonates and bicarbonates
to produce carbon dioxide gas.
2HCl (aq) + CaCO3 (s)
CaCl2 (aq) + CO2 (g) + H2O (l)
Aqueous acid solutions conduct electricity.
19
Properties of Bases
Have a bitter taste.
Feel slippery. Many soaps contain bases.
Cause color changes in plant dyes.
Aqueous base solutions conduct electricity.
Examples:
20
Arrhenius acid is a substance that produces H+ (H3O+) in water.
Arrhenius base is a substance that produces OH- in water.
21
Arrhenius Acid/Base definition is exclusive
to aqueous solutions
Hydronium ion, hydrated proton, H3O+
Greater electron
density
O
H
H
H
Lower electron density
A Brønsted acid is a proton (H+) donor
A Brønsted base is a proton acceptor
“proton” - the H+ ion consists of only a single proton
Bronsted definition is not exclusive to aqueous solutions
base
acid
acid
base
A Brønsted acid must contain at least one ionizable Hydrogen!
Ionizable – able to come off
Monoprotic acids (1 H+)
HCl
H+ + Cl-
HNO3
CH3COOH
H+ + NO3H+ + CH3COO-
Strong electrolyte, strong acid
Strong electrolyte, strong acid
Weak electrolyte, weak acid
Diprotic acids (2 H+)
H2SO4
H+ + HSO4-
Strong electrolyte, strong acid
HSO4-
H+ + SO42-
Weak electrolyte, weak acid
Triprotic acids (3 H+)
H3PO4
H2PO4HPO42-
H+ + H2PO4H+ + HPO42H+ + PO43-
Weak electrolyte, weak acid
Weak electrolyte, weak acid
Weak electrolyte, weak acid
24
Strong Acid (HCl)
Complete
dissociation
Weak Acid (HF)
Partial
dissociation
TedEd: The strengths and weaknesses of acids and base
www.youtube.com/watch?v=DupXDD87oHc
Generally follow the same solubility
rules for other ionic compounds.
Completely soluble would indicate
strong acid (unless organic acid)
*Most acids tend to be weak with the
short list of strong acids shown
Organic acids possess the Carboxylic
acid functional group and are weak
H
H
C
H
O
C
Carboxylic
OH
Acid
Example Acid or Base
Classify each substance as a Brønsted acid or base in water:
(a) HBr
Brønsted acid
(b)
Brønsted base
(c)
(d) C2H5COOH
Acid
& Base
Propanoic Acid, Organic acids will display the
ionizable H+ at the end (not grouped with other H’s)
(e) Glucose (C6H12O6)
Neither, Not an ionic compound
nor is a H shown on the end
Neutralization Reaction
*Know the products
acid + base
salt + water
HCl (aq) + NaOH (aq)
NaCl (aq) + H2O
H+ + Cl- + Na+ + OH-
Na+ + Cl- + H2O
H+ + OH-
H2O Net ionic equation
If we start with equal molar amounts of acid and base
we “neutralize” the solution and end up with water
Example Neutralization Reactions
Write molecular, ionic, and net ionic equations for each of the
following acid-base reactions:
(a) hydrobromic acid (aq) + barium hydroxide (aq)
2HBr(aq) + Ba(OH)2(aq)
BaBr2(aq) + 2H2O(l)
(b) sulfuric acid (aq) + potassium hydroxide (aq)
H2SO4(aq) + 2KOH(aq)
K2SO4(aq) + 2H2O(l)
Example Solutions
Solution
(a)Molecular equation:
2HBr (aq) + Ba(OH)2(aq) → BaBr2(aq) + 2H2O(l)
Ionic equation:
2H+(aq) + 2Br−(aq) + Ba2+(aq) + 2OH−(aq)
→ Ba2+(aq) + 2Br−(aq) + 2H2O(l)
Net ionic equation:
2H+(aq) + 2OH−(aq)
2H2O(l)
or
H+(aq) + OH−(aq) → H2O(l)
Both Ba2+ and Br− are spectator ions.
Example Solutions
(b) Molecular equation:
H2SO4(aq) + 2KOH(aq)
K2SO4(aq) + 2H2O(l)
Ionic equation:
Net ionic equation:
Note that because
is a weak acid and does not ionize
appreciably in water, the only spectator ion is K+.
Neutralization Reaction Producing a Gas
acid + Carbonate
2HCl (aq) + Na2CO3 (aq)
salt + water + CO2
2NaCl (aq) + H2CO3
Carbonic acid
degrades quickly
2HCl (aq) + Na2CO3 (aq)
2H+ + 2Cl- + 2Na+ + CO32-
2H+ + CO32-
2NaCl (aq) + H2O +CO2
2Na+ + 2Cl- + H2O + CO2
H2O + CO2
CaCO3 + 2HCl -> NaCl + CO2 + H2O
Antacid
*Gastric juices include HCl
Acid-Base Properties of Water
Water is a weak electrolyte (1 H+ in 107 H2O)
H
O
+ H
H
O
[H
H
H2O (l)
O
+
]
H
+ H
O
-
H
H+ (aq) + OH- (aq)
Auto-ionization
Solution Is
neutral
acidic
basic
[H+] = [OH-]
[H+] > [OH-]
[H+] < [OH-]
• Adding acid increases H+
• Adding base increases OH-
33
pH – A Measure of Acidity
pH describes the concentration of [H+] on a logarithmic scale.
pH = -log [H+]
Remember: A log scale is base 10:
pH 6 is 10x more acidic than pH 7
pH 5 is 100x more acidic than pH 7
pH 4 is 103x more acidic than pH 7
Solution Is
neutral
[H+] = [OH-]
[H+] = 1.0 x 10-7
pH = 7
acidic
[H+] > [OH-]
[H+] > 1.0 x 10-7
pH < 7
basic
[H+] < [OH-]
[H+] < 1.0 x 10-7
pH > 7
pH Meter
H+ is an electrolyte
pH meters measure electrical
conductance to determine concentration
and then calculate pH
Acid-Base reactions are colorless and can be difficult to monitor.
pH Indicators – substances that change colors at certain pH
ranges and can be used for titrations
The indicator changes color
as pH changes
Phenolphthalein
Colorless
in acid
Pink in
base
Chemistry In Action: Antacids and the Stomach pH Balance
NaHCO3 (aq) + HCl (aq)
NaCl (aq) + H2O (l) + CO2 (g)
Mg(OH)2 (s) + 2HCl (aq)
MgCl2 (aq) + 2H2O (l)
Acid-Base Reactions in Solution: Crash Course Chemistry #8
www.youtube.com/watch?v=ANi709MYnWg
Reduction-Oxidation Reactions (Redox)
(electron transfer reactions)
2Mg + O2
2Mg
2MgO
2Mg2+ + 4e- Oxidation half-reaction (lose e-)
O2 + 4e2O2Reduction half-reaction (gain e-)
2Mg + O2 + 4e2Mg2+ + 2O2- + 4e39
Redox Reaction reactants
Its charge is
“reduced” b/c it
gained an e-
Neutral Zn is insoluble,
but oxidized Zn+2 is soluble
Neutral metals are insoluble,
Ionized metals are soluble.
Write the Half-reactions of the given redox reaction.
0
+2 -2
Zn(s) + CuSO4(aq)
Zn(s)
Zn2+ + 2e-
Note: Lost e- shown in products
Cu2+ + 2e-
Cu(s)
Note: Gained e- shown in reactants
+2 -2
0
ZnSO4(aq) + Cu(s)
Zn is oxidized, lost electrons
Zn is the reducing agent
Cu2+ is reduced, gained electrons
Cu2+ is the oxidizing agent
Note: Sulfate (SO4-2) is neither oxidized or reduced so it is not shown
It could be said that:“Zn reduced Cu” or “Cu oxidized Zn”
Additional practice:
2Na(s) + Mg(NO3)2(aq) → Mg(s) + 2NaNO3(aq)
4Fe + 3O2 → 2Fe2O3
Oxidation numbers (ON’s)
• Describes the charge of an atom, in a compound, if
electrons were completely transferred between atoms
(Both ionic and molecular compounds).
• Used to predict how an atom will react with other atoms.
• For ionic compounds, ON’s are equal to the ion charge
as determined by each atom’s group number
• For molecules, ON’s change depending on what atoms
they bond to and follow a set of rules.
Ex. Cl has an ON of -1 when bonded to a metal (NaCl), but
not with non-metals. (Chlorate (ClO3-), Cl ON = +5)
*Note: An ON of +5 is not equal to a true ion of charge +5
43
Oxidation numbers (ON’s)
1. Free elements (uncombined state) and non-compounds have an
oxidation number of Zero.
Na, Be, K, Pb, H2, O2, P4 = 0
2. In monatomic ions, the oxidation number is equal to the charge on
the ion.
Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2
3. The oxidation number of oxygen is usually -2.
(In H2O2 and O22- it is –1).
*you will not be given any of these exceptions on quizzes or tests.
4. Hydrogen: ON is +1
*Unless bonded to metals in binary compounds.
In these cases –1 (LiH, NaH, etc.).
*you will not be given any of these exceptions on quizzes or tests.
5. Alkali metals are +1
Alkaline Earth Metals are +2
Fluorine is always -1. (Other Halogens can vary)
6. Sum of all ONs of each atoms in a compound/molecule (or ion)
is equal to the net charge.
CaBr2 Charge = 0
+2 -1
(+21) + (-12) = 0
SO4-2
Charge = -2
+6 -2
(+61) + (-24) = -2
Oxidation numbers of Transition metals
Alkali metals always have an oxidation number of +1
Alkaline Earth metals are always +2
• Transition metals (d group) can have several possible
oxidation states that are stable
Fe+3 can form FeCl3
Fe+2 can form FeCl2
Cu+2 can form CuCO3
Cu+1 can form Cu2CO3
• The oxidation state may change during a reaction.
K2Cr2O7 is orange with Cr+6
Cr2(SO4)3 is blue with Cr+3
Redox Chemistry in Action: Breath Analyzer
+6
3CH3CH2OH + 2K2Cr2O7 + 8H2SO4
Ethanol
Reduction of
+3 dichromate to Cr+3
3CH3COOH + 2Cr2(SO4)3 + 2K2SO4 + 11H2O
Acetic acid
47
Assign oxidation numbers to all the elements in the
following compounds and ion:
+1 -2
(a) Li2O
+1 +5 -2
(b) HNO3
+6 -2
(c)
+1 +6 -2
(d) HSO4-1
+2
+5 -2
(e) Ca3(PO4)2
Oxygen is always -2 and Li is +1 in a compound
(+1)•2 + (-2)•1 = 0
Oxygen is always -2 and H is +1 in a compound.
(+1)•1 + (N) •1 + (-2)•3 = 0 ; N must = +5
Oxygen is always -2 and the sum must equal -2
(Cr)•2 + (-2)•7 = -2 ; Cr must = +6
Oxygen is always -2 and H is +1 in a compound.
(+1)•1 + (S) •1 + (-2)•4 = -1 ; S must = +6
Oxygen is always -2 and Ca is +2 in a compound.
(+2)•3 + (P) •2 + (-2)•8 = 0 ; P must = +5
The Oxidation Numbers of Elements in
their Compounds
Red # signifies most common
oxidation state in compound
49
Types of Oxidation-Reduction Reactions
Combination Reaction
A+B
0
C
+3 -1
0
2Al + 3Br2
2AlBr3
Decomposition Reaction
C
+1 +5 -2
2KClO3
A+B
+1 -1
0
2KCl + 3O2
50
Types of Oxidation-Reduction Reactions
Combustion Reaction
A + O2
B
0
+4 -2
0
S + O2
SO2
Derailed sulfur train video
0
0
2Mg + O2
+2 -2
2MgO
51
Types of Oxidation-Reduction Reactions
Displacement Reaction
A + BC
0
+1
+2
Sr + 2H2O
+4
0
TiCl4 + 2Mg
0
AC + B
-1
Cl2 + 2KBr
0
Sr(OH)2 + H2
0
Hydrogen Displacement
+2
Ti + 2MgCl2
-1
Metal Displacement
0
2KCl + Br2
Halogen Displacement
How do we know what can displace what? Activity Series
52
The Activity Series for Metals
Shows the tendency for an atom to give away its electrons
(reduce) another atom (which ones is on top reduces the other).
Hydrogen Displacement Reaction
Al has greater
reduction potential
M + BC
MC + H2
M is metal
BC is acid or H2O
Ca + 2H2O
Ag + H2O
Ca(OH)2 + H2
AgOH + H2
Metal Displacement Reaction
Al + FeCl3
AlCl3 + Fe
Fe + AlCl3
FeCl3 + Al
53
The Activity Series for Halogens
F2 > Cl2 > Br2 > I2
I2 can reduce F2, but F2 can’t reduce I2
Halogen Displacement Reaction
0
-1
Cl2 + 2KBr
I2 + 2KBr
-1
0
2KCl + Br2
2KI + Br2
54
Types of Oxidation-Reduction Reactions
Disproportionation Reaction
The same element is simultaneously oxidized and
reduced.
reduced
Example:
+1
0
Cl2 + 2OHoxidized
Elements most likely to
disproportionate: each have at
least 3 oxidation states
-1
ClO- + Cl- + H2O
Redox Reactions Problems
Classify the following redox reactions and indicate changes in the
oxidation numbers of the elements:
(a)
(b)
(c)
(d)
(e)
CH4(g) + O2(g) → CO2(g) + H2O(l)
Redox Reactions: Crash Course Chemistry #10
www.youtube.com/watch?v=lQ6FBA1HM3s
Redox Solutions
(a) This is a decomposition reaction because one reactant is
converted to two different products. The oxidation number of N
changes from +1 to 0, while that of O changes from −2 to 0.
(b) This is a combination reaction (two reactants form a single
product). The oxidation number of Li changes from 0 to +1 while
that of N changes from 0 to −3.
(c) This is a metal displacement reaction. The Ni metal replaces
(reduces) the Pb2+ ion. The oxidation number of Ni increases from 0
to +2 while that of Pb decreases from +2 to 0.
(d) The oxidation number of N is +4 in NO2 and it is +3 in HNO2 and
+5 in HNO3. Because the oxidation number of the same element
both increases and decreases, this is a disproportionation reaction.
(e) Combination and combustion
Summary of Reactions Types
Precipitation: Two aqueous solutions mixing to forming a solid product.
Na2CO3(aq) + Ca(NO3)2(aq) → 2NaNO3(aq) + CaCO3(s)
Acid-Base Reactions: Transfer of H+ and/or OH- ions in H2O solution.
Neutralization: Reaction of Acid & Base to form a salt and water
2HNO3 + Ba(OH)2 → Ba(NO3)2 + 2H2O(l)
Acid + Metal: Oxidize to dissolve metal and produce Hydrogen gas
2HCl + Mg → MgCl2 + H2
2H+(aq) + Mg(s) → Mg+2(aq) + H2(g) (net ionic equation)
Redox: Reaction involving the transfer of electrons between atoms
0
0
+3 -2
Combination:
Combustion: 4Fe + 3O2 → 2Fe2O3
+2 +4 -2
+2 -2
+4 -2
Decomposition: CaCO3 → CaO + CO2
0
+1 -1
+1 -1
0
Displacement: Na(s) + AgCl(aq) → NaCl(aq) + Ag(s)
Collision Theory of Chemical Reactions
• The rate of a reaction increases as the number of
molecular collisions increases.
• Ways to increase collisions:
– Increase dynamic interactions (Liquid > Gas > Solid)
– Increase Concentrations (closer proximity)
– Increase Temperatures (faster movements)
• Any molecule in motion possesses Kinetic
energy, the faster it moves, the greater KE
• High energy collisions are needed to break
original bonds (Ea :Activation energy)
Ted-Ed: How to speed up chemical reactions (and get a date)
https://www.youtube.com/watch?v=OttRV5ykP7A
Disposing of excess WWII Sodium metal in 1947
2Na
+
2H2O →
2NaOH
+
H2(g)
60
Solution
Stoichiometry
The concentration of a solution is the amount of solute
present in a given quantity of solvent or solution.
M = molarity =
moles of solute
liters of solution
61
Measures of Concentration
The concentration of a solution is the amount of solute present in a
given quantity of solvent or solution.
Molarity (M)
moles of solute
M =
liters of solution
Includes solute
volume
Because density (volume) can change with temperature it is helpful to
express solvent by mass when sample undergoes temperature changes
Molality (m)
m =
moles of solute
mass of solvent (kg)
Excludes solute
mass
No volumetric measurements needed; all mass
(grams)
M =
moles of solute
liters of solution
(mL)
moles = Liters x Molarity
L =
moles of solute
Molarity
63
Problem
Calculate the Molarity and Molality of a H2SO4 solution
containing 24.4 g of sulfuric acid in 198 g of water at 50 °C.
M =
moles of solute
liters of solution
H2O at 70 °C,
0.97 g/mL
M =
0.249 moles
0.204 Liters of solution
= 1.22 M H2SO4
Molarity/Molality Problem
The definition of molality (m) is
The mass of water is 198 g, or 0.198 kg. Therefore,
Molarity will equal Molality at 4 °C, but Molarity will
decrease as temperature goes up and density goes down
Preparing a Solution of Known Concentration from solids
Volumetric Flask
Mix till dissolved
Bring to
desired volume
Molarity Problem
What is the Molar concentration of the Sodium ion [Na+] when
23.4 g of NaCl and 34.1 g of Na2O are dissolved in 0.60 L H2O?
23.4 g NaCl 1 mol
58.5 g NaCl
34.1 g Na2O 1 mol
62.8 g Na2O
= 0.40 mol NaCl = 0.40 mol Na+
= 0.54 mol Na2O = 1.08 mol Na+
(doubles from subscript)
0.40 mol Na+ + 1.08 mol Na+ = 1.48 mol Na+
M =
1.48 mol
0.60 L
= 2.5 M Na+
67
Molarity Problem
How many grams of potassium dichromate (K2Cr2O7)
are required to prepare a 250-mL solution whose
concentration is 2.16 M?
M =
mol
L
mol = ML
Dilution is the procedure for preparing a less concentrated
solution from a more concentrated solution (stock).
Dilution
Add Solvent
Moles of solute
before dilution (1)
M1V1
=
=
Moles of solute
after dilution (2)
M2V2
Number of moles does not change
69
Dilution Practice
Describe how you would prepare 500 mL
of a 1.75 M H2SO4 solution, starting with
an 8.61 M stock solution of H2SO4.
Keep in mind that in dilution, the concentration of the solution
decreases but the number of moles of the solute remains the same.
M1V1 = M2V2
Dilution Solution
Solution We prepare for the calculation by tabulating our data:
M1 = 8.61 M
V1 = ?
M2 = 1.75 M
V2 = 500 mL
Thus, we must dilute 102 mL of the 8.61 M H2SO4 solution with
water to give a final volume of 500 mL
Dilution Problem
Describe how you would prepare 300
mL of a 0.4 M H3PO4 solution, starting
with an 1.5 M stock solution of H3PO4.
M1V1 = M2V2
Bell Ringer
You have 250 mL of a 3.0 M Ba(OH)2 solution.
What is the concentration if we add 150 mL of
water to the solution?
M1 = 3.0 M
V1 = 250 mL
M2 = ?
V2 = 250 mL + 150 mL = 400 mL
Crash Course: Water and Solutions for Dirty Laundry
www.youtube.com/watch?v=AN4KifV12DA
Bell Ringer
1) How many grams of solid NaNO3 are needed to produce
125 mL of a 0.85 M NaNO3 solution?
2) What is the Molar concentration of the Sodium ion [Na+]
when 2.8 g of Na3PO4 and 4.5 g of Na2CO3 are dissolved in
85 mL?
3) How would you prepare 250 mL of a 0.65 M H2SO4
solution from a stock of 6.5 M H2SO4 solution?
4) You have 50 mL of 6.0 M NaF and 450 mL water are
added, what is the new Molarity?
74
Titrations
In a titration, a solution of accurately known concentration is
added gradually to another solution of unknown concentration
until the chemical reaction between the two solutions is
complete.
Standard solution – solution with known concentration to be
precisely added for comparison
Equivalence point – the point at which the reaction is complete
example) 1 mol H2SO4  2 mol NaOH
(obtained from balanced equation)
75
Titrations
Indicator – substance that changes color at (or near) the
equivalence point
Slowly add base
to unknown acid
UNTIL
the indicator
changes color
76
Titrations can be used in the analysis of:
Acid-base reactions
H2SO4 + 2NaOH
2H2O + Na2SO4
Redox reactions
5Fe2+ + MnO4- + 8H+
Mn2+ + 5Fe3+ + 4H2O
*Can involve color changes without indicator
77
Titration Steps
1) Write Balanced equation for stoichiometry
(mole-to-mole ratio)
2) Determine moles of Standard solution used.
(mol = M x L)
3) With stoichiometry, convert molesstandard to molesunknown
(train tracks)
4) Determine unknown Molarity using given volume (L)
(M = mol/L)
Alternative Equation:
MsVs = MuVu
Coefficient #
Coefficient #
Titration Problem #1
It takes 32 mL of 2.0M HCl standard to neutralize a 500. mL
solution of Ba(OH)2. What is the concentration of Ba(OH)2?
1) 1Ba(OH)2 + 2HCl → BaCl2 + 2H2O (Reacts 1:2)
2) mol HCl = 2.0M x 0.032 L = 0.064 mol HCl
3) 0.064 mol HCl 1 mol Ba(OH)2
2 mol HCl
4) M Ba(OH)2 = 0.032 moles
0.500 L
= 0.032 mol Ba(OH)2
= 0.064 M Ba(OH)2
Alternative:
MHVH = MOHVOH
Coefficient #
Coefficient #
2.032 = MOH500
2
1
MOH = 0.064
Titration Problem #2
How many milliliters (mL) of a 0.610 M NaOH
solution are needed to neutralize 20.0 mL of a
0.245 M H2SO4 solution?
2 NaOH + H2SO4
2 H2O + Na2SO4
1) For every 2 moles base added,
it neutralizes 1 mole acid
Titration #2 Solution
2) Next we calculate the number of moles of H2SO4 in a
20.0 mL solution:
Moles = M x L
0.245 M x 0.0200 L = 0.00490 mol H2SO4
3) From the Balanced Equation: 1 mol H2SO4  2 mol NaOH.
4.9 x 10-3 mol H2SO4
2 mol NaOH
= 9.80 × 10-3 mol
1 mol H2SO4
NaOH
4) L = 9.80 x 10-3 mol = 0.0161 L or 16.1 mL NaOH
0.61 M NaOH
Titration: Finding the Molar Mass of an Unknown
Lauric Acid is a short-chain fatty acid that is solid at room
temperature and monoprotic. We dissolve 0.022 grams into
500. mL of water and then titrate it with 0.010 M NaOH.
If it takes 11.0 mL of NaOH to neutralize the fatty acid, what
is the Molar Mass of Lauric Acid?
• Molar mass has the units grams/mole. We weighed out
the mass of the solid acid in grams. Titration can tell us
how many moles of acid are present in the same sample.
0.022 grams
= ? Molar mass
+
Moles H
• Because it is monoprotic, it will react 1:1 with NaOH.
Molar Mass Titration Solution
1) Given information states it reacts 1:1
2) (0.010 M NaOH) x (0.0110 L) = 1.1 x 10-4 mol NaOH
3) 1.1 x 10-4 mol NaOH 1 mol Lauric acid = 1.1 x 10-4 mol
Lauric Acid
1 mol NaOH
0.022 grams
4) Molar Mass =
1.1 x 10-4 moles
C11H23COOH
= 200 g/mol
Large component of
coconut oil
~ 3-6% of milk
Example: Redox Titration
A 16.42-mL volume of 0.1327 M KMnO4
solution is needed to oxidize 25.00 mL of a
FeSO4 solution in an acidic medium.
What is the concentration of the FeSO4
solution in molarity?
The net ionic equation is
need to
find
want to
calculate
given
Redox Titration Solution
Solution The number of moles of KMnO4 (in 16.42 mL) = M x L
(0.1327 M KMnO4) x (0.01642 L) = 2.179 x 10-3 mol
From the net ionic equation we see that 5 mol Fe2+  1 mol MnO4-
M =
moles of solute
liters of solution
1.090 x 10-2 mol
=
0.025 L
= 0.436 M FeSO4
Chemistry in Action: Metals from the Sea
Many metals are found
in the earth’s crust,
but it is cheaper to
“mine” from the sea
CaCO3 (s)
CaO (s) + CO2 (g)
Precipitation
Ca2+ (aq) + 2OH- (aq)
CaO (s) + H2O (l)
Slightly soluble
Mg2+ (aq) + 2OH- (aq)
Precipitation
Mg(OH)2 (s) + 2HCl (aq)
1.3 g of Magnesium/
Kg seawater
Mg(OH)2 (s)
MgCl2 (aq) + 2H2O (l)
Electrolysis of MgCl2 (redox)
Mg2+ + 2e2Cl-
MgCl2 (aq)
Mg
Cl2 + 2e-
Mg (s) + Cl2 (g)
86
Gravimetric Analysis
 Analytical technique based on the measurement of mass
• Precipitation: the analyte is precipitated out of solution by adding
another reagent to make it insoluble. Then filtered and weighed.
• Volatilization: the analyte is converted to a gas and removed. The
loss of mass from the starting material indicates the mass of gas.
87
Gravimetric Analysis
1. Dissolve unknown substance in water (if not already dissolved)
2. React unknown with precipitating reagent to form a solid precipitate
Reagent: chemical added to another substance to bring about a change
3. Filter, dry, and weigh precipitate.
4. Use chemical formula and mass of precipitate to determine amount of
unknown analyte (chemical of interest in experiment)
88
Gravimetric Analysis Problem #1
A sample of an unknown soluble compound contains Ba+2 and is
dissolved in water and treated with excess sodium phosphate.
If 0.411 g of Barium phosphate precipitates out of solution, what
mass of Barium in found in the unknown compound?
We need to find 1st find the Mass % of the analyte in the precipitated
compound using their respective molar masses
3 x 137.3 g Ba+2
601.9 g Ba3(PO4)2
x 100% = 68.4% Ba in Ba3(PO4)2
0.684 x 0.411 g = 0.253 g Ba+2
*It is not mandatory to convert to
percentage form. The mass
fraction can be used directly.
411.9 g Ba+2
+2
x
0.411g
Ba
(PO
)
=
0.253
g
Ba
3
4 2
601.9 g Ba3(PO4)2
Gravimetric Analysis Problem #2
We have 250 mL of a Copper (Cu+1) aqueous solution. We add
excess sodium carbonate (Na2CO3) to precipitate out 3.8 g of
Cu2CO3. What is the [Cu+1] Molarity of the solution?
Since we need to find moles of Cu+1 , it will be quicker to use train-tracks
instead of % composition (either would work).
3.8 g Cu2CO3 1 mol Cu2CO3 2 mol Cu+1
187.0 g Cu2CO3 1 mol Cu2CO3
M =
0.041 mol Cu+
0.250 L
= 0.16 M Cu+1
= 0.041 mole Cu+1
Gravimetric Analysis Problem #3
A 0.5662-g sample of an ionic compound containing
chloride ions and an unknown metal is dissolved in water
and treated with an excess of AgNO3.
If 1.0882 g of AgCl precipitate forms, what is the percent
by mass of Cl in the original compound?
35.45 g Clx 1.0882 g AgCl = 0.269 grams Cl
143.4 g AgCl
More Gravimetric Review Problems
1) An unknown ionic compound contains Carbonate (CO3-2).
To precipitate the carbonate, we add excess CaCl2 and collect
25.3 grams of CaCO3 precipitate. Calculate mass of Carbonate
present in the original compound.
2) 50 mL of a solution contains an unknown amount of Ni+
ions. We add excess Na3PO4 to precipitate out 5.67 grams
of Ni3PO4. What is the Molarity of Ni+?
3) 53.9 g of an unknown soluble compound contains Silver
(Ag). When dissolved and treated with excess Na2S a
precipitate of Ag2S is formed. What is the % Ag in the
original compound if 34.2 grams Ag2S is collected?
H2O reactions Exam Review Problems
1)Write the net ionic equation for: (NH4)2S and MgSO4
2)Write the labeled half reactions: 2N2O → 2N2 + O2
3)How many grams NH4NO3 needed for 250 mL of 3.5M?
4) Describe preparation of 300 mL 0.5M HCl (5M stock)
5)120 mL of H3PO4 is titrated with 35 mL 2.4 M
standard KOH, what is concentration of H3PO4?
6) 53.87 g mixture contains a quantity of lead (Pb). When
dissolved and treated with excess NaCl a precipitate of
PbCl2 is formed. What is the %Pb in the original
unknown if 45 grams PbCl2 is collected?