Transcript problem and and mass

Chemical Stoichiometry, Aqueous
Solutions and Solution Stoichiometry
Chapters 3, 4
1
Previous knowledge
you must know!!!
1. Mole concept.
2. Molar mass calculations.
3. Conversions between mass-moleparticles.
4. Limiting Reactant
5. Percent yield
6. Percent composition.
7. Empirical formula calculations.
8. Molecular formula calculations.
2
The Mole
Avogadro’s number: Provides a
connection between the number of moles
in a pure sample and the number of
particles or units in the sample.
1 mol = 6.022 x 1023 particles
atoms
molecules
formula units
ions
3
The Mole
Molar mass: Mass in grams of 1 mol of a
pure substance.


1 mol Ne atoms = 20.18 g = 6.022 x 1023
atoms
1 mol Cl- ions = 35.45 g = 6.022 x 1023 ions
Difference between amu and mass



1 amu = 1.66x10-24 g
1 carbon atom = 12.01 amu
1 mol carbon atoms = 12.01 g
4
Molar mass of a compound
Problem 1: Calculate the molar mass for the following
compounds:
a) Fe(NO3)2·6H2O
b) (NH4)3PO4
5
Converting moles to grams
Problem 2: How many moles of iron (II) nitrate hexahydrate
are there in 48.5 g of the salt?
Problem 3: How many grams of ammonium phosphate are
there in 1.87 moles of the salt?
6
Converting grams to particles
Problem 4: How many water molecules are contained within
56.0 g of a pure sample of water?
Problem 5: How many hydrogen atoms are contained within
23.6 g of pure sucrose, C12H22O11?
7
Conservation of Mass and Atoms in
Chemical Reactions

Fe2 O3 + 3 CO  2 Fe + 3 CO 2
reactants
yields
products
1 formula unit + 3 molecules
≠
1 mole
+ 3 moles
≠
159.7 g
+
=
5 atoms
84.0 g
243.7
+ 6 atoms
11 atoms
=
2 atoms
2 moles
111.7 g
+ 3 molecules
+ 3 moles
+ 132 g
243.7
2 atoms + 9 atoms
11 atoms
8
Predicting mass of products
Problem 6: What mass of carbon dioxide can be produced
by the reaction of 32.0 g of iron (III) oxide with excess
carbon monoxide?
9
Predicting mass of reactants
Problem 7: What mass of sodium bicarbonate is needed to
produce 10.6 g of sodium carbonate?
NaHCO3(aq)  Na2CO3(aq) + H2O(l) + CO2(g)
10
Limiting Reactant Concept
If you add more of one reactant you cannot
make more product. You must increase all
reactants stoichiometrically.
Look at a chemical limiting reactant situation.
Zn + 2 HCl ZnCl2 + H2
11
Limiting Reactant Concept
If quantities of two or more reactants is
known, then one reactant is used up and
some of the other(s) reactant(s) will be left
over.
Limiting Reactant: The reactant that is used
up limits how far the reaction will proceed.
Excess reactant: The reactant that is left over
when the reaction is complete.
12
Limiting Reactant Concept
Problem 8: Find the limiting reactant and the maximum mass
of sulfur dioxide that can be produced by the reaction of
38.1 g of carbon disulfide with 52.0 g of oxygen?
CS2
+
O2

CO2
+
SO2
13
Limiting Reactant Concept
Problem 9: Find mass of excess reactant that remains after
the reaction reached completion when 38.1 g of carbon
disulfide reacted with 52.0 g of oxygen?
CS2
+
O2

CO2
+
SO2
14
Percent Yields from Reactions
Theoretical yield is calculated by assuming
that the reaction goes to completion.

Determined from the limiting reactant calculation.
Actual yield is the amount of a specified pure
product made in a given reaction.

In the laboratory, this is the amount of product
that is formed in your beaker, after it is purified
and dried.
Percent yield indicates how much of the
product is obtained from a reaction.
actual yield
% yield =
 100%
theoretical yield
15
Percent Yields from Reactions
Problem 10: A 4.60 g sample of ethanol, C2H5OH, was boiled
with excess acetic acid, CH3COOH, to produce 6.40 g of ethyl
acetate, CH3COOC2H5. What is the percent yield?
CH3COOH + C2H5OH  CH3COOC2H5 + H2O
16
Solving Stoichiometry problems
without a calculator!!!
a) Calculate the moles of precipitate produced from the reaction of
0.0500 moles of BaCl2 with 0.0300 moles of Na2CO3.
b) Calculate the moles of the excess reactant.
Initial
BaCl2(aq) + Na2CO3(aq)  BaCO3(s) + 2 NaCl(aq)
0.0500
0.0300
0
0
Change -0.0300
End
0.0200
-0.0300
0
+0.0300
+0.0600
0.0300
0.0600
c) Sketch the picture of how the reactant vessel will look like in
terms of contents after the completion of the reaction .
17
ClNa+
Cl-
Ba2+
Na+
Cl-
Na+
Cl-
ClCl-
Na+
BaCO3
18
Problem 11: Calculate the mass of carbon dioxide produced by
the reaction of 2.5 moles of sodium carbonate with 5.3
moles of hydrochloric acid.
No calculator allowed
19
Derivation of Formulas from
Elemental Composition
Empirical Formula - smallest whole-number ratio of
atoms present in a compound

CH2 is the empirical formula for alkenes
Molecular Formula - actual numbers of atoms of each
element present in a molecule of the compound


Ethene – C2H4
Pentene – C5H10
We determine the empirical and molecular formulas of a
compound from the percent composition of the
compound given by the mass spec.

percent composition is determined experimentally
20
Molecular and Empirical Formulas
Common Name
Molecular Formula
Empirical Formula
hydrogen peroxide
H2O2
HO
ethylene
C2H4
CH2
dextrose
C6H12O6
CH2O
21
Problem 12: A compound contains 24.74% K, 34.76% Mn, and
40.50% O by mass. What is its empirical formula?
22
Problem 13: A sample of a compound contains 6.541g of Co
and 2.368g of O. What is empirical formula for this
compound?
23
Problem 14: A sample of caffeine was found to contain
49.5% carbon, 28.9% nitrogen, 16.5% oxygen, and 5.1%
hydrogen by mass. Find the empirical formula for caffeine.
Problem 15. The molar mass for caffeine is 194.2 g/mol. Find
the molecular formula for caffeine.
24
Combustion Analysis
 An organic compound is burnt in excess of
oxygen.
H2O absorber
O2 
Organic sample
CO2 absorbed
 All of the hydrogen from the organic sample
combines to form water and absorbed in the
water absorber.
 All of the carbon from the organic sample
combines to form CO2 and absorbed in the CO2
absorber.
 Oxygen is the only element that is added to the
25
sample and it must be calculated last.
Problem 16: A 2.04 g sample containing C, H, and O underwent
combustion analysis. Find the empirical formula of the
compound if 4.49 g of CO2 and 2.45 g of H2O were produced.
26
Some Other Interpretations of
Chemical Formulas
What mass of ammonium phosphate, (NH4)3PO4,
would contain 15.0 g of N?
molar mass of (NH 4 )3 PO 4  149.0 g/mol
1 mol N
15.0 g of N 
 1.07 mol N
14.0 g N
1 mol (NH 4 )3 PO 4
1.07 mol N 
 0.357 mol (NH 4 )3 PO 4
3 mol N
149.0 g (NH 4 )3 PO 4
0.357 mol (NH 4 )3 PO 4 
 53.2 g (NH 4 )3 PO 4
1 mol (NH 4 )3 PO 4
27
Problem 17: How many grams of manganese are there in
57.9 g of a pure sample of potassium permanganate?
28
Purity of Samples
The percent purity of a sample of a
substance is always represented as
mass of pure substance
% purity =
100%
mass of sample
mass of sample includes impurities
29
Purity of Samples
A bottle of sodium phosphate, Na3PO4, is 98.3% pure
Na3PO4. What are the masses of Na3PO4 and
impurities in 250.0 g of this sample of Na3PO4?
98.3 g Na 3 PO 4
unit factor
100.0 g sample
98.3 g Na 3 PO 4
250.0 g sample 
100.0 g sample
= 246 g Na 3 PO 4
250.0 g sample - 246 g Na 3 PO 4
= 4.00 g impurities
30
When gases involved….PV = nRT
Problem 18: The number of moles and mass of the gases can be
calculated if the volume, temperature and pressure is given.
Calcium carbonate decomposes to produce calcium oxide and
carbon dioxide. The decomposition of CaCO3 was accomplished
by heating the sample at 300°C.
130.1 mL of the gas produced was collected at 25.0°C and 751
mm Hg. Calculate the mass of calcium carbonate decomposed.
31
Chemical Equations
Law of Conservation of Matter



There is no detectable change in quantity
of matter in an ordinary chemical reaction.
Balanced chemical equations must always
include the same number of each kind of
atom on both sides of the equation.
This law was determined by Antoine
Lavoisier.
32
Law of Conservation of Matter
NH3 burns in oxygen to form NO &
water

2 NH3 + O 2 
 2 NO + 3 H 2 O
5
2
or

4 NH3 + 5 O 2 
 4 NO + 6 H 2 O
33
Reaction
Types
34
Combination
Reaction between metal and a non-metal produces
the binary ionic compound of most stable oxidation
state.
Example:
2 Mg + O2  2 MgO
(Mg is 2+ and O is 2-)
Reaction between two non-metals produces a binary
covalent compound. The compound depends on
conditions.
Example:
2 S + 3 O2 (excess)  2 SO3
S + O2 (limited)  SO2
Reaction of metal oxides with water produces metal
hydroxide.
Example:
CaO + H2O  Ca(OH)2
35
Combination
Reaction of non-metal oxides with water produces
the oxyacid of the same oxidation number.
Example:
SO3 + H2O  H2SO4
(oxidation number of S in SO3 and H2SO4 is 6)
SO2 + H2O  H2SO3
(oxidation number of S in SO2 and H2SO3 is 4)
Reaction of metal oxides and non-metal oxides
produces the ternary salt.
Example:
MgO + CO2  MgCO3
Na2O + SO2  Na2SO3
36
Decomposition

Binary compounds generally decompose into the elements
Example:

2 KClO3  2 KCl + 3 O2
Metal hydroxides decompose into the metal oxide and water
Example:

Na2CO3  Na2O + CO2
Metal chlorates decompose into metal chlorides and oxygen
Example:

2 HgO  2 Hg + O2
SO2  S + O2
Metal carbonates or hydrogen carbonate decompose into
metal oxide and CO2 or metal oxide, CO2 and H2O
Example:

see decomposition of metal bicarbonate
Ca(OH)2  CaO + H2O
Oxyacids decomposed into non-metal oxides and water
Example:
H2CO3  CO2 + H2O
H2SO4  SO3 + H2O
37
Aqueous solutions and
chemical reactions
Electrolytes
Substances that
dissociate into ions
when dissolved in
water.
A nonelectrolyte may
dissolve in water, but
it does not dissociate
into ions when it does
so.
Electrolytes
Strong electrolytes.
Substances that
dissolve in water
and separate into
ions.
Strong conductor
of electricity
Soluble ionic
compounds.
 Solubility rules
Strong acids.
Strong bases.
Weak electrolytes.
• Substances that
dissolve in water
moderately and
separate into ions
partially.
• Poor conductor of
electricity.
• Ionic compounds
with low solubility.
– Solubility rules
• Weak acids.
• Weak bases.
Non electrolytes.
• Substances that
may dissolve in
water but cannot
separate into ions.
• Non conductors of
electricity.
• Covalent
compounds.
Strong Acids
Strong Bases
Strong Electrolytes Are…
Strong acids
Strong bases
Soluble ionic salts. Memorize the simple
solubility rules:
+
 All NH4 salts are soluble.
 All NO3- , C2H3O2-, ClO4- salts are soluble.
 All group 1 salts are soluble.
Dissociation
When an ionic substance dissolves
in water, the solvent pulls the
individual ions from the crystal and
solvates them.
This process is called dissociation.
Water is a polar molecule. Its
positive ends (hydrogen sites) are
attracted by the negative ions in the
ionic compound, while the negative
ends (oxygen sites) are attracted to
the positive ions.
Ionic compounds dissolve in water
because the attraction of the ions to
water is stronger than the attraction
between the ions themselves.
45
Dissociation
NaCl(s)  Na+(aq) + Cl-(aq)
Na2CO3(s)  2 Na+(aq) + CO32-(aq)
Ba(NO3)2 (s)  Ba2+(aq) + 2 NO3-(aq)
(NH4)3PO4(s)  3 NH4+(aq) + PO43-(aq)
Problem 19: Complete the following dissociation equations
a) Ca(CH3COO)2 
b) (NH4)2SO4 
c) Fe(NO3)3 
d) K3PO3 
47
Molecular compounds in water
With the exception of strong acids, molecular
compounds are non electrolytes or weak electrolytes.
Non-electrolytes:
C12H22O11 (s)  C12H22O11 (aq)
CH3OH (l)  CH3OH (aq)
Weak electrolytes are weak acids:
CH3COOH(aq)
HNO2(aq)
H+(aq) + CH3COO-(aq)
H+(aq) + NO2-(aq)
48
Like dissolves like
Polar molecules dissolve in polar solvents
δ-
49
Double Replacement reactions
Precipitation reactions (must know
solubility rules).
Neutralization reactions or acid-base
reactions.
Gas forming reactions.
50
Precipitation Reactions
When one mixes ions
that form compounds
that are insoluble (as
could be predicted by
the solubility
guidelines), a
precipitate is formed.
Molecular Equation
AgNO3 (aq) + KCl (aq) 
AgNO3 (aq) + KCl (aq)  AgCl (s) + KNO3 (aq)
Ionic Equation
In the ionic equation all strong electrolytes (strong
acids, strong bases, and soluble ionic salts) are
dissociated into their ions.
This more accurately reflects the species that are
found in the reaction mixture.
Ag+ (aq) + NO3- (aq) + K+ (aq) + Cl- (aq) 
AgCl (s) + K+ (aq) + NO3- (aq)
Net Ionic Equation
To form the net ionic equation, cross out anything
that does not change from the left side of the
equation to the right.
Ag+(aq) + NO3-(aq) + K+(aq) + Cl-(aq)  AgCl (s) + K+(aq) + NO3-(aq)
Net Ionic Equation
To form the net ionic equation, cross out anything
that does not change from the left side of the
equation to the right.
The only things left in the equation are those things
that change (i.e., react) during the course of the
reaction.
Ag+(aq) + Cl-(aq)  AgCl (s)
Net Ionic Equation
Those things that didn’t change (and were
deleted from the net ionic equation) are called
spectator ions.
K+ and NO3-
Problem 20: Write the balanced complete and net ionic
equation for the reaction that takes place when an aqueous
solution of barium nitrate is added to an aqueous solution of
sodium sulfate.
57
Problem 21: Write the balanced complete and net ionic
equation for the reaction that takes place when an aqueous
solution of lead (II) nitrate and sodium iodide are mixed in a
beaker.
58
Acids:
Substances that
increase the
concentration of H+
when dissolved in
water (Arrhenius).
Proton donors
(Brønsted–Lowry).
Acids
There are only seven
strong acids:
•
•
•
•
•
•
•
Hydrochloric (HCl)
Hydrobromic (HBr)
Hydroiodic (HI)
Nitric (HNO3)
Sulfuric (H2SO4)
Chloric (HClO3)
Perchloric (HClO4)
Bases:
Substances that
increase the
concentration of
OH− when dissolved
in water (Arrhenius).
Proton acceptors
(Brønsted–Lowry).
Bases
The strong bases
are the soluble salts
of hydroxide ion:
•
•
•
•
Alkali metals
Calcium
Strontium
Barium
Neutralization Reactions
Generally, when solutions of an acid and a base are
combined, the products are a salt and water.
HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l)
Neutralization Reactions
When a strong acid reacts with a strong base, the net
ionic equation is…
HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l)
H+ (aq) + Cl- (aq) + Na+ (aq) + OH- (aq) 
Na+ (aq) + Cl- (aq) + H2O (l)
Gas-Forming Reactions
These metathesis reactions do not give the
product expected.
The expected product decomposes to give a
gaseous product (CO2 or SO2).
CaCO3 (s) + HCl (aq) CaCl2 (aq) + CO2 (g) + H2O (l)
NaHCO3 (aq) + HBr (aq) NaBr (aq) + CO2 (g) + H2O (l)
SrSO3 (s) + 2 HI (aq) SrI2 (aq) + SO2 (g) + H2O (l)
Gas-Forming Reactions
This reaction gives the predicted product, but
you had better carry it out in the hood, or
you will be very unpopular!
Just as in the previous examples, a gas is
formed as a product of this reaction:
Na2S (aq) + H2SO4 (aq)  Na2SO4 (aq) + H2S (g)
Complexation Reactions

Transition metal ions (such as Fe, Co, Ni, Cu, Zn,
Ag) and Al will react with excess of molecules or
ions with pairs of electrons in their Lewis structure
(such as NH3, H2O, OH-, CN-, Cl-, Br-, SCN-) to
form coordination compounds. The amount of
ligands is usually twice the charge of the metal
ion.
Example: Cu2+ + 4 NH3  [Cu(NH3)4]2+
(the amount of NH3 is twice the charge of Cu2+)
Fe3+ + 6 CN-  [Fe(CN)6]3-
(the amount of CN- is twice the charge of Fe3+)
67
Oxidation-Reduction Reactions
(REDOX)
An oxidation occurs
when an atom or ion
loses electrons. LEO
A reduction occurs
when an atom or ion
gains electrons. GER
Electrons are
transferred.
Single replacement
and combustion
reactions.
See video
A piece or iron is immersed in a solution of
copper(II) sulfate.
Reduction
Fe(s) + Cu2+(aq)  Fe2+(aq) + Cu(s)
Oxidation
Electrons are transferred from the iron to
the copper(II) ion.
Fe gets oxidized.
Cu2+ gets reduced.
69
Oxidation Numbers
To determine if an oxidation-reduction
reaction has occurred, we assign an
oxidation number to each element in a
neutral compound or charged entity.
Oxidation Numbers
1. Elements in their elemental form have an
oxidation number of 0.
2. In a binary compound, the oxidation number of
the second element is the same as its ionic
charge.
3. Oxygen has an oxidation number of −2, except
in the peroxide ion in which it has an oxidation
number of −1.
4. Hydrogen is −1 when bonded to a metal, +1
when bonded to a nonmetal.
Oxidation Numbers
5. Fluorine always has an oxidation number of
−1.
6. The other halogens have an oxidation number
of −1 when they are negative; they can have
positive oxidation numbers, however, most
notably in oxyanions.
7. The sum of the oxidation numbers in a neutral
compound is 0.
8. The sum of the oxidation numbers in a
polyatomic ion is the charge on the ion.
Problem 22: Identify the substance that gets oxidized and
the one that gets reduced in the following chemical reactions.
a. MnO4-(aq) + H2O2(aq) + H+  Mn2+(aq) + O2(g) + H2O(l)
b. Ca(s) + Ni2+(aq)  Ca2+(aq) + Ni(s)
c. Cu(s) + HNO3(aq)  Cu2+(aq) + NO2(g) + H2O(l)
d. KClO3(l) + HNO2(aq)  KCl(aq) + HNO3(aq)
Single Displacement Reactions
In displacement
reactions, ions
oxidize an element.
The ions, then, are
reduced.
Single Displacement

Reactive metals displace another metal ion from their compounds.
Example:

Reactive metals displace hydrogen from water forming the
hydroxides.
Example:

2 Na + 2 H2O  2 NaOH + H2
Reactive metals react with acids to produce the salt of the metal
and hydrogen.
Example:

2 Al + 3 Pb(NO3)2  2 Al(NO3)3 + 3 Pb
(aluminum forms ions with a charge of 3+)
Mg + 2 HCl  MgCl2 + H2
Halogens (Group 17) can displace another halogen from the
compound only if it is higher in the periodic table.
Example:
Cl2 + 2 KBr  2 KCl + Br2
Br2 + 2 KCl  No Reaction. Br2 is not
more reactive than Cl2.
75
Single Displacement Reactions
In this reaction,
silver ions oxidize
copper metal.
Cu (s) + 2 Ag+ (aq)  Cu2+ (aq) + 2 Ag (s)
Single Displacement Reactions
The reverse reaction,
however, does not
occur.
x Cu (s) + 2 Ag+ (aq)
Cu2+ (aq) + 2 Ag (s) 
Activity Series
Increasing ease of oxidation
Au
Pt
Ag
Cu
H
Fe
Zn
Mg
Ca
Ba
K
Li
An element will oxidize
anything that is below it
in this series.
Problem 23: Write net ionic equations and identify the
substance that gets oxidized and the one reduced.
a. A copper(II) nitrate solution is poured over solid zinc.
b. Chlorine gas is bubbled through an aqueous solution
containing potassium bromide.
c. Sodium metal is placed in water.
d. A solution of copper(I) chloride sits in a beaker for an
extended period of time.
e. Solid iron filings are added to a 0.5 M solution of iron(III)
chloride.
79
Combustion

Organic compounds (carbon containing
compounds) react with oxygen to form
carbon dioxide and water
Example:
2 C3H6O2 + 7 O2  6 CO2 + 6 H2O
2 C4H8O + 11 O2  8 CO2 + 8 H2O
80
Balancing redox reactions
Acidic solutions
1.
2.
3.
4.
5.
6.
7.
Divide the equation into 2 halves: the oxidation half and the
reduction half.
For each half balance the atoms other than H and O.
Add H+ whenever H is needed; add H2O whenever O is
needed.
Balance the charges. Add electrons if needed. Oxidation is
the loss of electrons so e- should be in the products, while
reduction id the gain of e-.
Electrons must cancel. Add both halves to cancel the
electrons. It may be necessary to multiply the halves to make
the electrons the same.
Cancel water molecules and H+ is they appear on both sides
of the equation.
Check final equation for balancing materially and electrically.
81
Common Reductions in Acidic
solutions
nitrate
NO3-  NO
permanganate
MnO4-  Mn2+
dichromate
Cr2O72-  Cr3+
82
Problem 24: Balance the following equation in acidic solution
MnO4-(aq) + Sn2+ (aq)  Mn2+ (aq) + Sn4+ (aq)
½ oxidation:
½ reduction
83
Balancing redox reactions
Basic solutions
1.
2.
3.
4.
5.
6.
7.
8.
Divide the equation into 2 halves: the oxidation half and the
reduction half.
For each half balance the atoms other than H and O.
Add H+ whenever H is needed; add H2O whenever O is
needed.
Balance the charges. Add electrons if needed. Oxidation is
the loss of electrons so e- should be in the products, while
reduction id the gain of e-.
Electrons must cancel. Add both halves to cancel the
electrons. It may be necessary to multiply the halves to make
the electrons the same.
Cancel water molecules and H+ is they appear on both sides
of the equation.
Check final equation for balancing materially and electrically.
Add (to both sides) as many OH- as H+ there are to cancel
and form H2O. Cancel the waters.
84
Problem 25: Balance the following equation in basic solution.
MnO4-(aq) + C2O42- (aq)  MnO2 (aq) + CO2 (aq)
½ oxidation:
½ reduction
85
Molarity
Two solutions can contain the same
compounds but be quite different because
the proportions of those compounds are
different.
Molarity is one way to measure the
concentration of a solution.
moles of solute
Molarity (M) =
volume of solution in liters
Concentrations of Solutions
Molarity changes with temperature.

If temperature increases, the volume
increases slightly so the molarity decreases.
number of moles of solute
molarity 
number of liters of solution
moles
M
L
mmol
M
mL
Mixing a Solution
See video
Concentrations of Solutions
Example: Calculate the molarity of a
solution that contains 12.5 g of sulfuric
acid in 1.75 L of solution.
12.5 g H 2SO 4 1 mol H 2SO 4

1.75 L sol' n
98.1 g H 2SO 4
0.0728 mol H 2SO 4

L
 0.0728 M H 2SO 4
Concentrations of Solutions
Example: Determine the mass of
calcium nitrate required to prepare 3.50
L of 0.800 M Ca(NO3)2 .
0.800 mol Ca(NO3 ) 2 164 g Ca(NO3 ) 2
3.50 L 

 459 g Ca(NO3 ) 2
L
1 mol Ca(NO3 ) 2
Problem 26: A 3.75g sample of NaCl is dissolved in water. The
total volume of the solution is 768mL. What is the molarity of
the solution?
Problem 27: How many mL of 0.245 M NaOH are needed to
deliver 1.75 moles of NaOH?
Dilution
Dilution of Solutions
The relationship M1V1 = M2V2 is appropriate for
dilutions, but not for chemical reactions.
Example 3: Suppose you needed to prepare 100.0 mL of 1.00 M
NH3 using 1.25 M NH3, distilled water, and a 100 mL graduated
cylinder. How would you do this?
M1 V1 = M2 V2
1.00 M * 100.0 mL = 1.25 M * V2
V2 = 80.0 mL
Pour 80.0 mL of 1.25 M NH3 into the graduated cylinder and add
distilled water until you have 100.00 of solution.
This question can be adapted to whatever other container or
accuracy on the concentration is required.
Problem 28: Describe how to prepare 100.00 mL of 3.00 M HCl
using 12.6 M HCl, distilled water, and the glassware below.
• 100 mL graduated cylinder
• 100 mL volumetric flask
• 25 mL graduated cylinder
• 25 mL buret
• 25 mL volumetric pipet
Dilution of Solutions
Common method to dilute a solution
involves the use of volumetric flask,
pipet, and suction bulb.
Problem 29: If 10.0 mL of 5.00 M NH4NO3 is added to enough
water to give 100. mL of solution, what is the concentration
of the solution?
Problem 30: What volume of 18.0 M sulfuric acid is required
to make 2.50 L of a 2.40 M sulfuric acid solution?
Mole fraction
Mole fraction = XA =
.
moles
moles A
total number of
Mole fraction does nor change with
temperature.
Problem 31: Find the mole fraction of KOH and H2O in a solution
that is prepared by dissolving 1.5 mol KOH in 1.0 kg H2O.
Using Solutions in Chemical
Reactions
Combine the concepts of molarity and
stoichiometry to determine the amounts
of reactants and products involved in
reactions in solution.
Using Molarities in
Stoichiometric Calculations
Using Solutions in Chemical
Reactions
Example: What volume of 0.500 M BaCl2 is
required to completely react with 4.32 g of
Na2SO4?
Na 2SO 4 + BaCl 2  BaSO 4 + 2 NaCl
1 mol Na 2SO 4 1 mol BaCl 2
1 L BaCl 2
4.32 gNa 2SO 4 


 0.0608 L
142 g Na 2SO 4 1 mol Na 2SO 4 0.500 mol BaCl 2
Problem 32: a) What volume of 0.200 M NaOH will react with
50.0 mL 0f 0.200 M aluminum nitrate, Al(NO3)3?
b) What mass of Al(OH)3 precipitates in (a)?
Using Solutions in Chemical
Reactions
Titrations are a method of determining the
concentration of an unknown solutions from the
known concentration of a solution and solution
reaction stoichiometry.
 Requires special lab glassware
 Buret, pipet, and flasks
 Must have an an indicator also
Titration
Titration
Titrations
Equivalence point: Is the exact moment when:
the moles of the titrant added = the moles of the titrated
The use of an indicator which changes colors around the
equivalence point.
When the indicator changes color the volume of the titrant can be
recorded.
Knowing the molarity and the volume of the titrant, the
moles of the titrant can be calculated.
Using Solutions in Chemical
Reactions
Example: What is the molarity of a KOH solution
if 38.7 mL of the KOH solution is required to
react with 43.2 mL of 0.223 M HCl?
Step 1: Convert to moles
KOH + HCl  KCl + H 2O
0.0432 L  0.223 M HCl = 0.00963 mol HCl
Using Solutions in Chemical
Reactions
Step 2: Convert to moles of the unknown.
KOH + HCl  KCl + H 2 O
1 mol KOH
0.00963 mol HCl 
 0.00963 mol KOH
1 mol HCl
Using Solutions in Chemical
Reactions
Step 3: Convert to molarity.
KOH + HCl  KCl + H 2O
0.00963 mol KOH
 0.249 M KOH
0.0387 L KOH
Problem 33: What is the molarity of a barium hydroxide
solution if 44.1 mL of 0.103 M HCl is required to react with 38.3
mL of the Ba(OH)2 solution?
Problem 34: What volume of 0.545 M of iron(II) nitrate is
required to react completely with 25.00mL of 0.200 M potassium
permanganate?
Problem 35: What is the molarity of Na+ ions present in a
solution prepared by mixing 10.00 mL of 0.150M NaCl with 25.00
mL of 0.0500M Na2SO4?
Problem 36: What is the molarity of Cl- ions present in a solution
prepared by mixing 100.00 mL of 0.100M MgCl2 with 125.00 mL
of 0.200M AlCl3?