Transcript Chapter 3: Calculations with Chemical Formulas

Vanessa Prasad-Permaul
Valencia Community College
CHM 1045



IONIC THEORY OF SOLUTIONS AND
SOLUBILITY RULES
CERTAIN SUBSTANCES PRODUCE
FREELY MOVING IONS WHEN THEY
DISSOLVE IN WATER
THOSE IONS CONDUCT AN
ELECTRICAL CURRENT IN AQUEOUS
SOLUTION

ELECTROLYTE: A SUBSTANCE THAT DISSOLVES IN
WATER TO GIVE AN ELECTRICALLY CONDUCTING
SOLUTION
In general, ionic solutions are electrolytes
NaCl(aq)
Na+(aq) + Cl-(aq)
some electrolytes are molecular substances
HCl (aq)
H+(aq) + Cl-(aq)

NONELECTROLYTE: A SUBSTANCE THAT
DISSOLVES IN WATER TO GIVE A NONCONDUCTING
OR VERY POORLY CONDUCTING SOLUTION

STRONG ELECTROLYTE: AN ELECTROLYTE THAT
EXISTS IN SOLUTION ALMOST ENTIRELY AS IONS
MOSTLY IONIC SUBSTANCES
NaCl(aq)

Na+(aq) + Cl-(aq)
WEAK ELECTROLYTE: AN ELECTROLYTE THAT
DISSOLVES IN WATER TO GIVE A RELATIVELY SMALL
PERCENTAGE OF IONS
GENERALLY MOLECULAR SUBSTANCES
WEAK ELECTROLYTES:
NH3(aq) + H2O(l)
NH4+(aq) + OH-(aq)
NH3(aq) + H2O(l)
NH4+(aq) + OH-(aq)
NH3(aq) + H2O(l)
NH4+(aq) + OH-(aq)
SOLUBILITY RULES FOR IONIC COMPOUNDS
EXERCISE 4.1
Determine whether the following compounds are
soluble or insoluble in water:
A)
B)
C)
D)
E)
NaBr
Ba(OH)2
CaCO3
Hg(NO3)2
AgCl
a.
b.
c.
d.
e.
According to the chart, all compounds that contain
sodium, Na+ are soluble, so NaBr is soluble in water.
According to Table 4.1, most compounds that contain
hydroxides, OH, are insoluble in water. However,
Ba(OH)2 is listed as one of the exceptions to this rule, so
it is soluble in water.
According to the chart, most compounds that contain
carbonate, CO32−, are insoluble. CaCO3 is not one of the
exceptions, so it is insoluble in water.
Mercuric nitrate or Hg(NO3)2 is soluble because all
forms of nitrates are soluble.
AgCl is insoluble. According to the chart, all Clcontaining compounds are soluble except Ag, Hg & Pb
MOLECULAR & IONIC EQUATIONS
Molecular equation: a chemical equation in which all
compounds are written as molecules (even if ions):
Ca(OH)2(aq) + Na2CO3(aq)
CaCO3(s) + 2NaOH(aq)
Complete Ionic Equations: a chemical equation in which
strong electrolytes are written as separate ions:
Ca2+(aq) + 2OH-(aq) + 2Na+(aq) + CO32-(aq)
CaCO3(s) + 2Na+(aq) + 2OH-(aq)
Spectator Ion: an ion that does not take part in an ionic
equation:
Ca2+(aq) + 2OH-(aq) + 2Na+(aq) + CO32-(aq)
CaCO3(s) + 2Na+(aq) + 2OH-(aq)
Net Ionic Equation: an ionic equation from which the
spectator ions have been cancelled:
Ca2+(aq) + CO32-(aq)
CaCO3(s)
EXERCISE 4.2
Write the complete net ionic equation for each of the
following:
2HNO3(aq) + Mg(OH)2(s)  2H2O(l) + MgNO3(aq)
*nitric acid is a strong electrolyte
PbNO3(aq) + Na2SO4(aq)  PbSO4(s) + 2NaNO3(aq)
2HClO4(aq) + Ca(OH)2(aq)  Ca(ClO4)2(aq) + 2H2O(l)
*perchloric acid is a strong electrolyte
HC2H3O2(aq) + NaOH(aq)  NaC2H3O2(aq) + H2O(l)
The resulting complete ionic equation is:
2H+(aq) + 2NO3−(aq) + Mg(OH)2(s) 
2H2O(l) + Mg2+(aq) + 2NO3−(aq)
The corresponding net ionic equation is:
2H+(aq) + Mg(OH)2(s)  2H2O(l) + Mg2+(aq)
The resulting complete ionic equation is:
Pb2+(aq) + 2NO3−(aq) + 2Na+(aq) + SO42−(aq) 
PbSO4(s) + 2Na+(aq) + 2NO3−(aq)
The corresponding net ionic equation is:
Pb2+(aq) + SO42−(aq)  PbSO4(s)

PRECIPITATION REACTIONS: Two ionic
solutions mix and a solid ionic substance
(precipitate) forms.

ACID-BASE REACTIONS: an acidic substance
reacts with as basic substance. This reaction
involves the transfer of a proton between reactants.
 Neutralization
 Acid-Base with Gas Formation

OXIDATION-REDUCTION REACTIONS: a
reaction in which there is a transfer of electrons
between the reactants.

PRECIPITATION REACTIONS: Two ionic solutions
mix and a solid ionic substance (precipitate) forms.
Molecular Equation:
MgCl2(aq) + 2AgNO3(aq)
2AgCl(s) + Mg(NO3)2(aq)
Complete Ionic Equation:
Mg2+(aq) + 2Cl-(aq) + 2Ag+(aq) + 2NO3- (aq)
2AgCl(s) + Mg2+(aq) + 2NO3-(aq)
Net Ionic Equation:
2Cl-(aq) + 2Ag+(aq)
2AgCl(s)
EXERCISE 4.3
•
•
For each of the following, decide whether a precipitation
reaction occurs. If it does, write the balanced molecular
equation and then the net ionic equation. If no
precipitation reaction occurs, write NPR.
•
Aqueous solution of sodium chloride and iron(II)
nitrate are mixed
•
Aqueous solution of aluminum sulfate and
sodium hydroxide are mixed
You mix an aqueous solution of sodium iodide and lead
(II) acetate. If a reaction occurs, write the balanced
molecular equation, the complete ionic equation and the
net ionic equation.
The formulas of the compounds are NaI which is soluble and
Pb(C2H3O2)2 is also soluble. Exchanging anions, you get sodium
acetate, NaC2H3O2 which is soluble, and lead(II) iodide, PbI2 which is
insoluble and will form a precipitate.
The balanced molecular equation is:
Pb(C2H3O2)2(aq) + 2NaI(aq)  PbI2(s) + 2NaC2H3O2(aq)
To get the net ionic equation, first get the complete ionic equation:
Pb2+(aq) + 2C2H3O2-(aq) + 2Na+(aq) + 2I-(aq) 
PbI2(s) + 2Na+(aq) + C2H3O2-(aq)
The final result is:
Pb2+(aq) + 2I−(aq)  PbI2(s)
COMMON ACIDS AND BASES
Acid–Base Neutralization: A process in which an
acid reacts with a base to yield water plus an ionic
compound called a salt.
The driving force of this reaction is the formation of
the stable water molecule.
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)

Arrhenius Acid: A substance which dissociates in
water to form hydrogen ions (H+).
HNO3(aq)

H+(aq) + NO3-(aq)
Arrhenius Base: A substance that dissociates in (or
reacts with) water to form hydroxide ions (OH–).
NaOH(s)
Na+(aq) + OH-(aq)
Limitations: Has to be an aqueous solution and doesn’t
account for the basicity of substances like NH3.
NH3(aq) + H2O(l)
NH4+(aq) + OH-(aq)


Brønsted Acid: Can donate protons (H+) to
another substance.
Brønsted Base: Can accept protons (H+)
from another substance. (NH3)
20
H+
NH3(aq) + H2O(l)
base
NH4+(aq) + OH-(aq)
acid
HNO3(aq)
HNO3(aq) + H2O(l)
H+(aq) + NO3-(aq)
H3O+(aq) + NO3-(aq)
H+
HNO3(aq) + H2O(l)
acid
base
H3O+(aq) + NO3-(aq)

Strong acid: strong electrolyte - almost
completely dissociates in water:
HCl, H2SO4, HNO3, HClO4, HI
HCl(aq) + H2O(l)
H3O+(aq) + Cl-(aq)



Weak acid: weak electrolyte - does not dissociate
well in water:
HF, HCN, CH3CO2H
+ H2O(l)
H3O(aq) + CN-(aq)


HF(aq)

Strong base: strong electrolyte - almost completely
dissociates in water:
 Metal hydroxides
 NaOH(s)
Na+(aq) + OH-(aq)

Weak base: does not dissociate well in water:
 NH3(aq) + H2O(l)
NH4+(aq) + OH-(aq)
STRONG ACIDS AND BASES
Other Weak bases – trimethyl ammonia N(CH3)3,
C5H5N pyridine, ammonium hydroxide NH4OH,
H2O water
EXERCISE 4.4
Label each of the following as a strong or weak acid or
base:
a) H3PO4
b) HClO
c) HClO4
d) Sr(OH)2
a. H3PO4 is not listed as a strong acid in the table so
it is a weak acid.
b. Hypochlorous acid, HClO, is not one of the
strong acids listed in the table, so we assume that
HClO is a weak acid.
c. As noted in the table, HClO4 is a strong acid.
d. As noted in the table, Sr(OH)2 is a strong base.
NEUTRALIZATION REACTIONS: A reaction of an
acid and a base that results in an ionic compound
(salt)and possibly water:
2HCl(aq) + Ca(OH)2(aq)
acid
base
HCN(aq) + KOH(aq)
acid
CaCl2(aq) + 2H2O(l)
base
salt
KCN(aq) + H2O(l)
salt
EXERCISE 4.5
•Write the molecular equation and the complete
ionic equation and the net ionic equation for the
neutralization of nitrous acid by sodium hydroxide,
both in aqueous solution.
•Write the molecular equation and the complete
ionic equation and the net ionic equation for the
neutralization of hydrocyanic acid by lithium
hydroxide (both in aqueous solution).
HCN(aq) + LiOH(aq)  LiCN(aq) + H2O(l)
Note that LiOH (a strong base) and LiCN (a soluble ionic
substance) are strong electrolytes; HCN is a weak
electrolyte (it is not one of the strong acids in the table
HCN(aq) + Li+(aq) + OH-(aq)  Li+(aq) + CN-(aq) + H2O(l)
After eliminating the spectator ions (Li+ and CN−), the net
ionic equation is:
HCN(aq) + OH−(aq)  H2O(l) + CN−(aq)
EXERCISE 4.6
Write the molecular equation, the complete ionic
equation and the net ionic equation for the
successive neutralization of each of the acidic
hydrogens of sulfuric acid with potassium
hydroxide
The first step in the neutralization is described by the
following molecular equation:
H2SO4(aq) + KOH(aq)  KHSO4(aq) + H2O(l)
The corresponding net ionic equation is:
H+(aq) + OH−(aq)  H2O(l)
The reaction of the acid salt KHSO4 is given by the
following molecular equation:
KHSO4(aq) + KOH(aq)  K2SO4(aq) + H2O(l)
The corresponding net ionic equation is:
HSO4−(aq) + OH−(aq)  H2O(l) + SO42−(aq)
Acid-Base Reactions with Gas Formations
Na2CO3(aq) + 2HCl(aq)
carbonate
acid
Na2CO3(aq) + 2HCl(aq)
CO32-(aq) + 2H+(aq)
2NaCl(aq) + H2O(l) + CO2(g)
salt
gas
2NaCl(aq) + H2CO3(aq)
H2O(l) + CO2(g)
EXERCISE 4.7
 Write the molecular equation and the net ionic
equation for the reaction of zinc sulfide with
hydrochloric acid.
 Write the molecular equation and the net ionic
equation for the reaction of calcium carbonate
with nitric acid.
First, write the molecular equation for the exchange
reaction, noting that the products of the reaction would be
soluble Ca(NO3)2 and H2CO3. The carbonic acid
decomposes to water and carbon dioxide gas.
The molecular equation for the process is:
CaCO3(s) + 2HNO3(aq)  Ca(NO3)2(aq) + H2O(l) + CO2(g)
The corresponding net ionic equation is
CaCO3(s) + 2H+(aq)  Ca2+(aq) + H2O(l) + CO2(g)
Oxidation-Reduction Reactions: reactions involving
a transfer of electrons from one species to another (or
in which atoms change oxidation states).
Fe(s) + CuSO4(aq)
Fe(s)
FeSO4(aq) + Cu(s)
Net ionic equation:
+ Cu2+(aq)
Fe2+(aq) + Cu(s)
Iron loses electrons = oxidized
Copper gains electrons = reduced
Oxidation Number: the actual charge of the atom if it
exists as a monatomic ion (or a hypothetical charge
assigned to the atom in the substance by simple rules).
2Ca(s) + O2(g)
0
0
2Ca(s) + O2(g)
2CaO(s)
+2
-2
2CaO(s)
*The concept of oxidation numbers (states) was developed as a simple way
of keeping track of electrons in a reactions
Rules for Assigning Oxidation Numbers
1.
2.
3.
4.
5.
6.
7.
8.
Free elements are assigned an oxidation state of zero.
The sum of the oxidation states of all that atoms in a species must be
equal to the net charge on the species.
The alkali metals (Li, Na, K, Rb, and Cs) in compounds are always
assigned an oxidation state of +1.
Fluorine in compounds is always assigned an oxidation state of -1.
The alkaline earth metals (Be, Mg, Ca, Sr, Ba, and Ra) and also Zn and
Cd in compounds are always assigned an oxidation state of +2.
Hydrogen in compounds is assigned an oxidation state of +1.
Oxygen in compounds is assigned an oxidation state of -2.
Halogen in compounds is assigned an oxidation state of -1.
EXERCISE 4.8
Use the oxidation number rules to obtain the
oxidation number of the chlorine atom in each of the
following:
A) HClO4
B) ClO3-
Obtain the oxidation numbers of the atoms in
each of the following:
A) Potassium dichromate
B) Permanganate ion
A) For potassium dichromate, K2Cr2O7:
2 x (oxidation number of K) + 2 x (oxidation number of Cr) + 7
x (oxidation number of O) = 0
For oxygen, the oxidation number is −2 (rule 3), and for potassium
ion, the oxidation number is +1 (rule 2)
[2 x (+1)] + 2 x (oxidation number of Cr) + [7 x (−2)] = 0
Therefore,
2 x oxidation number of Cr = − [2 x (+1)] − [7 x (−2)] = +12
or, oxidation number of Cr = +6.
B) For the permanganate ion, MnO4−:
(Oxidation number of Mn) + 4 x (oxidation number of O) = −1
For oxygen, the oxidation number is −2 (rule 3).
(oxidation number of Mn) + [4 x (−2)] = −1
Therefore,
Oxidation number of Mn = −1 − [4 x (−2)] = +7

Redox reactions are those involving the
oxidation and reduction of species.
Oxidation and reduction must occur together. They
cannot exist alone.
Fe2+ + Cu0  Fe0 + Cu2+
Half-reaction: one of two parts of an oxidationreduction reaction. One part involves a loss of
electron, the other a gain of electrons:
Reduced: Iron gained 2 electrons
Oxidized: Copper lost 2 electrons
Fe2+ + 2 e  Fe0
Cu0  Cu2+ + 2e
reduction
Fe2+ + Cu0  Fe0 + Cu2+
oxidizing reducing
agent
agent
oxidation


Fe2+ gains electrons, is reduced, and we call it an oxidizing
agent
 Oxidizing agent is a species that can gain electrons and
this facilitates in the oxidation of another species.
(electron deficient)
Cu0 loses electrons, is oxidized, and we call it a reducing
agent
 Reducing agent is a species that can lose electrons and
this facilitates in the reduction of another species.
(electron rich)
Some Common Oxidation-Reduction Reactions:
1. Combination Reaction
2. Decomposition Reaction
3. Displacement Reaction
4. Combustion Reaction
1. Combination Reaction: two substances combine
to form a third substance:
2Na(s) + Cl2(g)
2Sb(s) + 3Cl2(g)
2NaCl(s)
2SbCl3(s)
2. Decomposition Reaction: a single compound
reacts to give two or more substances.
2HgO(s)
2KClO3(s)
2Hg(l) + O2(g)
MnO2
2KCl(s) + 3O2(g)
3. Displacement Reaction (Single Replacement
Reaction): an element reacts with a compound
displacing another element from that compound
Cu(s) + 2 AgNO3(aq)
Zn(s) + 2H+
Cu(NO3)2(aq) + 2Ag(s)
ZnCl2(aq)
+
H2(g)
4. Combustion Reaction: a substance reacts with
oxygen, usually with a rapid release of heat to
produce a flame.
2C4H10(g) + 13O2(g)
4Fe(s) + 3O2(g)
8CO2(g) + 10H2O(g)
2Fe2O3(s)


Half-Reaction Method: Allows you to focus
on the transfer of electrons. This is
important when considering batteries and
other aspects of electrochemistry.
The key to this method is to realize that the
overall reaction can be broken into two
parts, or half-reactions. (oxidation half and
reduction half)
Balance for an acidic solution:
MnO4–(aq) + Br–(aq)  Mn2+(aq) + Br2(aq)
1. Determine oxidation and reduction halfreactions:
Oxidation half-reaction: Br–(aq)  Br20(aq)
Reduction half-reaction: MnO4–(aq)  Mn2+(aq)
2. Balance for atoms other than H and O:
Oxidation: 2 Br–(aq)  Br2(aq)
Reduction: MnO4–(aq)  Mn2+(aq)
3. Balance for oxygen by adding H2O to the
side with less oxygen
Oxidation: 2 Br–(aq)  Br2(aq)
Reduction: MnO4–(aq)  Mn2+(aq) + 4 H2O(l)
4. Balance for hydrogen by adding H+ to the
side with less hydrogens
Oxidation: 2 Br–(aq)  Br2(aq)
Reduction: MnO4–(aq) + 8 H+(aq)  Mn2+(aq) + 4
H2O(l)
5. Balance for charge by adding electrons (e–):
Oxidation: 2 Br–(aq)  Br2(aq) + 2 e–
Reduction: MnO4–(aq) + 8 H+(aq) + 5 e–  Mn2+(aq) + 4 H2O(l)
6. Balance for numbers of electrons by multiplying:
Oxidation: 5[2 Br–(aq)  Br2(aq) + 2 e–]
Reduction: 2[MnO4–(aq) + 8 H+(aq) + 5 e–  Mn2+(aq) + 4
H2O(l)]
7. Combine and cancel to form one equation:
Oxidation: 10 Br–(aq)  5 Br2(aq) + 10 e–
Reduction: 2 MnO4–(aq) + 16 H+(aq) + 10 e– 2 Mn2+(aq) + 8
H2O(l)
2 MnO4–(aq) + 10 Br–(aq) + 16 H+(aq) 2 Mn2+(aq) + 5 Br2(aq) + 8
H2O(l)
We will not be balancing in basic solutions!!
(until CHM 1046)
EXERCISE 4.9
Use the half reaction method to balance the equation:
Ca(s) + Cl2(g)
CaCl2(s)
Identify the oxidation states of the elements.
Break the reaction into two half-reactions, making sure that both
mass and charge are balanced.
Ca  Ca2+ + 2e−
Cl2 + 2e−  2Cl−
Since each half-reaction has two electrons, it is not necessary to
multiply the reactions by any factors to cancel them out. Adding
the two half-reactions together and canceling out the electrons,
you get:
Ca(s) + Cl2(g)  CaCl2(s)
SOLUTE: The substance that dissolves in liquid.
SOLVENT: The liquid that the solute dissolves in.
MOLAR CONCENTRATION:
MOLARITY(M) = moles of solute
liters of solution
EXERCISE 4.10:
A sample of NaNO3 weighing 0.38-g is placed in a 50.0mL volumetric flask. The flask is then filled with DI water
to mark and carefully mixed. What is the molarity of the
solution?
A sample of NaCl weighing 0.0678-g is placed in a 25.0mL volumetric flask. Enough water is added to dissolve
the NaCl, and the flask is filled to mark and carefully
mixed. What is the molarity of the solution?
Convert mass of NaCl (molar mass, 58.44 g) to moles of
NaCl.
Then divide moles of solute by liters of solution. Note
that 25.0 mL = 0.0250 L.
0.0678 g NaCl x
1 mol NaCl
58.44g
= 1.160 x 10−3 mol NaCl
Molarity = 1.160 x 10-3mol = 0.04641 = 0.0464 M NaCl
0.0250L
EXERCISE 4.11
An experiment calls for the addition to the reaction vessel
of 0.814-g of sodium hydroxide(aq). How many mL of
0.150M NaOH should be added?
How many mL of 0.163M NaCL are required to give
0.0958-g of NaCl?
EXERCISE 4.12
How many moles of NaCl should be put in a 50.0mL
volumetric flask to give 0.15M NaCl solution? How many
grams of NaCl is this?
EXERCISE 4.11: Convert grams of NaCl (molar mass, 58.44 g) to
moles NaCl and then to volume of NaCl solution.
0.0958 g NaCl x mol NaCl x
L NaCl x 1000mL = 10.06
58.44g
0.163mol
1L
= 10.1 mL NaCl
EXERCISE 4.12: One (1) liter of solution is equivalent to 0.15 mol
NaCl. The amount of NaCl in 50.0 mL of solution is:
50.0 mL x 1L x 0.150mol = 0.00750 mol NaCl
1000mL
1L
Convert to grams using the molar mass of NaCl (58.44 g/mol).
0.00750 mol NaCl x
58.44g NaCl
mol
= 0.438 = 0.44 g NaCl
Mi x Vi = Mf x Vf
Where Mi = initial concentration (molarity)
Vi = initial volume
Mf = final concentration (molarity)
Vf = final volume
EXERCISE 4.13
There is a solution of 14.8M NH3. How many mL of this
solution is needed to give 100.0mL of 1.00M NH3
solution?
There is a solution of 1.5M sulfuric acid. How many mL
of this acid is needed to prepare 100.0mL of 0.18M
sulfuric acid?
Use the rearranged version of the dilution formula from
the text to calculate the initial volume of 1.5 M sulfuric
acid required:
Mi x Vi = Mf x Vf
1.5M H2SO4 x Vi = 0.18M H2SO4 x 100.0mL
Vi = 0.18M x 100.0mL
1.5M
Vi = 12.0 = 12 mL