Download Part One: Ions in Aqueous Solution A. Electrolytes and Non

CHAPTER FOUR: CHEMICAL REACTIONS
Part One: Ions in Aqueous Solution
A. Electrolytes and Non-electrolytes.
1.
Pure water does not conduct electric current appreciably. It is the ions dissolved in the
water that conduct charge.
2.
Electrolyte = substance that dissolves in water forming ions
Chapter 4
Page 1
3.
Classification of water-soluble substances.
Water-soluble
Substances
Electrolytes
-dissociate to some extent
in water to form ions
-solution conducts electricity
Strong
Electrolytes
-dissociate to large extent
-conduct well
Nonelectrolytes
-do not dissociate into
ions in water
-solution is non-conductive
Weak
Electrolytes
-dissociate to small extent
-conduct slightly
HAc (acetic acid)
strong
acids
HCl, HNO3,
H2SO4
a.
strong
bases
most soluble
salt
NaOH
NaCl, CaCl2
Dissociation = process in which a solid ionic compound separates into ions in
solution.
H O
NaCl 
→ Na + + Cl −
2
b.
Ionization = process in which a molecular compound separates to form ions.
HO
HCl 
→ H + + Cl −
2
c.
When one of the ions is H+, the substance is an acid.
HO
HCl 
→ H + + Cl −
2
Chapter 4
Page 2
d.
When OH- are produced, substance is a base.
HO
NaOH 
→ Na + + OH −
2
e.
When neither ion is H+ or OH-, substance is a salt.
H O
KBr 
→K + + Br −
2
B.
Strong and weak acids.
1.
Strong acids ionize completely in dilute aqueous solution.
Example: HCl is a gas when pure, dissolves to large extent in water, and fully ionizes.
H O
H O
HCl(g) 
→HCl(aq) 
→ H+ (aq) + Cl − (aq)
2
2
2.
Know these:
3.
H+ ions exist in aqueous solution attached to H2O (i.e., hydrated).
H+ (aq) really means H3O+
H +
H O
hydronium ion
H
Chapter 4
Page 3
4.
Weak acids ionize only slightly.
Example: acetic acid
CH3COOH(aq)
dominant form
≈ 99.5%
Abbreviation: HAc(aq)
H+ (aq) + CH3COO-(aq)
≈ 0.5%
H+ (aq) + Ac- (aq)
5.
Know these:
6.
Note that we have introduced the idea of reversible reactions. ( )
Reaction is continually happening to appreciable extent in both directions to produce a
balance in which both Reactants and Products are present.
C.
Bases = most commonly metal hydroxides. (Mx(OH)y)
1.
Strong soluble bases:
H O
NaOH(s) 
→ Na + (aq) + OH − (aq)
2
dissociate completely
Chapter 4
Page 4
a.
Insoluble bases: metal hydroxides which have low solubility in H2O, produce low
amount of OH-(aq).
Cu(OH)2, Zn(OH)2, Fe(OH)2, Fe(OH)3
b.
Weak bases: soluble in H2O but ionize only slightly.
Example:
NH3 (aq) + H2O(l)
molecular
compound
NH4+(aq) + OH- (aq)
D. Solubility rules for ionic compounds in aqueous solution.
1.
Compounds of Group IA and NH4+ are soluble.
2.
Nitrates, NO3-; acetates, CH3COO-; chlorates, ClO3-; and perchlorates, ClO4-, are
soluble.
3.
Chlorides, Cl-, are soluble EXCEPT AgCl, Hg2Cl2, and PbCl2. Bromides, Br-, and
iodides, I-, similar to chlorides, with some exceptions. As these halide ions (Cl-, Br-, I-)
increase in size, solubilities decrease.
4.
Sulfates, SO42-, are soluble EXCEPT PbSO4, BaSO4, HgSO4, CaSO4 , SrSO4 and
Ag2SO4
Carbonates, CO32-, are insoluble EXCEPT those of the Group IA metals and NH4+.
MgCO3 is moderately soluble.
5.
6.
Phosphates, PO43-, and arsenates, AsO43-, are insoluble EXCEPT those of the Group
IA metals and NH4+.
7.
Sulfides, S2-, are insoluble EXCEPT Group IA and IIA metals and NH4+.
8.
Metal hydroxides, OH-, are insoluble EXCEPT Group IA metals and Group IIA
metals below Mg.
Chapter 4
Page 5
E.
Writing Reactions in Aqueous Solution. (Section 4.2)
1.
Three ways of writing equations. (illustrated by neutralization of Mg(OH)2 with HCl
in aqueous solution.
a.
Molecular equations = show complete formulas for all compounds. (best for
stoichiometric calculations)
2HCl(aq) + Mg(OH)2(s) → MgCl2 (aq) + 2H2O
(insoluble)
(soluble)
b.
Complete ionic equations = show predominant form of compounds as ions if they
strongly dissociate. (the bridge to the net ionic equation)
2[H+(aq) + Cl-(aq)] + Mg(OH)2(s) → [Mg2+ (aq) + 2Cl-(aq)] + 2H2O
c.
Net ionic equation = complete ionic equation after “spectator ions” have been
canceled out.
2H+(aq) + Mg(OH)2(s) → Mg+ (aq) + 2H2O
[This is the essence of the reaction.]
2.
3.
To get the net ionic equation:
a.
Write down the full molecular equation first.
b.
For each compound, ask: Is it a strong electrolyte? If not, keep full formula. For
example, AgCl(s).
c.
Cancel out ions that appear on both sides. These are “spectator ions.”
Problem: Write down all 3 types of equations for the reaction of sodium sulfide with
silver nitrate to produce silver sulfide and sodium nitrate.
a.
Molecular equation:
Na2S(aq) + 2AgNO3(aq) → Ag2S(s) + 2NaNO3(aq)
↑
↑
↑
all Na salts
all nitrates
insoluble
soluble
soluble
sulfide
Chapter 4
Page 6
b.
Complete ionic equation:
[2Na+(aq) + S2-(aq)] + 2[Ag+(aq) + NO3-(aq)] →
Ag2S(s) + 2[Na+(aq) + NO3-(aq)]
c.
Net ionic equation:
S2-(aq) + 2Ag+(aq) → Ag2S(s)
(Simple formation of insoluble silver sulfide.)
Part Two: Types of Chemical Reactions
There are millions. We will look at three important types: a) precipitation; b) acid-base; and c)
oxidation-reduction.
Three driving forces for ionic reaction to happen, all in which there is a removal of ions
from solution:
a.
Formation of insoluble solid precipitate. (e.g. Ag2S)
b.
Formation of a very stable nonionized molecule. (e.g. H2O)
c.
Formation of a gas that escapes. (e.g., H2)
A. Precipitation Reactions. (Metathesis Reaction I) (Section 4.3)
1.
An insoluble solid (precipitate) forms and then settles out of solution.
2.
Driving force is strong attraction between cations and anions, resulting in removal of
ions from solution.
3.
Example: addition of sodium phosphate solution to a calcium chloride solution. What
happens? Pictorially:
Chapter 4
Page 7
Na+
Ca2+
Na+
Ca2+
PO43-
Cl-
Na+
Cl-
+
PO43-
Cl-
solid Ca3(PO4)2
Ca3(PO4)2 ppts!
2Na3PO4 (aq) + 3CaCl2(aq) → Ca3(PO4)2(s) + 6NaCl(aq)
Net ionic equation:
2PO43- (aq) + 3Ca2+(aq) → Ca3(PO4)2(s)
4.
Additional example: Potassium chromate solution added to lead nitrate solution. What
happens?
K+
Pb2+
soluble
insoluble
CrO42-
NO3-
K2CrO4(aq) + Pb(NO3)2(aq) → PbCrO4(s) + 2KNO3(aq)
Net ionic equation:
CrO42-(aq) + Pb2+(aq) → PbCrO4(s)
yellow solid
B.
Acid/Base Reactions. (Metathesis Reaction II) (Section 4.4)
1.
General form of neutralization between acid and base:
acid + base → salt + H2O
Chapter 4
Page 8
2.
Driving force here is formation of H2O from H+ and OH-.
3.
Example - strong acid + strong base:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Total ionic reaction:
[H+(aq) + Cl-(aq)] + [Na+(aq) + OH-(aq)] →
[Na+(aq) + Cl-(aq)] + H2O
Net ionic reaction for strong acid + strong base:
H+(aq) + OH-(aq) → H2O(l)
4.
Example - weak acid + strong base:
HAc(aq) + NaOH(aq) → NaAc(aq) + H2O(l)
Total ionic reaction:
HAc(aq) + [Na+(aq) + OH (aq)] → [Na+(aq) + Ac (aq)] + H2O(l)
Net ionic reaction for weak acid + strong base:
HAc(aq) + OH-(aq) → Ac-(aq) + H2O(l)
5.
Example of Acid-Base reaction with Formation of a Gas (a Driving Force for
Reaction)
reaction of insoluble carbonate salt with strong acid:
2HCl(aq) + CaCO3(s) → H2CO3(aq) + CaCl2(aq)
limestone
carbonic acid
seashells
(weak acid)
chalk
Chapter 4
Page 9
Net ionic reaction:
2H+(aq) + CaCO3(s) → H2CO3(aq) + Ca2+(aq)
i.e. calcium carbonate dissolves in acid.
driving force: further decomposition of H2CO3 to CO2(g)
H2CO3(aq) → CO2(g) + H2O(l)
escapes
(origin of bubbles)
C.
Oxidation-Reduction (Redox) Reactions. (introduction) (Section 4.5)
1.
Many reactions involve transfer of electrons between species, called oxidationreduction reactions, abbreviated redox reactions.
2.
To recognize these, we first need to introduce a concept called the oxidation number
or oxidation state.
3.
“Oxidation” originally referred to combining with Oxygen, resulting in increase in
oxidation number.
D. Oxidation Numbers. (keeping track of electrons)
1.
Oxidation number (ox#), or oxidation state, of an element:
a.
In an ionic compound, oxidation number of an element = number of e- gained or
lost by an atom of that element.
Example: NaCl compound consists of Na+ and Cl- ion pair.
ox# of Na = +1
Na+ said to be in “+1 oxidation state”
ox# of Cl = -1
CaCO3
ox# of Ca = +2
Chapter 4
Page 10
b.
In a molecular compound, ox# is assigned according to rules that aid in balancing
redox equations. Meaning is different.
Example: CO2
ox# of O is -2
ox# of C is +4
(+4)(-2)
CO2
Example: CO
ox# of O is -2
ox# of C is +2
(+2)(-2)
CO
C is said to be less fully oxidized to CO than in CO2.
2.
Rules for assigning ox#’s:
a.
Ox#’s always assigned on a per atom basis.
b.
Ox#’s of any free uncombined element is zero.
Examples: O2, S8, Cu(s)
Chapter 4
Page 11
c.
Ox#’s of an element in a simple monatomic ion equals the charge of the ion.
Na+, Na is +1
d.
Sum of ox#’s in a compound = 0.
e.
Sum of ox#’s in polyatomic ion = charge of ion.
CO3-2
C is +4; O is -2
sum = +4 + 3(-2) = -2
f.
H is always +1 except when combined with a metal in a metal hydride: NaH,
CaH2.
(+1)(-1)
NaH
g.
Group IA are always +1.
h.
Group IIA are always +2.
i.
Group IIIA are always +3, except in exotic compounds.
j.
Group VA are -3 in binary compound with a metal or with hydrogen.
(+1)(-3)
Na3N
(-3)(+1)
NH3
-sodium nitride
-ammonium
Otherwise, most often +3 or +5.
k.
l.
Chapter 4
Oxygen is -2 except in OF2 (+2) and in peroxides. (-1).
(+1)(-1)
H2O2
-example of exception
(+1)(-1)
Na2O2
-sodium peroxide
Halogens are -1 in binary compounds with metals or H.
Page 12
3.
Examples of assigning ox# in some compounds. (Hint: 1st do elements having hard
and fast rules, then others by elimination.)
a.
Sodium hypochlorite (Clorox bleach): NaOCl
(+1)
-do Group IA first always; Na
(-2)
-do O next; O
-Cl has to be what to sum to zero? +1
(+1) (-2)(+1)
Na O Cl
After bleach does its thing, Cl ends up as NaCl. Cl is -1 there.
b.
Potassium chromate: K2CrO4
(+1)
-do Group IA first always; K
(-2)
-do O; O
-Cr has to be what to sum to zero? +6
(+1) (+6)(-2)
K2 Cr O4
c.
Sulfuric acid: H2SO4
H is +1, except in hydrides.
O is -2, except in peroxides and OF2
S is what to make sum = 0? +6
(+1)(+6)(-2)
H2 S O4
d.
Sulfurous acid: H2SO3
(+1)(+4)(-2)
H2 S O3
Note: S is (+4) here and (+6) in sulfuric acid.
Chapter 4
Page 13
4.
“Oxidation” originally referred to combining with Oxygen, resulting in increase in
ox#.
Example - Combustion of methane is an oxidation of the Carbon:
(-4)(+1) (0) (+4)(-2)
CH4 + 2O2 → CO2 + 2H2O
(-4)
(+4)
{C} → {C} oxidation of C
5.
Oxidation = algebraic increase in ox# with corresp. loss of e-.
6.
Reduction = algebraic decrease in ox# with corresp. gain of e-.
7.
Oxidation and reduction always occur simultaneously.
Oxygen (0) → (-2)
Hydrogen (0) → (+1)
8.
9.
(0)
(0)
(+1)(-2)
2H2 + O2 → 2 H2O
reduced
oxidized
Oxidizing agent:
a.
Oxidizes other species.
b.
Is itself reduced.
Example: Immerse copper wire in silver nitrate solution. Copper is dissolved as the
Cupric ion and silver is deposited as metal.
AgNO3(aq) + Cu(s) → Cu(NO3)2(aq) + Ag(s)
Net ionic equation:
Ag+(aq) + Cu(s) → Cu2+(aq) + Ag(s)
What species is oxidized?
Cu (0) → (+2)
What species is reduced?
Ag (+1) → (0)
What is the oxidizing agent?
AgNO3 or Ag+(aq)
Chapter 4
Page 14
E.
F.
Many Oxidation-Reduction Reactions also fit under other categories, such as:
1.
Combination reactions.
2.
Decomposition reactions.
3.
Displacement reactions.
4.
Combustion reactions.
Combination Reactions.
1.
Combination reaction = reaction in which 2 or more substances combine to form a
third substance.
2.
Element + Element  Compound
3.
a.
2 Mg(s) + O2  2 MgO (metal + nonmetal  ionic cmpd)
b.
P4(s) + 6 Cl2(g)  4 PCl3(s) (nonmetal+nonmetal  molec cmpd)
Compound + Element  Compound
PCl3(s) + Cl2(g)  PCl5(s)
4.
Compound + Compound  Compound
CaO(s) + CO2(g)  CaCO3(s) (this one is NOT redox)
G. Decomposition Reactions.
1.
Chapter 4
Example: 2 KClO3  2 KCl + 3O2
Page 15
H. Displacement Reactions.
1.
Reaction in which one element displaces another from a compound.
M + AX → MX + A
2.
The more readily a metal M displaces another element and forms a positive ion, the
more active that metal is.
3.
The metal doing the displacing is being oxidized.
4.
Table 4.6 is Activity Series:
Chapter 4
Page 16
5.
[More Active Metal +
Salt of Less Active Metal]
→
[Less Active Metal +
Salt of More Active Metal]
Example: Zinc metal is placed in copper(II) sulfate solution. Copper metal falls to
bottom. Resulting solution contains zinc sulfate, ZnSO4.
Formula unit equation:
CuSO4(aq) + Zn(s) → Cu(s) + ZnSO4(aq)
Net ionic equation:
Cu2+ (aq) + Zn(s) → Cu(s) + Zn2+(aq)
Displacement reaction: the more active metal, zinc, displaces ions of less active metal,
copper, from aqueous solution.
6.
[Active Metal + Nonoxidizing Acid] → [Hydrogen + Salt of Acid]
a.
Example:
Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)
strong acid
soluble salt
Net ionic equation:
Zn(s) + H+(aq) + HSO4-(aq) → Zn2+(aq) + SO42-(aq) + H2(g)
b.
HNO3 is the common oxidizing acid. It reacts with active metals to produce
oxides of nitrogen, but not hydrogen, H2.
c.
Very active metals can even displace hydrogen from water.
Example:
2K(s) + 2H2O(l) → 2[K+(aq) + OH-(aq)] + H2(g)
7.
[Active Nonmetal +
→ [Less Active Nonmetal +
Salt of Less Active Nonmetal]
Salt of More Active Nonmetal]
a.
Example:
Cl2(g) + 2[Na+(aq) + Br-(aq)] → 2[Na+(aq) + Cl-(aq)] + Br2(l)
chlorine
sodium bromide
sodium chloride
bromine
Chapter 4
Page 17
b.
Each halogen will displace less active (heavier) halogens from their binary salts;
i.e., the order of increasing activities is:
(least active) I2 < Br2 < Cl2 < F2 (most active)
I.
Combustion Reactions.
1.
Combustion reaction = a reaction in which a substance reacts with oxygen, sometimes
with rapid release of heat to product a flame.
2.
Oxygen goes from 0 to -2 oxidation state.
3.
Of the combusted material, all Carbon ends up as CO or CO2, all Hydrogen ends up in
H2O.
4. Example: Combustion of propane:
___ C3H8(g) + ___ O2(g) 
J.
Balancing Simple Redox Reactions.
1.
The following redox reaction looks balanced, but what is wrong?
Zn(s) + Ag+(aq)  Zn2+(aq) + Ag(s)
The charge is not balanced.
2.
Balance using the Half-Reaction method:
(separate into two hypothetical half-reactions, one an oxidation, the other a reduction,
by specifically putting electrons in the equations.
Zn(s)  Zn2+(aq) + 2 eoxidation
+
2[Ag (aq) + 1 e-  Ag(s) ]
reduction
+
2+
Zn(s) + 2Ag (aq)  Zn (aq) + 2Ag(s)
3.
More complicated cases are handled in Chapter 19, in connection with
electrochemical cells.
Chapter 4
Page 18
Part Three: Working with Solutions
A. Molar Concentrations of Solution. (Section 4.7)
1.
Many reactions carried out in solution for convenience.
2.
Solution = homogeneous mixture of 2 or more substances.
3.
Solvent = the dispersing medium. (substance usually present in greatest abundance)
4.
Solute = substance dissolved in the solvent. (usually present in least amount)
5.
Aqueous solutions most common. (water is solvent)
6.
Concentrations expressed in various ways:
a.
Mass Percent of solute:
% solute =
b.
mass of solute
× 100%
mass of solution
Molarity (M) or molar concentration:
€
number of moles of solute
molarity =
number of liters of solution
B.
Mass Percent of Solute.
1.
€
Problem involving
percent solute:
A saline solution is 5.00% NaCl. Calculate the mass of NaCl present in 400 g of
solution.
5.00% means →
5.00 g NaCl
100 g solution
5.00 g NaCl
× 400g solution = 20.0g NaCl present
100 g solution
€
(% solute) ×
€
Chapter 4
€
100
mass solution = mass solute
Page 19
2.
Variation involving density:
A saline solution is 5.00% NaCl. Calculate the mass of NaCl present in 400 mL of
solution. The density of the solution is 1.10 g/mL.
400 mL solution x 1.10 g/mL = 440 g solution
Then work like previous example:
5.00 g NaCl
× 440g solution = 22.0g NaCl present
100 g solution
3.
€
Another variation: What volume of a 30% glucose solution contains 50 grams of
glucose? The density of solution is 1.15 g/mL.
50 g glucose ---> vol of solution required
50g glu ×
100g soln 1mL soln
×
= 145mL soln
30g glu 1.15g soln
100g soln
is the % solute by mass
30g glu
€
1mL soln
is the density of solution
1.15g soln
€
C. € Molarity, or Molar Concentration (M).
1.
Definition:
molarity =
2.
moles of solute
Liters of solution
Preparation of solutions of desired concentration:
€
Chapter 4
Page 20
What mass of NaOH would you weigh out to prepare 2.00 L of 0.100 M NaOH?
0.100 M reads 0.100 molar meaning 0.100 moles NaOH per Liter of solution.
0.100 moles NaOH
× 2.00 L solution = 0.200 moles NaOH
1 Liter of solution
€
0.200 moles NaOH ×
40.0 g NaOH
mol NaOH
= 8.0 g NaOH
€
Figure 4.18
3.
Important relations:
M=
n
V
molarity =
moles solute
Volume of solution
so: n = MV or:
€
(moles of solute present) = (conc in molarity) x (Volume in Liters)
4.
Problem: What mass of HCl is present in a 2.5 L jug of concentrated HCl solution
(11.6 M)?
a.
First compute moles HCl present:
moles
Chapter 4
= conc x volume
= 11.6 mol/L x 2.5 L
= 29.0 mol
Page 21
b.
Now convert to grams HCl:
29.0 mol x 36.45 g HCl/mol
= 1,057 g HCl
5.
Problem: Calculate the molarity of a solution in which 10.0 g NaOH are dissolved in
4.000 L solution.
 10.0 g 


moles NaOH
 40.0 g/mol 
molarity =
=
Vol solution (L)
4.000 L
= 0.0625 mol/L
= 0.0625 M
€
6.
Commercial phosphoric acid is 85% H3PO4 by mass, and its specific gravity is 1.69.
Calculate its molarity.
molarity =
moles H 3PO 4
Vol solution (L)
85 g H 3PO 4 1.69 g soln 10 3 mL 1 mol H 3PO 4
×
×
×
100g soln
1 mL soln
L
97.9 g H 3PO 4
= 14.7 mol/L
85% H 3PO 4 =
D. Dilution of Solutions. (Section 4.8)
€
1.
Dilution is simply the addition of solvent to an existing solution. Moles of solute
remains fixed.
2.
Therefore:
moles solute before = moles solute after
n1 = n2
M 1 V 1= M 2 V 2
molarity x volume = molarity x
before
before
after
Chapter 4
volume
after
Page 22
3.
Since V is on both sides, can use V in any units as long as consistent.
4.
Dilution problem:
If 10.0 mL of 3.0 M NaCl is diluted to 250.0 mL, what is the new molarity of NaCl
solution?
V 1M 1 = V 2M 2
(10.0 mL) x (3.0 M) = (250.0 mL) x M2
M2 =
E.
10.0 mL × 3.0 M
= 0.12 M
250 mL
Volumetric Analysis (Section 4.10)
€
(Solution Stoichiometry)
1.
Carrying out reactions in solution:
Problem: What volume of 0.100 M HCl(aq) would exactly neutralize 400 mL of
0.200 M NaOH(aq)?
moles HCl = moles NaOH
(at neutralization)
nHCl = nNaOH
MHCl VHCl = MNaOH VNaOH
0.100M VHCl = 0.200M × 0.400 L
VHCl =
0.200 M × 0.400 L
= 0.800 L = 800 mL
0.100 M
€
Chapter 4
Page 23
F.
Method of Titration:
1.
Titration = process in which a solution of one reactant (the titrant) is carefully added
to a solution of another reactant. Volume of titrant required for complete reaction is
measured.
2.
Complete reaction is determined by a color change in the indicator.
a.
Example: the indicator phenolphthalein is colorless in acidic solution and pink in
basic solution.
b.
The indicator is placed in acidic reactant solution. The base is then added until
the first drop produces a pink color that persists.
c.
When stoichiometric amounts have reacted, we call this the end point or
equivalence point of the titration.
d.
Example: reacting H2SO4 with NaOH, the equivalence point is when:
moles of NaOH added = 2 x (moles of H2SO4 present)
e.
Example: reacting HCl with NaOH, equivalence point is when:
moles of HCl = moles of NaOH
Chapter 4
Page 24
3.
Problem: 20.0 mL of HCl solution are required to neutralize 10.0 g of NaOH. What is
the molarity of the HCl solution?
4.
In the neutralization of phosphoric acid with sodium hydroxide, what volume of 0.200
M NaOH is required to neutralize 400 mL of 0.50 M H3PO4?
1 mole H3PO4 reacts with 3 mol NaOH.
0.400 L H3PO4 x 0.50 mol/L H3PO4 = 0.200 mol H3PO4
Requires 3 x 0.200 mol = 0.600 mol NaOH
0.600 mol = MV = 0.200 M x V
V=
0.600 mol
0.200 M
= 3.00 L
€
Chapter 4
Page 25
5.
Problem: 50.0 mL of H2SO4 of unknown concentration is titrated with a 0.100 M
NaOH titrant solution. The end point is reached after 40.0 mL of titrant are added.
Calculate molarity of H2SO4 solution.
H2SO4 + 2 NaOH → Na2SO4 + H2O
end point when:
(moles of NaOH) = 2 x (moles of H2SO4)
VbMb = 2 x (VaMa)
40.0 mL x 0.100 M = 2 x (50.0 mL x Ma)
Ma = 0.0400 M H2SO4
6.
In the neutralization of 2.00 g of MgO with HCl, 25.0 mL of an HCl solution are
required. What is the molarity of the HCl solution?
2 HCl + MgO  MgCl2 + H2O
Chapter 4
Page 26