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SOLUTIONS
1
Definitions
The solvent is the dissolving agent
-- i.e., the most abundant component of the solution
- The solute is the component that is dissolved
-- i.e., the least abundant component of the solution
A solution is a system in which one or more substances are
homogeneously mixed or dissolved in another substance
- homogeneous mixture
-- uniform appearance
-- similar properties throughout mixture
2
Concentration of solutions
concentration -- the amount of solute dissolved in a given
quantity of solvent or solution
There are many different types of concentration units:
molarity
mass %
volume %
mass/volume %
parts per million (ppm)
parts per billion (ppb)
mole fraction
molality (not to be confused with molarity )
we will focus on this
molarity -- number of moles of solute per liter of
solution
Molarity =
moles of solute
liters solution
Concentration units based on
the number of moles of solute
molarity has units of moles per liter
mol
L
which can be abbreviated as M
Example: Preparation of a 1 molar solution of NaCl
Add 1 mole of NaCl
to an empty 1-liter
volumetric flask
Add water to
completely
dissolve NaCl
Add water until the
fill mark is reached
and mix thoroughly
1 mole NaCl =
58.44 g NaCl
Molarity =
1.000 mol
1.000 L
= 1.000 M
water
fill
line
1000.0-ml
flask
5
Example: Preparation of a 1 molar solution of NaCl
Add 0.250 mole of
NaCl to an empty
250-ml volumetric flask
Add water to
completely
dissolve NaCl
Add water until the
fill mark is reached
and mix thoroughly
0.250 mole NaCl
= 14.61 g NaCl
Molarity =
0.2500 mol
0.2500 L
= 1.000 M
water
fill
line
Note: Dissolving 1 mole of solute to make 1 liter of solution is not the only
way to prepare a solution with a concentration of 1 M ( i.e., 1 mol / L )
250.0-ml
flask
6
Example: Preparation of a 0.5 molar solution of NaCl
Molarity =
0.5000 mol
1.000 L
= 0.5000 M
fill
line
1000.0-ml
flask
Molarity =
0.2500 mol
0.5000 L
= 0.5000 M
fill
line
500.0-ml
flask
0.5 mole NaCl =
29.22 g NaCl
0.250 mole NaCl
= 14.61 g NaCl
What is the molarity of a solution made by dissolving 2.00 g of
potassium chlorate in enough water to make 150. ml of solution?
Step 1: Start with the definition of molarity:
Molarity =
moles of solute
liters solution
Step 2: Determine the number of moles of solute
Molar mass of KClO
3
= 39.10 + 35.45 + 3(16.00) = 122.6 g / mol
2.00 g KClO
3
1 mole KClO
3
122.6 g KClO
3
= 0.0163 moles KClO
3
moles of solute
liters solution
Step 4: Convert moles KOH to grams KOH
Molar mass of KOH = 39.10 + 16.00 + 1.008 = 56.11 g / mol
0.225 moles KOH
56.11 g KOH
1 mole KOH
= 12.6 g KOH
Molarity =
How many grams of potassium hydroxide are required to
prepare 500. ml of 0.450 M KOH solution?
13
moles HNO
3
L
Calculate the number of moles of nitric acid in 325 ml of
16 M HNO
3
Step 1: Start with the definition of molarity:
Molarity =
moles of solute
liters solution
Step 2: Plug known values into molarity equation and solve for
unknown (moles of solute)
16 M HNO
3
x moles of HNO
3
0.325 L
=
x moles of HNO
3
0.325 L
16 = (0.325 L)
(0.325 L)
5.2 moles HNO
3
= x
14
Concentrations of ions in aqueous solutions
When an ionic compound dissolves, the concentrations of the
individual ions depend on the chemical formula of the compound
Example: What are the concentrations of Na
+
and Cl
-
ions in
a 0.125 M aqueous solution of NaCl?
0.125 mol NaCl
L solution
0.125 mol Na
L solution
0.125 mol Cl
L solution
0.125 M NaCl =
Na
+
Cl
-
Na
+
Cl
-
Each NaCl formula unit produces
1 Na
+
ion and 1 Cl
-
ion in solution
= 0.125 M Na
+
= 0.125 M Cl
-
Concentrations of ions in aqueous solutions
Example: What are the concentrations of Mg
2+
and Cl
-
ions
in a 0.125 M aqueous solution of MgCl 2
0.125 mol MgCl 2
L solution
0.125 mol Mg
2+
L solution
0.250 mol Cl
L solution
0.125 M MgCl 2
Mg
2+
Cl
-
Cl
-
Each MgCl 2
formula unit produces
1 Mg
2+
ion and 2 Cl
-
ions in solution
Cl
-
Mg
2+
Cl
-
= 0.125 M Mg
2+
= 0.250 M Cl
-
When an ionic compound dissolves, the concentrations of the
individual ions depend on the chemical formula of the compound
A solution is made by dissolving 9.82 g of copper (II) chloride (CuCl 2
) in enough
water to make 600. mL of solution. What is the molarity of Cl
-
ions in solution?
Step 1: Determine the number of moles of solute
Molar mass of CuCl
2
= 63.55 + 2(35.45) = 134.45 g / mol
9.82 g CuCl
2
1 mole CuCl
2
134.45 g CuCl
2
= 0.0730 moles CuCl
2
Step 2: Determine molarity of solute
Molarity =
0.0730 moles CuCl
2
0.600 L
= 0.122 M CuCl
2
17
Step 3: Determine the ion to solute ratio
CuCl
2
dissociates to give one Cu
2+
ion and two Cl
ions
Step 4: Determine molarity of the ion
0.122 M CuCl
2
CuCl
2
Cu
2+
2 moles Cl
ions
1 mole CuCl
2
2 moles Cl
ions
1 mole CuCl
2
= 0.244 M Cl
-
ions
A solution is made by dissolving 9.82 g of copper (II) chloride (CuCl 2
) in enough
water to make 600. mL of solution. What is the molarity of Cl
-
ions in solution?
18
Dilutions
More solvent is added:
-- volume of the solution increases
No additional solute is added
-- number of moles of solute stays the same
Net result: The molarity of the solution decreases
Molarity =
moles of solute
liters solution
(unchanged)
Dilution: Reducing the concentration of a solution by adding
more solvent to the solution
Na
NaNO
3
solution
NO
3
Moles = 1.0 mol
Volume = 1.0 L
Molarity =
1.0 mol
1.0 L
= 1.0 M
Solubility rules
AN IONIC COMPOUND
IS SOLUBLE IN WATER
IF IT CONTAINS THE
FOLLOWING IONS:
EXCEPTIONS
Ammonium ion (NH 4
) none
Alkali metal (Group IA) ions
(Li
, Na
, K
)
none
Nitrate (NO 3
)
Acetate (C 2
H 3
O 2
)
none
Halides (Cl
, Br
, I
)
Compounds containing
Ag
, Pb
2+
, Hg 2
2+
Sulfate (SO 4
2-
)
Compounds containing
Ag
, Pb
2+
, Ca
2+
, Sr
2+
, Ba
2+
25
Solubility rules
AN IONIC COMPOUND IS
NOT SOLUBLE IN WATER
IF IT CONTAINS THE
FOLLOWING IONS:
EXCEPTIONS
Carbonate (CO 3
2-
)
Phosphate (PO 4
3-
)
Compounds containing
Li
, Na
, K
, NH 4
(soluble)
Hydroxide (OH
)
Compounds containing
Li
, Na
, K
, NH 4
(soluble)
Compounds containing
Ca
2+
, Ba
2+
, Sr
2+
(slightly soluble)
Sulfide (S
2-
)
Compounds containing
Li
, Na
, K
, NH 4
(soluble)
Compounds containing
Ca
2+
, Ba
2+
, Sr
2+
(soluble)
26
Sample problems
Are the following compounds soluble or insoluble in water?
NaCl
(NH
4
3
PO
4
CaCO
3
MgSO
4
BaSO
4
soluble
soluble
insoluble
soluble
insoluble
Double-displacement reactions
double-displacement reaction -- two ionic compounds exchange
partners ( i.e., cations and anions) to form two different compounds
General form:
AB + CD AD + BC
Precipitation reactions are a type of double-displacement reaction
cations
switch
places
two new
compounds
are formed
Reactions of aqueous solutions: Precipitation reactions
In a precipitation reaction , an insoluble solid (called a precipitate) is
formed when reactants in aqueous solution ( i.e., dissolved in water) are
combined
BaCl
2
( aq ) + 2 AgNO
3
( aq ) 2 AgCl ( s ) + Ba(NO
3
2
( aq )
insoluble precipitate
indicated by (s) after its formula
Precipitation reactions are a type of double-displacement reaction
29
Precipitation reactions
Most precipitation reactions occur when the anions and cations of two
aqueous ionic compounds switch partners
General form:
AB + CD AD + BC
To predict whether a precipitation reaction will occur:
look at the potential products of the reaction ( i.e., make the anions and
cations switch partners)
determine whether either product is an insoluble solid
ZnCl
2
( aq ) + 2 KOH ( aq ) Zn(OH)
2
+ 2 KCl
Example: Will a precipitation reaction occur when aqueous zinc chloride
and potassium hydroxide are mixed?
Use solubility rules to determine if either of these is an insoluble solid
( aq ) ( s )
30
+ HCl
+ CuSO
4
Will the following reactions take place?
insoluble precipitate
CoCl
3
( aq ) + H
2
S ( aq ) Co 2 S 3
insoluble precipitate
( s ) ( aq )
Na 2
SO
4
( aq ) + CuBr
2
( aq ) NaBr
no reaction
( aq ) ( aq )
Pb(NO
3
2
( aq ) + KI ( aq ) PbI
2
( s ) ( aq )
+ KNO
3
Pb
2 +
( aq ) + 2 I
( aq ) PbI
2
( s )
Pb
2 +
( aq ) + 2 NO 3
( aq ) + 2 K
( aq ) + 2 I
( aq ) PbI
2
( s ) + 2 K
( aq ) + 2 NO
3
( aq )
Complete ionic equation
All soluble strong electrolytes are shown as ions
-- aqueous substances are shown as separate cations and anions
Pb(NO
3
2
( aq ) + 2 KI( aq ) PbI
2
( s ) + 2 KNO
3
( aq )
For chemical reactions involving aqueous solutions,
three types of equations can be written
Molecular equation
Formulas written for all reactants & products do not show their ionic character
-- i.e., aqueous substances are shown as neutral compounds
Net ionic equation
Includes only the substances that undergo change
-- ions that are present but do not react (spectator ions) are not shown
Solution of sugar in water
(no charge carriers)
sugar in a non-electrolyte
Electrolytes
electrolyte -- a substance that forms ions when dissolved in
water, resulting in a solution that conducts electricity
Strong and weak electrolytes
Strong electrolytes are solutes that exist in solution
completely or nearly completely as ions
Weak electrolytes are solutes that dissociate only partially to
form ions in solution
-- exist primarily as non-dissociated molecules in solution,
with only a small fraction in the form of ions
- Nearly all soluble ionic compounds are strong electrolytes
- Strong acids and bases are strong electrolytes
- Weak acids and bases are weak electrolytes
We will talk about strong/weak acids and bases shortly
