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Chapter 4
Solutions and Solution Stoichiometry
Solutions
Homogeneous mixtures are called solutions.
The component of the solution that changes state is called the solute.
The component that keeps its state is called the solvent.
If both components start in the same state, the major component is the solvent.
Gas in gas Air (O2 in N2)
Gas in liquid Carbonated water (CO2 in H2O)
Gas in solid H2 in palladium metal
Liquid in liquid Gasoline, tequila
Liquid in solid Dental amalgam (Hg in Ag)
Solid in liquid Salt water (NaCl in H2O)
Solid in solid Sterling silver (Cu in Ag)
Kinds of Solutions
Solvent and Solute in an Aqueous Solution
Solvent and Solute in an Aqueous Solution
The Solution Process
What Happens When a Solute Dissolves?
There are attractive forces between the solute particles holding them together.
There are also attractive forces between the solvent molecules.
When we mix the solute with the solvent, there are attractive forces between the solute particles and the solvent molecules.
If the attractions between solute and solvent are strong enough, the solute will dissolve.
Table Salt Dissolving in Water
In the case of an ionic compound
each ion is attracted to the surrounding water molecules and pulled off and away from the
crystal.
Sugar Dissolving in Water
molecular compounds
do not dissociate when
they dissolve
In the case of a molecular compound,
each molecule is surrounded by water
molecules and pulled off and away from the
crystal.
Electrolytes and
Nonelectrolytes
Salt vs. Sugar Dissolved in Water
ionic compounds
dissociate into ions when
they dissolve
molecular compounds
do not dissociate when
they dissolve
Ionic compounds dissociate into ions when
they dissolve.
Molecular compounds do not dissociate into ions
when they dissolve.
Electrolytes and Nonelectrolytes
Materials that dissolve in water to form a solution
that will conduct electricity are called
electrolytes.
Materials that dissolve in water to form a solution
that will not conduct electricity are called
nonelectrolytes.
ionic compounds
covalent compounds
AcidsAcids are molecular compounds that ionize when they
dissolve in water. The molecules are pulled apart by water.
The percentage of molecules that ionize varies.
Acids that ionize virtually 100% are called strong acids.
HCl(aq) ➜ H+(aq) + Cl−(aq)
Acids that only ionize a small percentage are called weak acids.
HF(aq) ⇔ H+(aq) + F−(aq)
Strong and Weak Electrolytes
Strong electrolytes are materials that dissolve completely as ions.
1) ionic compounds and strong acids 2) The solutions conduct electricity well.
Weak electrolytes are materials that dissolve mostly as molecules, but partially as ions.
1) weak acids 2) The solutions conduct electricity, but not well.
• When ionic compounds dissolve in water, the
anions and cations are separated from each
other. This is called dissociation.
Na2S(aq) → 2 Na+(aq) + S2-(aq)
• When compounds containing polyatomic ions
dissociate, the polyatomic group stays together
as one ion
Na2SO4(aq) → 2 Na+(aq) + SO42−(aq)
• When strong acids dissolve in water, the
molecule ionizes into H+ and anions
H2SO4(aq) → 2 H+(aq) + SO42−(aq)
Dissociation vs IonizationWhen ionic compounds dissolve in water, the
anions and cations are separated from each other. This is called dissociation.
When compounds containing polyatomic ions dissociate, the polyatomic group stays together as one ion.
When strong acids dissolve in water, the molecule ionizes into H+ and anions.
• When ionic compounds dissolve in water, the
anions and cations are separated from each
other. This is called dissociation.
Na2S(aq) → 2 Na+(aq) + S2-(aq)
• When compounds containing polyatomic ions
dissociate, the polyatomic group stays together
as one ion
Na2SO4(aq) → 2 Na+(aq) + SO42−(aq)
• When strong acids dissolve in water, the
molecule ionizes into H+ and anions
H2SO4(aq) → 2 H+(aq) + SO42−(aq)
• When ionic compounds dissolve in water, the
anions and cations are separated from each
other. This is called dissociation.
Na2S(aq) → 2 Na+(aq) + S2-(aq)
• When compounds containing polyatomic ions
dissociate, the polyatomic group stays together
as one ion
Na2SO4(aq) → 2 Na+(aq) + SO42−(aq)
• When strong acids dissolve in water, the
molecule ionizes into H+ and anions
H2SO4(aq) → 2 H+(aq) + SO42−(aq)
Write the equation for the process that occurs when the following strong electrolytes dissolve in water
CaCl2
HNO3
(NH4)2CO3
CaCl2(aq) ➜ Ca2+(aq) + 2 Cl−(aq)
HNO3(aq) ➜ H+(aq) + NO3−(aq)
(NH4)2CO3(aq) ➜ 2 NH4+(aq) + CO3
2−(aq)
The Real Picture of Acid Ionization
HCl + H2O ➜ H3O+ + Cl−
hydronium ion
HCl ➜ H+ + Cl−
[ ]H Cl
H
HO O
H
H
H Cl+ +
+ -
strong acid water
chloride ion
Solubility
Solubility of Ionic Compounds
Some ionic compounds, such as NaCl, dissolve very well in water at room temperature.
Other ionic compounds, such as AgCl, dissolve hardly at all in water at room temperature.
Compounds that dissolve in a solvent are said to be soluble, where as those that do not are said to be insoluble.
Solubility Rules Compounds that Are Generally Soluble in Water
Compounds Containing the Following Ions are Generally
Soluble
Exceptions Insoluble
Li+, Na+, K+, NH4+ none
NO3-, C2H3O2 - none
Cl -, Br -, I - Ag+, Hg22+, Pb2+
SO4 - Ag+, Ca2+, Sr2+, Ba2+, Pb2+
Solubility Rules Compounds that Are Generally Insoluble in Water
Compounds Containing the Following Ions are Generally
Insoluble
Exceptions Soluble or Slightly Soluble
OH- Li+, Na+, K+, NH4+
Ca2+, Sr2+, Ba2+
S2- Li+, Na+, K+, NH4+
Ca2+, Sr2+, Ba2
CO32-, PO43-, CrO42- Li+, Na+, K+, NH4+
SO4 - Li+, Na+, K+, NH4+
Practice – Determine if each of the following is soluble in water
KOH
AgBr
CaCl2
Pb(NO3)2
PbSO4
KOH is soluble because it contains K
AgBr is insoluble; most bromides are soluble, but AgBr is an exception
CaCl2 is soluble; most chlorides are soluble, and CaCl2 is not an exception
Pb(NO3)2 is soluble because it contains NO3−
PbSO4 is insoluble; most sulfates are soluble, but PbSO4 is an exception
Solution Concentrations
Solution Concentration Qualitative Description
Dilute solutions have a small amount of solute compared to
solvent.
Concentrated solutions have a large amount of solute
compared to solvent.
Quantitative Descriptions of Solutions
One method for describing a solution is to quantify the amount of solute in a given amount of solution.
Solution Concentration - Molarity
Moles of solute per 1 liter of solution
Used because it describes how many molecules of solute in each liter of solution
mol LMolarity =
moles of solute
Liters of solution
How to prepare 1.000 L of a 1.000 M NaCl solution
First add 1 mole of NaCl to a 1.000 liter volumetric flask
25.5 g KBr x
= 0.21429 mol KBr
1.00 mol KBr119.00 g KBr
1. Find the molarity of a solution that has 25.5 g KBr dissolved in 1.75 L of solution.
gKBr
molKBr
g
mol
molKBr
LKBr soln
= M
0.214
molarity, M =
= 0.12245 M KBr
0.214 mol KBr1.75 L KBr soln
0.122
2. What Is the molarity of a solution containing 3.4 g of NH3 (MM 17.03) in 200.0 mL of solution?
g NH3 mol NH3
mL sol’n L sol’n
M
Using Molarity in CalculationsMolarity shows the relationship between the moles
of solute and liters of solution.
If a sugar solution concentration is 2.0 M, then 1.0 liter of solution contains 2.0 moles of sugar.
Questions one might ask: What is the molarity of....?
How many liters contain....? How many moles....?
How many grams of....? How would I prepare....?
2.0 mol sugar1.0 L soln
1.0 L soln2.0 mol sugar
0.255 mol NaOH x = 2.04 L NaOH soln1.00 L NaOH soln0.125 mol NaOH
3. How many liters of 0.125 M NaOH contain 0.255 mol NaOH?
mol NaOH L NaOH soln
mol
L
Molarity=moles of solute
Liters of solution
4. Determine the mass of CaCl2 (MM = 110.98) in 1.75 L of 1.50 M solution.
L CaCl2 soln mol CaCl2 g CaCl2
mol
L
mol
g
1.75 L CaCl2 soln x x
= 291.32 g CaCl2
1.50 mol CaCl21.00 L CaCl2 soln
110.98 g CaCl21.00 mol CaCl2
5. How would you prepare 250.0 mL of a 1.000 M solution CuSO4·5 H2O (MM 249.69)? How many grams of CuSO4·5 H2O
are required ?
Dissolve 62.42 g of CuSO4·5H2O in enough water to total 250.0 mL
mL soln
mol CuSO4 · 5H2O
g CuSO4 · 5H2O
L soln
mol
L
mol
g
mL
L
250.0 mL soln x x
x = 62.42 g CuSO4 · 5H2O
1.000 L soln1000 mL soln
1.000 mol CuSO4 · 5H2O1.000 L soln
249.69 g CuSO4 · 5H2O1.000 mol CuSO4 · 5H2O
Dilution
DilutionOften, solutions are stored as concentrated stock solutions.
To make solutions of lower concentrations from these stock solutions, more solvent is added.
The amount of solute doesn’t change, just the volume of solution.
moles solute in solution 1 = moles solute in solution 2
( ) ·L1 = ( ) ·L2molL1
molL2
M1·V1 = M2·V2
6. What is the concentration of a solution prepared by diluting 45.0 mL of 8.25 M HNO3 to 135.0 mL?
M1V1 = M2V2
M1V1V2
= M2
( )(45.0 mL)(135.0 mL)
molL
8.25= 2.75 =2.75 Mmol
L
7. To what volume should you dilute 0.200 L of 15.0 M NaOH to make 3.00 M NaOH?
M1V1 = M2V2
M1V1M2
= V2
( )(0.200 L)( )
molL
15.0molL
3.00= 1.00 L
8. How would you prepare 200.0 mL of 0.25 M NaCl solution from a 2.0 M solution?
M1V1 = M2V2
Dilute 25 mL of 2.0 M stock solution up to 200.0 mL.
M2V2M1
= V1
( )(0.200 L)( )
molL
0.25molL
2.0 = 0.025 L
You have a bottle of 1 molar stock solution
But you WANT a 1 millimolar solution.
What will you do?
What will save your job? SERIAL DILUTION will!
“It’s a snap ! I’ll add 1.00 mL of
the stock solution to 999 mL of
water.”
“But we only need 100
microliters of the solution !!”
“Do you have any idea how much it
would cost to dispose of a liter of
that stuff?
It’s toxic !!”
Serial Dilution1.00 mL
+ 9.00 mL water
1.00 mL 1.00 mL 1.00 mL
+ 9.00 mL water
+ 9.00 mL water
+ 9.00 mL water
1.00 M 0.100 M 0.0100 M 0.00100 M 0.000100 M
stock solution
Solution Stoichiometry
Moles of A
Moles of B
Liters of a Solution of A
Liters of a Solution of B
Solution StoichiometryBecause molarity relates the moles of solute to the liters of solution, it can be
used to convert between amount of reactants and/or products in a chemical
reaction.
coefficients
mol L
mol L
mol
L
mol
mol
9. What volume of 0.150 M KCl is required to completely react with 0.150 L of 0.175 M Pb(NO3)2 in the reaction:
2 KCl(aq) + Pb(NO3)2(aq)➜ PbCl2(s) + 2 KNO3(aq)
L Pb(NO3)2 soln
mol KCl
mol Pb(NO3)2
L KCl soln
mol
L
0.150 L Pb(NO3)2 soln x x
x = 0.350 L KCl soln
0.175 mol Pb(NO3)21.00 L Pb(NO3)2 soln
2 mol KCl1mol Pb(NO3)2
1.00 L KCl soln0.150 mol KCl
10. 43.8 mL of 0.107 M HCl is needed to neutralize 37.6 mL of Ba(OH)2 solution. What is the molarity of the base?
2 HCl(aq) + Ba(OH)2(aq)➜ BaCl2(aq) + 2 H2O(aq)
L HCl soln
mol Ba(OH)2
molHCl
L Ba(OH)2 soln
mol Ba(OH)2
Molarity of Ba(OH)2 soln
0.0438 L HCl soln x
x = 0.00234 mol Ba(OH)2
0.107 mol HCl1.00 L HCl soln
1 mol Ba(OH)22 mol HCl
0.00234 mol Ba(OH)20.0376 L Ba(OH)2 soln
= 0.0623 M Ba(OH)2
The Big Picture of Stoichiometry
Moles of A
Moles of B
Mole to
Mole Ratio*
Liters of a Solution of A
Liters of a Solution of BMM
Grams of A
Grams of B
Molar Mass
Particles of A
Particles of B
Avogadro’s Number
*from balanced equation
