Download U6B _13-14

Section 6.4—Solubility &
Precipitation
How can we make sure everything that’s added to the sports drink will dissolve?
A Review of DoubleReplacement Reactions
Double Replacement Reactions
The cations from two compounds replace
each other.
NaCl + AgNO3  AgCl + NaNO3
Two ionic compounds switch ions
Double Replacement Reactions
General format of a double replacement reaction:
A
X
B
Z
A
Z
B
X
Products of a Double Replacement
1
CaCl2
Combine the cation of the first reactant with the anion of the
second reactant
+ AgNO3
Products of a Double Replacement
2
CaCl2
Combine the cation of the second reactant with the anion of the
first reactant
+ AgNO3
Products of a Double Replacement
3
Remember to write cations first …
& balance charges with subscripts when writing formulas
Only leave subscripts that are in the original compound
there if they are a part of a polyatomic ion!
CaCl2
CaCl2
+ AgNO3
+
AgNO3
Ca(NO3)2
+
AgCl
Precipitation Reactions
Precipitation Reactions
A precipitation reaction is when 2 soluble
substances are mixed together and they
form an insoluble substance. This is
called a precipitate.
Reactants
2 soluble
chemicals:
NaOH
Cu(NO )
NaOH and Cu(NO3)2
3 2
Precipitation Reactions(DR Rxns)
Na+1
OH-1
Cu+2
NO3 -1
Cu(OH)2(S)
Products: 1 soluble chemical: NaNO3
1 insoluble chemical (the precipitate): Cu(OH)2
Na+1
NO3 -1
Solubility Rules
Solubility Rules Table
Use the table on the reference sheet!
Insoluble = Precipitate
Let’s Practice #1
NaNO3
Example:
Decide whether
each is soluble
or not
Fe(C2H3O2)2
CaBr2
Ba(OH)2
Cu(OH)2
Let’s Practice #1
Example:
Decide whether
each is soluble
or not
NaNO3
Soluble
Fe(C2H3O2)2
Soluble
CaBr2
Soluble
Ba(OH)2
Soluble
Cu(OH)2
Not Soluble
Let’s Practice #2
Example:
Write the
products for
this reaction &
predict the
precipitate
Remember to indicate compounds that
dissolve with “aq” for “aqueous” and
compounds that don’t dissolve with
“s” for “solid”
AgNO3 (aq) + NaCl (aq) 
Let’s Practice #2
Example:
Write the
products for
this reaction
AgNO3 (aq) + NaCl (aq) 
AgCl (s) + NaNO3 (aq)
precipitate
Let’s Practice #3
Example:
Write the
products for
this reaction &
identify the
precipitate
Remember to indicate compounds that
dissolve with “aq” for “aqueous” and
compounds that don’t dissolve with
“s” for “solid”
BaCl2 (aq) + K2CO3 (aq) 
Let’s Practice #3
Example:
Write the
products for
this reaction
BaCl2 (aq) + K2CO3 (aq) 
KCl (aq) + BaCO3 (s)
precipitate
Net Ionic Reactions
Shows the details of aqueous reactions that involve ions in aqueous solution
Molecular Equation: the typical equation you are use to writing
keeping all molecules together
Complete Ionic Equation: shows all the particles in a solution as they really
exist, as IONS or MOLECULES.
 Anything aqueous needs to be split apart into the cation and anion
 Anything solid stays intact
 Coefficients need to be multiplied by subscripts to determine the exact
amount of each cation and anion.
Spectator ions: ions that do not participate in a reaction; they are identical on both
sides of the equation & are crossed out!
Net Ionic Equation: the final equation showing the major players. All
spectator ions have been removed.
NET IONIC REACTIONS for
Precipitation Reactions
Molecular equation:
KI(aq) + AgNO3(aq)  AgI(s) + KNO3(aq)
Complete Ionic equation:
K+1 + I-1 + Ag+1+ NO3-1  AgI + K+1 + NO3-1
Spectator ions: ions that do not participate in a reaction; they are
identical on both sides of the equation & are crossed out!
Net Ionic equation:
I-1 + Ag+1  AgI
NET IONIC REACTIONS for
Precipitation Reactions
Molecular equation:
2 NaOH(aq) + CuCl2(aq)  2 NaCl(aq) + Cu(OH)2(s)
Complete Ionic equation:
2 Na+1 + 2 OH-1 + Cu+2 + 2 Cl-1  2 Na+1 + 2 Cl-1 + Cu(OH)2
Net Ionic equation:
2 OH-1 + Cu+2  Cu(OH)2
Take Home Practice:
Predict products and balance Iron (III) chloride reacts with
sodium hydroxide
Molecular equation:
1 FeCl3(aq) + 3 NaOH(aq)  1 Fe(OH)3(s) +3 NaCl(aq)
Complete Ionic equation:
3 Na+1 + 3 OH-1 + Fe+3 + 3 Cl-1  3 Na+1 + 3 Cl-1 + Fe(OH)3
Net Ionic equation:
3 OH-1 + Fe+3  Fe(OH)3
Section 6.5—Stoichiometry
How can we determine in a lab the concentration of electrolytes?
What do those coefficients really
mean?
The coefficient of the balanced chemical
equation tells how many moles of each
substance is used in the reaction.
For every 2 moles
of H2…
2
2
and 2 moles of H2O are
produced
2 H2 + O2  2 H2O
No coefficient = 1
1 mole of O2 is
need to react…
Mole Ratio
Is a conversion factor that relates 2
substances in moles; must use a
balanced chemical equation to create it
2 H2 + O2  2 H2O
Examples of Mole Ratio’s
2mol H2
1 mol O2
1 mol O2
2 mol H2O
2 mol H2O
2 mol H2
What is stoichiometry?
Stoichiometry – Calculations using the mole
ratio from the balanced equation and
information about one compound in the
reaction to determine information about
another compound in the equation.
Example: What is the mole ratio of
chlorine to sodium?
2 Na + Cl2  2 NaCl
2mol Na
1 mol Cl2
1 mol Cl2
2 mol NaCl
2 mol Na
2 mol NaCl
Stoich (Mole-Mole) : 1 step problem
using the mole ratio
Example:
If 4.2 mole of H2 reacts
completely with O2, how
many moles of O2 are
needed?
2 H2 + O2  2 H2O
Stoichiometry with Moles
Example:
If 4.2 mole of H2 reacts
completely with O2, how
many moles of O2 are
needed?
2 H2 + O2  2 H2O
4.2 mole H2
1
mole O2
2
mole H2
From balanced equation:
2 mole H2  1 mole O2
= ________
2.1
mole O2
Stoich (Mole-Mole)
Example:
If 0.67 moles of potassium nitrate
reacts, how many moles of oxygen
are produced?
2KNO3  2KNO2 + O2
0.67 mole
KNO3
1
mole O2
2
Mole
KNO3
From balanced equation:
2 mole KNO3  1 mole O2
= ________
0.34 mole O2
But we can’t measure moles in lab!
We can’t go to the lab and count or
measure moles…so we need a way to
work in measurable units, such as
grams and liters!
Molecular mass gives the grams = 1 mole of a compound!
Stoich( Mole-Mass): 2 step problem
use mole ratio & then molar mass
conversion factors
Example:
How many grams of AgCl will be
precipitated if
0.45 mole AgNO3 is reacted as follows:
2 AgNO3 + CaCl2  2 AgCl + Ca(NO3)2
Stoichiometry with Moles & Mass
Example:
How many grams of AgCl will be
precipitated if
0.45 mole AgNO3 is reacted as follows:
2 AgNO3 + CaCl2  2 AgCl + Ca(NO3)2
From balanced equation:
2 mole AgNO3  2 mole AgCl
Molar Mass of AgCl:
1 mole AgCl = 143.32 g
0.45 mole AgNO3
2
mole AgCl
2
mole AgNO3
143.32 g AgCl
1
mole AgCl
= ________
64
g AgCl
Stoich( Mass- Mol): 2 step problem
use molar mass & then mole ratio
conversion factors
Example:
If 4.42 g of H2 reacts, how
many moles of NH3 are
produced?
N2 + 3H2  2NH3
4.42 g H2
1
2.02
Mole H2
g H2
From balanced equation:
3 mole H2  2 mole N2
2 mole NH3
3 mole H2
=1.46 mole NH3
Stoich( Mass-Mass): 3 step problem
use molar mass, then mole ratio &
then molar mass conversion factors
(Honors Only)
Example:
How many grams Ba(OH)2 are
precipitated from 14.5 g of NaOH in the
following reaction:
2 NaOH + BaCl2  Ba(OH)2 + 2 NaCl
Stoichiometry with Mass (Honors)
Example:
How many grams Ba(OH)2 are
precipitated from 14.5 g of NaOH in the
following reaction:
2 NaOH + BaCl2  Ba(OH)2 + 2 NaCl
Molar Mass of NaOH:
1 mole NaCl = 40.00 g
From balanced equation:
2 mole NaOH  1 mole Ba(OH)2
Molar Mass of Ba(OH)2:
1 mole Ba(OH)2 = 171.35 g
14.5 g NaOH
1 mole NaOH
40.00 g NaOH
1
mole Ba(OH)2 171.35 g Ba(OH)2
2
mole NaOH
1
mole Ba(OH)2
31.1
= ________
g Ba(OH)2
Stoichiometry with Mass (Honors)
Example:
How many grams of HCl are needed to
produce 65.0 g of magnesium chloride:
__Mg + ____HCl  ____MgCl2 + __H2
Balance the equation and fill
in the missing information:
65 g MgCl2 1 mole MgCl2
g MgCl2
mole HCl
mole MgCl2
g HCl
mole HCl
=
g HCl
What about the stoichiometry of
gases? Recall
Molar Volume of a Gas – at STP
1 mole of any gas = 22.4 liters
Stoichiometry with Gases
1 mol= 22.4 L @STP
Example:
If you need react 1.5 g of zinc completely,
what volume of hydrogen gas will be
produced at STP?
2 HCl (aq) + Zn (s)  ZnCl2 (aq) + H2 (g)
Stoichiometry with Gases
Example:
If you need react 1.5 g of zinc completely,
what volume of hydrogen gas will be
produced at STP?
2 HCl (aq) + Zn (s)  ZnCl2 (aq) + H2 (g)
Molar volume of a gas:
1 mole H2 = 22.4 L
From balanced equation:
1 mole Zn  1 mole H2
1.5 g Zn
1
Molar Mass of Zn:
1 mole Zn = 65.39 g
mole Zn
1
mole H2
65.39 g Zn
1
mole Zn
22.4
1
L H2
mole H2
0.51
= ________
L H2
Stoichiometry with Gases
Example: How many moles of water will
be produced from the complete
combustion of 7.3 L of oxygen gas?
Assume STP
C3H8 + 5O2 → 3CO2 + 4H2O
Molar volume of a gas:
1 mole O2 = 22.4 L
From balanced equation:
4 mole H2O  5 mole O2
7.3 L O2
Molar Mass of H:
1 mole H2O = 18 g
1 Mole O2
4
mole H2 O
22.4 L L O2
5
Mole O2
= _0.26 mole
H2O
Keeping all these equalities straight!
TO GO BETWEEN
USE THE EQUALITY
Grams & moles
Molar mass (g)= 1 mole
Particles & Moles
1 mol = 6.02 x 1023 particles
Moles & liters of a gas at
STP
1 mole = 22.4 L at STP
2 different chemicals in a
reaction
Coefficient ratio(MOLE
RATIO) from balanced
equation
You Try!
Given the UNBALANCED EQUATION:
__MgCO3  __MgO + __CO2, how many
liters of CO2 gas are produced from the
reaction of 15 grams of MgCO3? Assume
STP!
Section 6.5b
Percent Yield
Percent Yield
A “Yield” is a product
Actual Yield(A): the actual amount of
product you produce in the lab
Theoretical Yield(T): the amount of
product you should produce if nothing
went wrong; use the balanced
chemical equation to calculate this
amount.
Percent Yield: ratio of actual yield to
theoretical yield
Percent Yield
%yield =
A x 100
T
Lets Practice in steps:
1a. If 4.20 moles H2 reacts completely with oxygen, how
many grams of H2O are produced?
2 H2 + O2  2 H2O
This is a mol-mass problem.
Your answer is the theoretical yield of water?
If 4.20 moles H2 react completely with oxygen how
many grams of H2O are produced?
2 H2 + O2  2 H2O
From balanced equation:
2 mole H2O  1 mole O2
4.2 mol H2
2 moleH2O
2 mol H2
Molar Mass of O2:
1 mole H2O = 18.02g
18.02 grams H2O
1 mole H2O
75.7 g
= ______g H2O
This is the
theoretical
yield.
What is the percent yield if 60.0 grams of H2O
are produced?
A= 60.0 g
T= 75.7 g
%yield = A x 100
T
%yield =
60.0x 100
75.7
79.3% yield
You have precipitated 8.50 g of Ba(OH)2. If you
start with 4.57 grams of NaOH, what is the % yield.
2 NaOH + BaCl2  Ba(OH)2 + 2 NaCl
From balanced equation:
2 mole NaOH  1 mole Ba(OH)2
4.57 g NaOH
1 molNaOH
40.00g NaOH
Molar Mass of Ba(OH)2: 173.25 g
Molar Mass of NaOH: 40.00 g
1 molBa(OH)2 171.35g Ba(OH)2
2 mol NaOH
1mol Ba(OH)2
= 9.79 g Ba(OH)2
This is the
theoretical
yield.
If 9.78 grams are obtained in the experiment,
what is the percent yield?
A= 8.50 g
%yield =
T= 9.79 g
%yield =
A x 100
T
8.50 x 100
9.79
86.8% yield
Section 6.5c
Titrations
Titrations—Using Stoichiometry
Titration – A technique where the
addition of a known volume of a known
concentration solution to a known
volume of unknown concentration
solution to determine the concentration.
•Use a buret to titrate unknown
concentration of solutions.
Titrations—Using Stoichiometry
The titrant is the known concentration
in the buret and the analyte is the
unknown concentration in the flask.
Formula: nMaVa = nMbVb
na= number of H+ in the acid
nb= number of OH- in the base
Ma= molarity of acid
Mb= molarity of base
V= volume
End Point vs. Equivalence Point
Equivalence Point (or Stoichiometric
Point)
– When there are no reactants left over—they
have all been reacted and the solution
contains only products
-the point where the acid and the base are
equal in equal moles
moles acid = moles base
Importance of Indicators
Indicators – Paper or liquid that change color based
on pH level.
End Point: point at which the indicator in the solution
changes color
It signals the equivalence point and the stop of the
titration
Always select an indicator that has a pH
value close to that of the pH of the
equivalence point of the titration.
Titration Process
Titration Problem #1
How many liters of 0.10 M NaOH is
needed to react with 0.125 L of 0.25 M
HCl?
NaOH + HCl  H2O + NaCl
Titration Problem #2
What is the molarity of a Ca(OH)2 solution
if 30.0 ml of the solution is neutralized by
20.0 ml of a 0.50 M solution of HCl?
Ca(OH) 2 + 2HCl  2H2O + CaCl2
Titration Problem #3
What volume of 2.0M solution of NH4OH is
needed to neutralize 50.0 ml of a 0.50M
solution of H2SO4?
2 NH4OH + H2SO4  2H2O + (NH4) 2SO4
Titration Curves
Strong Base - Strong
Acid
Weak Base - Strong Acid
 Shows the changes of pH
during a titration
 Identifies the pH of the
equivalence point
Strong Base - Weak Acid
Weak Base - Weak Acid
Titration curve for
Titrating a strong acid with a strong base
pH is always = 7
The titration curve graph shows the pH of the
equivalence point. Take the vertical region and cut
the length in half and then look to what pH value
aligns to that point.
Titration curve for
Titrating a strong base with an strong acid
pH is always = 7
Titration curve for
Titrating a weak acid with an strong base
pH is >7