Transcript VI: Solution Concentration, Stoichiometry

VII: Aqueous Solution Concentration,
Stoichiometry
LECTURE SLIDES
•Molarity
•Solution Stoichiometry
•Titration Calculations
•Dilution Problems
Kotz & Treichel: Sections 5.8 -5.9
Concentrations of Compounds in Aqueous
Solutions
(Chapter 5, Section 5.8, p. 213)
Usually the reactions we run are done in aqueous
solution, and therefore we need to add to our study
of stoichiometry the concentration of compounds in
aqueous solution.
We will utilize a very useful quantity known as
“molarity,” the number of moles of solute per liter
of solution.
Concentration(molarity) = # moles solute
L solution
If we placed 1.00 mol NaCl (58.4 g) in a 1 L volumetric
flask, dissolved it in water, swirled to dissolve and
diluted the solution to the 1.00 liter mark you would
have a solution that contains 1.00 mol NaCl per liter
of solution. This may be represented several ways:
Concentration(molarity) = 1.00 mol NaCl / L soln
= 1.00 M NaCl = [1.00] NaCl
Chemists call this a “1.00 molar solution”
Calculating Molar Amounts
TYPICAL PROBLEMS:
•What is the molarity of a solution made by dissolving
25.0 g of BaCl2 in sufficient water to make up a
solution of 500.0 mL?
•How many g of BaCl2 would be contained in 20.0 mL of
this solution?
•How many mL of this solution would deliver 1.25 g of
BaCl2?
•How many mol of Cl- ions are contained in 10.00 ml of
this solution?
What is the molarity of a solution made by dissolving
25.0 g of BaCl2 in sufficient water to make up a
solution of 500.0 mL?
25.0 g BaCl2
500.0 mL soln
=
? mol/ L BaCl2 (= ? M BaCl2)
Molar mass, BaCl2:
1Ba = 1 X 137.33g = 137.33
2Cl = 2 X 35.45g = 70.90
208.23 g/mol
Mass of solute
25.0 g BaCl2
500.0 mL
Molar mass,
solute
1 mol BaCl2
208.23 g BaCl2
Volume of soltn
1000 mL
= .240 mol / L BaCl2
1L
= .240 M BaCl2
Conversion
to L
How many g of BaCl2 would be contained in 20.0 mL of
this solution? (.240 M BaCl2)
Question:
20.0 mL soln = ? g BaCl2
Relationships:
1000 mL = 1 L
1 L soln = .240 mol BaCl2
1 mol BaCl2 = 208.23 g BaCl2
20.0 mL soln
= ? g BaCl2
mL  L
 mol
g
Molarity
20.0 mL soln
1L
.240 mol BaCl2
1000 mL
1L soln
conversion
Molar Mass
208.23 g BaCl2
1 mol BaCl2
= 1.00 g BaCl2
•How many mL of this solution would deliver 1.25 g of
BaCl2?
1.25 g BaCl2 = ? mL soln
.240 mol BaCl2 = 1 L soln
208.23 g BaCl2 = 1 mol BaCl2
1.25 g BaCl2
1 mol BaCl2
208.23 g BaCl2
Molar Mass
1L soln
.240 mol BaCl2
Molarity
1000 mL
= 25.0 mL
1 L soln
•How many mol of Cl- ions are contained in 10.0 ml of
this solution?
Note: BaCl2(aq) ---> Ba2+(aq) + 2 Cl- (aq)
10.0 mL soln = ? Mol Cl.240 mol BaCl2 = 1000 mL soln
1 mol BaCl2 = 2 mol Cl-
10.0 mL soln
.240 mol BaCl2
2 mol Cl-
1000 mL soln
1 mol BaCl2
= .00480 mol Cl-
GROUP WORK 7.1: Molarity Problems
If 35.00 g CuSO4 is dissolved in sufficient water to
makeup 750. mL of an aqueous solution,
a) what is the molarity of the solution?
b) how many mL of the solution will deliver 10.0 g of
CuSO4?
c) How many moles of sulfate ion (SO42-) will be
delivered in 10.0 mL of the solution?
1 Cu = 1 X 63.55 = 63.55
1 S = 1 X 32.07 = 32.07
4 O = 4 X 16.00 = 64.00
159.62 g/mol
STOICHIOMETRY OF REACTIONS IN AQUEOUS
SOLUTION: Chapter 5, Section 5.9
Let’s use our favorite reaction to add another
dimension to calculating from balanced equations:
How many ml of 3.00 M HCl solution would be required
to react with 13.67 g of Fe2O3 according to the following
balanced equation:
Fe2O3(s)
159.70 g/mol
13.67 g
+ 6HCl(aq) ---> 3H2O +
36.46 g/mol
3.00 M
? mL
18.02 g/mol
2FeCl3(aq)
162.20 g/mol
Fe2O3(s)
+ 6HCl(aq) ---> 3H2O +
159.70 g/mol
3.00 M
13.67 g
? mL
2FeCl3(aq)
13.67 g Fe2O3 = ? mL soln
1000 mL soln = 3.00 mol HCl
6 mol HCl = 1 mol Fe2O3
159.70 g Fe2O3 = 1 mol Fe2O3
13.67 g Fe2O3
= ? mL soltn
Pathway: g Fe2O3 ---> mol Fe2O3---> mol HCl --- > mL soln
Balanced
Equation
13.67 g Fe2O3
1 mol Fe2O3
6 mol HCl
159.70 g Fe2O3
1 mol Fe2O3 3.00 mol HCl
1000 mL soln
Molarity
Molar
Mass
= 171 mL soln
Group Work 7.2: Solution Stoichiometry
How many g of Fe2O3 would react with 25.0 mL of 3.00 M HCl?
Fe2O3(s)
159.70 g/mol
?g
+ 6HCl(aq) ---> 3H2O +
2FeCl3(aq)
3.00 M
25.0 mL
25.0 mL soltn
= ? g Fe2O3
mL  mol HCl  mol Fe2O3  g Fe2O3
Limiting Reagent, Solutions
Suppose you mixed 20.00 mL of .250 M Pb(NO3)2 solution
with 30.00 mL of .150 KI solution. How many g of PbI2
precipitate might you theoretically obtain?
Pb(NO3)2 (aq) + 2 KI (aq) ---> PbI2 (s) + 2 KNO3 (aq)
.250 M
.150 M
461.0 g/mol
20.00 mL
30.00 mL
?g
1Pb = 1 X 207.2 = 207.2
2 I = 2 X 126.9 = 253.8
461.0 g/mol
“A”
“B”
“C”
Pb(NO3)2 (aq) + 2 KI (aq) ---> PbI2 (s) + 2 KNO3 (aq)
.250 M
20.00 mL
.150 M
30.00 mL
461.0 g/mol
?g
Pathway: mL  mol “A”  mol “C”  g “C”
mL  mol “B”  mol “C”  g “C”
Low
number
wins!
Pb(NO3)2 (aq) + 2 KI (aq) ---> PbI2 (s) + 2 KNO3 (aq)
.250 M
20.00 mL
.150 M
30.00 mL
461.0 g/mol
?g
20.00 ml soltn A .250 mol A 1 mol C 461.0 g C = 2.3050 g C
1000 mL 1 mol A 1 mol C
= 2.30 g C
30.00 mL soltn B .150 mol B 1 mol C 461.0 g C = 1.037 g C
1000 mL
2 mol B 1 mol C
= 1.04 g C
Excess reagent:
How many mL of solution A, .250 M Pb(NO3)2, would be
required to react with 30.00 mL of solution B, .150 M KI?
Pb(NO3)2 (aq) + 2 KI (aq) ---> PbI2 (s) + 2 KNO3 (aq)
.250 M
? mL
.150 M
30.00 mL
Pathway:
mL “B” soltn  mol “B”  mol “A”  mL “A” soltn
30.00 mL B soltn
= ? mL A soltn
Pb(NO3)2 (aq) + 2 KI (aq) ---> PbI2 (s) + 2 KNO3 (aq)
.250 M
? mL
.150 M
30.00 mL
equation
Molarity
Molarity
30.00 mL soltn .150 mol B
1 mol A 1000 mL soltn
1000 mL soltn 2 mol B .250 mol A
=
9.00 mL soltn A
Group Work 7.3: Solution Stoichiometry #2
How many mL of .150 M KI solution would be required
to react with 20.00 mL of .250 M Pb(NO3)2 ?
Pb(NO3)2 (aq) + 2 KI (aq) ---> PbI2 (s) + 2 KNO3 (aq)
.250 M
20.00 mL
.150 M
? mL
Summary
1. Molarity, M: useful description of a solution; gives
number of moles of solute in 1 L of solution. Useful
to convert mL of solution to moles of solute in equation
situations.
2. Molarity of solution can also yield moles per liter of
the ions produced when the solute ionizes in water,
using the formula of the solute as conversion factor
3. Adds another way to calculate mass or moles
from volume (compare to density, #grams of substance
or solution in 1 mL or 1 cm3 of solution or substance).
TITRATION CALCULATIONS
Titration: Procedure in which measured increments
of one reactant are added to a known amount
of a second reactant until some indicator signals
that the reaction is complete.
This point in the reaction is called “the equivalence
point.”
Indicators include many acid/base dyes, potentiometers,
color change in one reagent...
Titrations are run with buret and Erlenmeyer flasks as
seen on the CD ROM or demo just observed...
Two types of problems are typically encountered:
•Standardizing an acid or a base solution
•(determining the exact molar concentration of
an acid or base solution)
• Determining amount of acidic or basic material
in a sample ( or other substances detectable by
some indicator)
Ex 5.13 type: Standardizing an acid solution:
Suppose a pure, dry sample of Na2CO3 weighing 0.379 g
is dissolved in water and titrated to the equivalence point
with 35.65 mL of HCl solution. What is the molarity of
the HCl solution?
Na2CO3(aq) + 2HCl(aq) -----> 2 NaCl(aq) + H2O + CO2(g)
105.99 g/mol
0.379 g
35.65 mL soln
M= ? = #mol HCl / L soltn
2 Na = 2 X 22.99 = 45.98
1 C = 1 X 12.01 = 12.01
3 O = 3 X 16.00 = 48.00
105.99 g/mol
Calculations for
Standardization
1) Calculate moles of solute from balanced equation,
using information about known reagent
2) Calculate M, using volume of solution required
for titration of known to equivalence point
Na2CO3(aq) + 2HCl(aq) -----> 2 NaCl(aq) + H2O + CO2(g)
105.99 g/mol
0.379 g
35.65 mL soln
M=? = Mol HCl / L soltn
Step One: Calculate Moles of HCl from equation
.379 g Na2CO3 = ? moles HCl
0.379 g Na2CO3
1 mol Na2CO3
2 mol HCl
105.99 g Na2CO3 1 mol Na2CO3
= .00715 mol HCl
Na2CO3(aq) + 2HCl(aq) -----> 2 NaCl(aq) + H2O + CO2(g)
105.99 g/mol
0.379 g
35.65 mL soln
.00715 mol
M=?
Calculated
in Step One
Step Two: Calculate molarity:
M, HCl soln = ? =
# mol HCl
L soln
.00715 mol HCl 1000 mL
35.65 mL soltn 1 L
=
= .201 mol HCl = .201 M HCl
L soltn
Group Work 7.4:
Suppose a pure, dry sample of Na2CO3 weighing 0.437 g
is dissolved in water and titrated to the equivalence point
with 39.85 mL of HCl solution. What is the molarity of
the HCl solution?
Na2CO3(aq) + 2HCl(aq) -----> 2 NaCl(aq) + H2O + CO2(g)
105.99 g/mol
0.437 g
39.85 mL soln
M= ? = #mol HCl / L soltn
#1: Find moles of HCl from g Na2CO3
#2: Find Molarity (moles HCl / volume soltn)
Standardization of a Base:
Let’s use the 0.201 M HCl which we standardized in
the first problem to determine the exact molar
concentration of a sodium hydroxide solution
(“standardize the base”!):
Problem:
If titrating 25.00 mL of NaOH to a phenolphthalein
endpoint required 31.25 mL of .201 M HCl, what is the
molar concentration of the base?
If titrating 25.00 mL of NaOH to a phenolphthalein
endpoint required 31.25 mL of .201 M HCl, what is the
molar concentration of the base?
NaOH(aq) + HCl(aq) ---> H2O(l) + NaCl(aq)
25.00 mL
M=?
31.25 mL
.201 M
Steps:
1) calculate moles NaOH reacted
2) calculate M, moles NaOH/ volume soln
NaOH(aq) +
25.00 mL
M=?
Step 1)
HCl(aq) ---> H2O(l) + NaCl(aq)
31.25 mL
.201 M
31.25 mL HCl soln = ? mol NaOH
31.25 mL HCl soln .201 mol HCl
1000 mL soln
Step 2)
1 mol NaOH
= .00628 mole NaOH
1 mol HCl
Calculate molarity:
.00628 mole NaOH
1000 mL
25.00 mL soln
1L
= .251 mol NaOH = .251 M NaOH
L
Group Work 7.5
A 30.0 mL sample of vinegar requires 39.35 mL of .843 M
NaOH solution for titration to the equivalence point.
What is the molar concentration of the acetic acid in
vinegar?
CH3CO2H(aq) + NaOH(aq) ----> H2O + NaCH3CO2(aq)
30.0 mL
39.35mL
.843 M
M=?mol CH3CO2H /L
Step 1) solve for moles, CH3CO2H
Step 2) solve for M CH3CO2H
Let’s do another base solution, standardizing it with
Potassium acid phthalate, KHC8H4O4, a popular solid for
this purpose. (Structure, next slide!) It reacts with strong
bases according to the following net ionic equation:
KHC8H4O4 (aq) + NaOH (aq) ------> H2O(l) + KNaC8H4O4 (aq)
H
H
O
C
C
C
C
C
C
H
O
H
O: K
C
C
H
O
H
H2O
NaOH
+
+
H
H
O
C
C
C
C
C
C
O:
O:
C
C
H
O
Na
K
H
H
H
O
C
C
C
C
C
C
H which acts
like an acid
O
O:
C
C
H
O
H
K
"Hydrogen Phthalate"
Anion HC8H4O4:-
Salt cation
Group Work 7.6
If a .896 g sample of potassium acid phthalate is
dissolved in water and titrated to the equivalence
point with 16.95 mL of NaOH solution, what is the
molarity of the NaOH soln?
1K = 1 X 39.10
8C = 8 X 12.01
5H = 5 X 1.008
4O = 4 X 16.00
= 39.10
= 96.08
= 5.04
= 64.00
204.22 g/mol
KHC8H4O4 (aq) + NaOH (aq) ---> H2O (l) + K NaC8H4O4 (aq)
204.22 g/mol
.896 g
16.95 mL
M=? mol NaOH /L
“Redox” titration
You wish to determine the weight percent of copper in
a copper containing alloy. After dissolving a sample of
an alloy in acid, an excess of KI is added, and the Cu2+
and I- ions undergo the reaction:
2Cu2+ (aq) + 5 I- (aq) -----> 2 CuI(s) + I3 - (aq)
The I3 - which is produced in this reaction is titrated
with sodium thiosulfate according to the equation:
I3 - (aq) + 2 S2O3 2- (aq) -----> S4O6 2- (aq) + 3 I- (aq)
If 26.32 mL of 0.101 M Na2S2O3 is required for titration
to the equivalence point, what is the wt % of Cu in
.251 g alloy?
2Cu2+ (aq)
+ 5 I- (aq) -----> 2 CuI(s)
63.55 g/mol
0.251 g alloy
g Cu =?
% Cu, alloy=?
I3 - (aq) +
? mol
+ I3 - (aq)
produced
by reaction
2 S2O3 2- (aq) -----> S4O6 2- (aq) + 3 I- (aq)
26.32 mL
.101 M Na2S2O3
I3 - (aq) +
? mol
2 S2O3 2- (aq) -----> S4O6 2- (aq) + 3 I- (aq)
26.32 mL
.101 M Na2S2O3
= .00133 mol I3 -
1000 mL 1 L soln
26.32 mL soln 1 L
1 mol Na2S2O3
.101 mol Na2S2O3 1 mol S2O3 2-
2 mol S2O3 21 mol I3 -
2Cu2+ (aq)
+ 5 I- (aq) -----> 2 CuI(s)
63.55 g/mol
g=?
% Cu, alloy=?
+ I3 - (aq)
.00133 mol
0.251 g alloy
.00133 mol I3 - 2 mol Cu2+
1 mol I3 -
1 mol Cu
63.55 g Cu
1 mol Cu2+
1 mol Cu
% Cu, alloy=? = 1.69 g Cu
X 100 = 67.3%
0.251 g alloy
= .169 g Cu
DILUTION PROBLEMS
Suppose you would like to make up a more “dilute”
solution (less moles/L) from a more “concentrated”
one (more moles /L). This is how you might proceed:
a) figure out how many total moles of solute you
want in the desired volume of the dilute solution
b) figure out what volume of the more concentrated
solution will deliver this number of moles
c) measure out the concentrated solution and add
water to make up the desired volume of the dilute
solution.
(DEMO!)
Describe how you might makeup 750. mL of 1.00 M
H2SO4 from a concentrated sulfuric acid solution
which is 12.0 M H2SO4 .
a) Calculate desired number of moles in dilute soln:
750 mL soln
1.00 mol H2SO4
=
.750 mol H2SO4
1000 mL soln
b) Calculate volume of concentrated solution which will
contain this number of moles:
.750 mol H2SO4
1000 mL soln
12.0 mol H2SO4
= 62.5 mL soln
c) Measure out 62.5 mL of the concentrated soln and
carefully add it to sufficient water to make up 750. mL
of solution.
NEVER ADD WATER TO CONCENTRATED
ACIDS;
ALWAYS ADD CON ACIDS TO WATER,
SLOWLY...
Formula for Dilution Problems:
Since we can calculate moles as follows:
mL X # moles
1000 mL
= moles of solute
and since:
moles, con soln = moles, dil soln
We can say:
mL, con X M, con = mL, dil X M, dil
and do the last problem “the easy way”:
Describe how you might makeup 750. mL of 1.00 M
H2SO4 from a concentrated sulfuric acid solution
which is 12.0 M H2SO4 .
mL, con X M, con = mL, dil X M, dil
? mL X 12.0 M H2SO4 = 750. mL X 1.00 M H2SO4
mL con = 750 mL dil X 1.00 M H2SO4
12.0 M H2SO4
mL con = 62.5 mL
Punch line the same: measure out 62.5 mL con acid
and dilute to 750. mL.
Group Work 7.7: Dilution Problem
How many mL of 2.35 M AgNO3 solution are required to
makeup 2.00 L of .100 M AgNO3 solution?
Describe how you would makeup this new solution.
mL con X M con = mL dil
?
2.35 M
2.00L
X
M dil
.100 M