Chem 105
Wed 14 Oct 2009
Clicker question on molarity
Solution concentration
pH definition
Titration calculations
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1
How many moles of perchlorate ions are in 0.40 L of
2.0 M calcium perchlorate (aq)?
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3
2
.2
5.
12
1.
6
4.
19
.8
3.
30
.6
2.
0.20
0.40
0.60
0.80
1.60
.4
1.
2
How many moles of perchlorate ions are in 0.40 L of
2.0 M calcium perchlorate (aq)?
Ca2+
1.
2.
3.
4.
5.
0.20
0.40
0.60
0.80
1.60
ClO4-
Ca(ClO4)2
 2.0 mol Ca(ClO 4 )2  2.0 mol ClO 4 − 
 = 1.6 mol ClO 4 −

0.40 L aq solution × 

 L aq solution  1 mol Ca(ClO 4 )2 
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1. Make a cartoon:
?
0.177 M Fe(NO3)3
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dried
m=4.29 g
5
How many moles is 4.29 g Fe(NO3)3 ?
 1 mol Fe(NO 3 )3
4.29 g Fe(NO 3 ) 3 × 
 241.86 g Fe(NO3 ) 3

 = 0.017738 mol Fe(NO 3 )3

What volume of aq Fe(NO3)3 contains 0.017738 mol Fe(NO3)3?
 1 L aq Fe(NO3 )3 
 = 0.10021 L aq Fe(NO 3 )3
0.017738 mol Fe(NO3 )3 × 
 0.177 mol Fe(NO3 )3 
What is that volume in mL?
 1000 mL aq Fe(NO 3 ) 3 
 =
0.10021L aq Fe(NO 3 ) 3 × 
L aq Fe(NO 3 ) 3


= 100.21 mL aq Fe(NO 3 ) 3 = 1.00 × 10 2 mL aq Fe(NO 3 ) 3
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  
what you start w ×    = ? final answer
  

 
  = ? mL aq Fe(NO3 )3
4.29 g Fe(NO3 )3 × 
g
Fe(NO
)
3 3  

 1 mol Fe(NO3 ) 3
4.29 g Fe(NO 3 ) 3 × 
 241.86 g Fe(NO3 ) 3
 1 L aq Fe(NO3 ) 3

 0.177 mol Fe(NO3 ) 3
 1000 mL aq Fe(NO3 ) 3

L aq Fe(NO 3 ) 3


 = ? mL aq Fe(NO3 ) 3

= 100.21 mL aq Fe(NO3 ) 3 = 1.00 × 10 2 mL aq Fe(NO 3 ) 3
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pH definition - board
[
pH = −log 10 H3O +
]
Take antilog of both sides :
10
−pH
[
= 10
log H3O +
[
10 −pH = H3O+
]
]
Hint: find the “log” and “10x” buttons on your calculator and practice using them.
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9
Ba(OH)2 + 2 HClO4 2 H2O + Ba(ClO4)2
Neutralization: moles OH- = mole H+
Use 29.0 mL of 0.177 M aq HClO4
Start w 30.8 mL aq Ba(OH)2
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Ba(OH)2 + 2 HClO4 2 H2O + Ba(ClO4)2
Neutralization: moles OH- = moles H+
M Ba(OH) 2 = ? =
Use 29.0 mL of
0.177 M aq
HClO4
M Ba(OH)2 =
mol Ba(OH) 2
L Ba(OH) 2
mol Ba(OH)2
0.0308 L Ba(OH)2
 0.177 mol HClO4
0.0290L HClO4 × 
L HClO4

Start w 30.8 mL
aq Ba(OH)2
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 1 mol Ba(OH)2 

 = ? mol Ba(OH)2
2
mol
HClO
4 

= 2.5665 × 10 −3 mol Ba(OH)2
11
mol Ba(OH) 2 2.5665 × 10 −3 mol Ba(OH) 2
=
M Ba(OH) 2 =
L Ba(OH) 2
0.0308 L Ba(OH) 2
= 0.08333
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mol Ba(OH) 2
= 0.08333 M = 0.0833 M
L
12
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JK’s method of solving these chemical analysis problems is to first calculate the
mass of carbon, hydrogen, and other elements if any. Then convert to moles of
C, H, and other. Then determine the mole ratio.
CxHy
+ O2
4.875g
excess
CO2
+
H2O
15.30 g
6.264 g
Notice that this approach involves first multiplying by the atomic weight of the element, then dividing by it, to obtain the #of moles of the element in
the product. This serves to illustrate that there are usually several ways to solve these problems. As long as the final answer is correct, and the
method is clearly recorded including units and significant figures, then the method would be marked correct on a chemistry exam paper or lab
notebook entry.
 1 mol CO2  1 mol C  12.01 g C 


 = 4.175 g C
15.30 g CO2 × 
44.01
g
CO
1
mol
CO
1
mol
C

2 
2 

 1 mol H2O  2 mol H  1.008 g H 


 = 0.6985 g H
6.264 g H2O × 
18.01g
H
O
1
mol
H
O

2 
2  mol H 
0.3476 mol C
=1
0.3476 mol C
0.6930 mol H 2 mol H
=
0.3476 mol C 1 mol C
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 1 mol C 
 = 0.3476 mol C
4.175 g C × 
12.01g
C


 1 mol H 
 = 0.6930 mol H
0.6985 g H × 
1.008
g
H


“CH2”
16
A molecule with the formula CH2 would have a molar mass
of 14.02 g/mol. But the observed mass for this particular
molecule was 70.13 g/mol, therefore the formula must be
C5H10 with a molar mass of 5 x 14.02 (=70.1 g/mol)
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