1. Molarity calculations

2. ( ) And the final answer is? What is the final answer?
Calculate the molarity concentration for 80 g of
NaOH in 4.00 L of solution.
And the final answer is?
What is the final answer?
How many liters of solution?
Is this the only conversion factor we need?
How many moles of NaOH?
Now consider the definition of molarity?
What number should we write first?
Next comes a conversion factor
What number goes in the numerator?
What number goes in the denominator?
Calculate the number of moles of NaOH
Let’s do it stepwise first.
( )
1 mol NaOH
80. g NaOH
2.0 moles NaOH =
40. g NaOH
molarity = moles solute
L solution
2.0 mols NaOH =
.50 M NaOH
4.00 L solution

3. ( ) Is this all we need? Put this on one side of the equation
What volume of 1.0 M HF contains 20. g of HF?
Is this all we need?
Put this on one side of the equation
What is the unknown variable?
What is the final answer?
Substitute in the values
What goes in the numerator?
And the final answer is?
What is the first number we write?
How many moles are in 20 g of HF?
What goes in the numerator?
Next comes a conversion factor?
What goes in the denominator?
Next consider the definition of molarity
Let’s do this stepwise again.
( )
1 mole HF
20. g HF =
1.0 mole HF
20. g HF
molarity = moles solute
liters of solvent
L of solvent
moles solute
L of solvent =
1.0 mole HF =
1.0 L soln
molarity
1.0 M HF

4. The Concept of Titration
NaOH
NaOH
What happens with this volume of
NaOH is added?
What volume of NaOH solution
is needed to react with all of
the HF’s?
How many NaOH’s are needed
to react with all of the HF’s?
NaOH
NaOH
NaOH
5 NaOH’s are needed.
This volume is needed.
NaOH
NaOH
NaOH
Na+ F-
H2O
Na+ F-
H2O
Na+ F-
H2O HF
Na+ F-
H2O
Na+ F-
H2O

5. AgNO3 + MgCl2  AgCl(s)+ Mg(NO3)2 2
Given that 100 mL of 2.0 M AgNO3(aq) reacts
completely with 1.00 M MgCl2(aq), what volume
of MgCl2(aq) is needed for complete precipitation? 2
AgNO MgCl2  AgCl(s)+ Mg(NO3)2 2
Let’s look at this problem graphically.
AgNO3
MgCl2
100 mL
Mg2+
100 mL
Ag+

6. AgNO3 + MgCl2  AgCl(s)+ Mg(NO3)2 2
Given that 100 mL of 2.0 M AgNO3(aq) reacts
completely with 1.00 M MgCl2(aq), what volume
of MgCl2(aq) is needed for complete precipitation? 2
AgNO MgCl2  AgCl(s)+ Mg(NO3)2 2
What is the denominator of the second conversion factor?
What is the numerator of the second conversion factor?
What is the numerator of the third conversion factor?
What is the denominator of the first conversion factor?
What is the denominator of the third conversion factor?
What is the numerator of the first conversion factor?
What is the second step?
What is the third step?
And the final answer…?
What is the first step?
Write the units of the desired quantity at the right.
It must be liters MgCl2
Again, it must be moles MgCl2
Again, it must be moles of AgNO3.
It must be moles MgCl2
Write the given quantity at the left.
It must be L AgNO3.
It must be moles of AgNO3.
Comes from doing the calculation.
Balance the chemical equation.
1 mol MgCl2
1 L MgCl2
2.0 mol AgNO3
.100 L AgNO3 =
1 L AgNO3
2 mol AgNO3
1.00 mol MgCl2
.100
L MgCl2

7. AlCl3 + KOH  Al(OH)3(s) + KCl 3 3
Given that 50. mL of .20 M AlCl3(aq) reacts
completely with .40 M KOH, what volume
of KOH is required for complete precipitation?
AlCl KOH  Al(OH)3(s) KCl 3 3
What is in the numerator of the first conversion factor?
What is in the denominator of the first conversion factor?
What is in the numerator of the second conversion factor?
What is in the denominator of the third conversion factor?
What is in the numerator of the third conversion factor?
What is in the denominator of the second conversion factor?
And the final answer comes from…?
What is the second step?
What is the third step?
What is the first step?
Balance the equation.
It must be mol KOH.
Again, it must be mol KOH.
It must be mol AlCl3.
Write the given quantity on the left.
It must be L AlCl3(aq).
It must be L KOH.
Doing the calculation.
Write the units of the desired quantity on the right.
.20 mol AlCl3
3 mol KOH
1.0 L KOH
.050 L AlCl3
1 L AlCl3
1 mol AlCl3
.40 mol KOH =
.075
L KOH