Announcements
•E X A M # 1 A N S W E R K E Y P O S T E D
•O W L :
•CHAPTER
•EXAM
4 PART 1 HOMEWORK
GRADES POSTED (LEFT SIDE “COURSE
GRADES” LINK)
Solution Concentration
 Solution
 Solute
 Solvent
 Concentration
 Qualitative
Dilute
 Concentrated


Quantitative

Molarity=# moles of solute per 1L of solution
What do goldfish have to do with concentration?
10
goldf
2
baby
food
cups
x

20
gol
baby
food
cup
Solution Concentration Examples
 How many moles of NH3 are in 575mL of 0.12mol/L
solution?
What you “have”
Conversion factor
 A 0.56M solution of NH3 contains 22 mol NH3. How
many Liters of solution are there?
What volume of a 0.23M solution of K2SO4 contains 0.50 moles?
0.115L
2. 0.115mL
3. 2.174L
4. 2.174mL
1.
How would you make 250mL of 1.00M NH3 from 15M NH3?
1.
2.
3.
4.
Take 16.7mL of 15M NH3 and dilute it
to 250mL with water
Take 60mL of 15M NH3 and dilute it
to 250mL with water
Take 0.0167mL of 15M NH3 and
dilute it to 250mL with water
Take 3.75mL of 15M NH3 and dilute
it to 250mL with water
Stoichiometry in Solution
Grams or mL reactant A
Grams or mL reactant B
Molar mass
or
Solution
concentration
Molar mass
or
Solution
concentration
Moles reactant A
Moles reactant B
Balanced
equation
and
Stoichiometry
Titration Example- Part 1
GOAL: To determine the molar mass of an unknown acid (H2A)
H2A + 2NaOH  Na2A + 2H2O
1. Standardizing an NaOH solution using a “primary standard”
A solution of sodium hydroxide is “standardized” by titration using solid oxalic
acid, H2C2O4, as the acid. It is found that 24.5 mL of the NaOH solution is
needed to react completely with 2.05 g of oxalic acid.
What is the concentration of the NaOH solution?
H2C2O4(aq) + 2 NaOH  2 H2O(l) + Na2C2O4(aq)
Titration Example- Part 2
2. Determining the molar mass of an unknown diprotic acid
The NaOH solution above was used to titrate an acid of unknown formula,
but known to be diprotic: H2A, where we don’t know what “A” is. A 1.046
gram sample of the unknown acid is titrated using 31.4 mL of the NaOH
solution. What is the molar mass of H2A?
H2A(aq) + 2 NaOH  2 H2O(l) + Na2A(aq)