thursday, march 6, 2008 discussion of molarity lab results introduce section 15.2b -- dilutions...
TRANSCRIPT
Thursday, March 6, 2008
Discussion of Molarity Lab Results
Introduce Section 15.2b -- Dilutions
Homework: Pg. 555, #31a-d, 32, 33, 34
Concentrated Solution
To save time and space, usually buy solutions that are concentrated- stock solutions
Add solvent to get the molarity you wantDilution- process of adding more solvent
to a solution
Dilutions
Only water is added to the solutionsFigure out how much water needs to be
addedMoles of solute after dilution = moles of
solute before dilution
What happens?
( )
moles of soluteM
volume L
Moles of solute stays the sameVolume of water increasesSo Molarity decreases
Example
Prepare 500. mL of 1.00M acetic acetic acid, HC2H3O2 from a 17.5 M stock solution of acetic acid. What volume of stock solution is required?
1st- find the number of moles of acetic acid needed in the final solution
Vdilute x Mdilute = moles solute
Example (continued)
Only source of acetic acid is the stock solution
So moles of solute in dilute solution must equal moles in concentrated solution
This is always true!
Example (continued)
1500. 0.500
1000
L solutionmL solution L solution
mL solution
2 3 22 3 2
1.000.500 0.500
mol HC H OL solution mol HC H O
L solution
Now we need to find the volume of 17.5 M acetic acid that contains 0.500 mol of HC2H3O2
This is the unknown volume V
Volume x molarity = moles
Solving for V gives
V = 0.0286 L or 28.6 mL of solution
To make 500 mL of 1.00M acetic acid solution, we take 28.6 mL of 17.5 M acetic acid and dilute it to a total volume of 500 mL
2 3 22 3 2
17.5( ) 0.500
mol HC H OV in Liters mol HC H O
L solution
Because moles of solute remain the same before and after dilution,
M1 x V1 = moles of solute = M2 x V2
1 is initial conditions, 2 is final conditions
Preparing a dilution
Example
What volume of 18 M H2SO4 would be needed to prepare 1.5 L of 0.10 M H2SO4?
V x 18 M = 1.5 L x 0.10 MV= 8.3 mL
Common Stock Solutions
Stock Solutions
Sulfuric acid 18 M
Nitric acid 16 M
HCl 12 M
Practice Problem Exercise 15.8
What volume of 12M HCl must be taken to prepare 0.75L of 0.25M HCl?
16 mL