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