1. Solutions and Molarity

2. Solutions The solute is the substance dissolved in the solution
The solvent is the substance in which the solute is dissolved
Aqueous Solution is a solution in which the solvent is water

3. Heterogeneous Mixture:
A mixture in which the composition is not uniform throughout
Sand and water
A salad
Oil and water
Homogeneous Mixture:
A mixture in which the composition is uniform throughout
May be solids, liquids and/or gases
Also called a solution
Salt and water
Ice tea mix and water

4. Solutions What is an ionic substance?
A bond between a metal and a non-metal
The metal gives up one or more electrons to the non-metal and becomes positively charged
The non-metal accepts one or more electrons and becomes negatively charged.
The two ions are held together by electrostatic attraction
[Na+][Cl-]

5. Solutions Ionic substances readily dissociate (come apart) in water.
The positive and negative ions become attracted to individual water molecules
Salt dissolving
Ionic equation for salt dissolving in water: NaCl -> Na+ + Cl-

6. Solutions
An electrolyte is a substance, when dissolved in water, conducts electricity
A strong electrolyte is a solute that produces many ions in water (sodium chloride)
A weak electrolyte is a solute that produces only a few ions in water
A non-electrolyte does not dissociate into ions in water and does not conduct electricity (sucrose)

7. Conversions 1000 ml = 1 L (liter)
To convert from ml to L, divide by 1000
To convert from L to ml, multiply by 1000
350 ml = ? L
350 ml = .350 L
1.7 L = ? ml
1.7 L = 1700 ml

8. Molarity Molarity is an expression of solution concentration
Molarity = moles of solute/liter of solution
Steps for calculating molarity
Convert volume from ml to L (if necessary)
Determine the molarity by dividing the number of moles of solute by the solution volume in liters

9. Volumetric flasks

10. Molarity 100.5 ml/1000 = .1005 L Molar mass of glucose = 180.16 g/mol
5.10 g/ g/mol = .03 mol glucose
.03 mol/.1005 L = .29 mols per liter
.29 M

11. Molarity Practice Problem #1: .15 M Practice Problem #2: .13 M

12. Preparation of Solutions
Convert ml to L (if necessary)
Determine molar mass of solute
Set up a proportion
Solve to get number of moles of solute needed
Multiply that number by the molar mass of the solute.

13. Preparation of Solutions
Example 1:
0.1 M means .1 mol/1 Liter
Convert to L -> ml/1000 = 0.10 L
Molar mass NaOH = 40 g/mol
Proportion -> .1mol/1 L * x mol/.1 L
X = .01 mol of NaOH
.01 * 40.0 g/mol = 0.4 g

14. Preparation of Solutions Alternate
Convert ml to L (if necessary)
Determine molar mass of solute
Multiply all the numbers:
Liters x mols/liter x molar mass
Example 1:
0.10 L x .1 mol/1 Liter x 40.0 g/mol = 0.4 g

15. Preparation of Solutions
Practice Problems
11.1 g CaCl2
80.0 g NaOH
30.0 g NaOH

16. Dilutions
Solutions you work with in the lab (dropper bottles) are made by diluting premixed concentrated stock solutions.
By adding additional solvent to concentrated solutions decreases the ratio of solute to solvent particles.
Use the following relationship to calculate the volume of solvent you may need
M1V1 = M2V2
M1V1 represent the molarity and volume of the stock solution
M2V2 represent the molarity and volume of the dilute solution
Remember to convert ml to L !

17. Dilutions Example problem 2.00 M * V1 = 0.300 M * 0.50 L
Solve for V1 = 0.150/2.00
V1 = .075 L or 75.0 ml

18. Dilutions Practice Problems V1 = 0.125 L or 125 ml
M2 = .7 M