2. Concentration/Molarity/Molar Concentration A measure of the amount of solute that is dissolved in a measured amount of solvent. Dilute: a solution that contains a small amount of solute. Concentrated: a solution that contains a large amount of solute.

3. Percent solutions Percent means parts per one hundred. percent by volume = volume of solute volume of solution X 100 percent by mass = mass of solute volume of solution X 100 grams of solute dissolved in 100 mL solution

4. Examples 10.0 mL of acetic acid is diluted to a total volume of 200 mL. What is the percent by volume of acetic acid? = = 5.00 % (v/v) % (v/v ) = volume of solute volume of solution X 100 0.010 L 0.200 L X 100

5. Parts per million Very dilute concentrations can be recorded as parts per million. (ppm) One ppm can be thought of as one drop in a bathtub of water. 1 ppm = 1 mg/L

6. Examples What mass of chlorine is present in 15.0 L of solution, if the solution is 6.00 ppm of chlorine? 1 ppm = 1 mg/L 6 ppm = 6 mg/L 6 mg 15.0 L = 1 L x x =x = (6 mg) (15.0 L) 1 L = 90.0 mg

7. Examples The label on a bottle of sparkling water lists the dissolved minerals as 440 ppm. What mass of minerals is present in a 200 ml glass of water? 1 ppm = 1 mg/L 440 ppm = 440 mg/L 440 mg 0.200 L = 1 L x x =x = (440 mg) (0.200 L) 1 L = 88.0 mg

8. Concentration concentration = # of mol of solute (mol) # of litres of solution (L) c = n V Units  mol/L or M

9. Examples What is the concentration of a solution if 5.6 mol of NaCl are dissolved in 20.8 L of water? WS 20/5 - 1 c = n V c = 5.6 mol 20.8 L c = 0.27 M

10. Practice  P.205 # 1 – 6  P. 208 # 7 – 12  P. 210 # 13 - 16

11. Calculating ionic Concentrations In solution electrolytic compounds exist as free, separate ions. NaCl (aq) really means Na + (aq) and Cl - (aq) In chemical reactions involving such a solution the ions react independantly.

12. Calculating ionic Concentrations To calculate the ionic concentration of ions in solution: – Step 1: write a balanced dissociation equation – Step2: use a mole ratio from the equation to determine the ion concentration

13. Calculating ionic Concentrations What is the concentration of each ion in 0.23 M Al 2 (SO 4 ) 3 solution? Balanced dissociation eqaution: – Al 2 (SO 4 ) 3(s) → 2 Al 3+ (aq) + 3 SO 4 2- (aq) Use mole ratio: 0.23 M Al2(SO4)3 1 mol Al2(SO4)3 2 mol Al 3+ = 0.46 M

14. Practice  P. 212 # 17 - 19 Section 5.3 Questions  P.214 # 1 – 17