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# 5.3.1

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### 5.3.1

1. 1. Concentration
2. 2. Concentration/Molarity/Molar Concentration <ul><li>A measure of the amount of solute that is dissolved in a measured amount of solvent. </li></ul><ul><li>Dilute: a solution that contains a small amount of solute . </li></ul><ul><li>Concentrated: a solution that contains a large amount of solute. </li></ul>
3. 3. Percent solutions <ul><li>Percent means parts per one hundred. </li></ul>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. 4. Examples <ul><li>10.0 mL of acetic acid is diluted to a total volume of 200 mL. What is the percent by volume of acetic acid? </li></ul>= = 5.00 % (v/v) % (v/v ) = volume of solute volume of solution X 100 0.010 L 0.200 L X 100
5. 5. Parts per million <ul><li>Very dilute concentrations can be recorded as parts per million. (ppm) </li></ul><ul><li>One ppm can be thought of as one drop in a bathtub of water. </li></ul><ul><li>1 ppm = 1 mg/L </li></ul>
6. 6. Examples <ul><li>What mass of chlorine is present in 15.0 L of solution, if the solution is 6.00 ppm of chlorine? </li></ul><ul><li>1 ppm = 1 mg/L </li></ul><ul><li>6 ppm = 6 mg/L </li></ul>6 mg 15.0 L = 1 L x x = (6 mg) (15.0 L) 1 L = 90.0 mg
7. 7. Examples <ul><li>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? </li></ul><ul><li>1 ppm = 1 mg/L </li></ul><ul><li>440 ppm = 440 mg/L </li></ul>440 mg 0.200 L = 1 L x x = (440 mg) (0.200 L) 1 L = 88.0 mg
8. 8. Concentration <ul><li>concentration = # of mol of solute (mol) </li></ul><ul><li># of litres of solution (L) </li></ul><ul><li>c = n </li></ul><ul><li> V </li></ul><ul><li>Units  mol/L or M </li></ul>
9. 9. Examples <ul><li>What is the concentration of a solution if 5.6 mol of NaCl are dissolved in 20.8 L of water? </li></ul>WS 20/5 - 1 c = 0.27 M c = n V c = 5.6 mol 20.8 L
10. 10. Homework <ul><li>Practice </li></ul><ul><li>P.205 # 1 – 6 </li></ul><ul><li>P. 208 # 7 – 12 </li></ul><ul><li>P. 210 # 13 - 16 </li></ul>
11. 11. Calculating ionic Concentrations <ul><li>In solution electrolytic compounds exist as free, separate ions. </li></ul><ul><li>NaCl (aq) really means Na + (aq) and Cl - (aq) </li></ul><ul><li>In chemical reactions involving such a solution the ions react independantly. </li></ul>
12. 12. Calculating ionic Concentrations <ul><li>To calculate the ionic concentration of ions in solution: </li></ul><ul><ul><li>Step 1: write a balanced dissociation equation </li></ul></ul><ul><ul><li>Step2: use a mole ratio from the equation to determine the ion concentration </li></ul></ul>
13. 13. Calculating ionic Concentrations <ul><li>What is the concentration of each ion in 0.23 M Al 2 (SO 4 ) 3 solution? </li></ul><ul><li>Balanced dissociation eqaution: </li></ul><ul><ul><li>Al 2 (SO 4 ) 3(s) -> 2 Al 3+ (aq) + 3 SO 4 2- (aq) </li></ul></ul><ul><li>Use mole ratio: </li></ul>0.23 M Al2(SO4)3 1 mol Al2(SO4)3 2 mol Al 3+ = 0.46 M
14. 14. Homework <ul><li>Practice </li></ul><ul><li>P. 212 # 17 - 19 </li></ul><ul><li>Section 5.3 Questions </li></ul><ul><li>P.214 # 1 – 17 </li></ul>