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1 12.5 Molarity and Dilution Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings.

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Presentation on theme: "1 12.5 Molarity and Dilution Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings."— Presentation transcript:

1 1 12.5 Molarity and Dilution Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

2 2 Molarity (M) Molarity (M) is A concentration term for solutions. The moles of solute in 1 L solution. moles of solute liter of solution

3 3 Preparing a 1.0 Molar Solution A 1.00 M NaCl solution is prepared By weighing out 58.5 g NaCl (1.00 mol) and Adding water to make 1.00 liter of solution. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

4 4 Add GPS Guide to Calculating Molarity Basic 2e p 394

5 5 What is the molarity of 0.500 L NaOH solution if it contains 6.00 g NaOH? STEP 1 Given 6.00 g NaOH in 0.500 L solution Need molarity (mol/L) STEP 2 Plan g NaOH mol NaOH molarity STEP 3 Conversion factors 1 mol NaOH = 40.00 g 1 mol NaOH and 40.00 g NaOH 40.00 g NaOH 1 mol NaOH Calculation of Molarity

6 6 STEP 4 Calculate molarity. 6.00 g NaOH x 1 mol NaOH = 0.150 mol 40.00 g NaOH 0.150 mol = 0.300 mol = 0.300 M NaOH 0.500 L 1 L Calculation of Molarity (cont.)

7 7 What is the molarity of 325 mL of a solution containing 46.8 g of NaHCO 3 ? 1) 0.557 M 2) 1.44 M 3) 1.71 M Learning Check

8 8 3) 1.71 M 46.8 g NaHCO 3 x 1 mol NaHCO 3 = 0.557 mol NaHCO 3 84.01 g NaHCO 3 0.557 mol NaHCO 3 = 1.71 M NaHCO 3 0.325 L Solution

9 9 What is the molarity of a solution if 225 mL contains 34.8 g KNO 3 ? 1)0.0775 M 2)1.53 M 3)15.5 M Learning Check

10 10 2) 1.53 M 34.8 g KNO 3 x 1 mol KNO 3 = 0.344 mol KNO 3 101.11g KNO 3 M = mol = 0.344 mol KNO 3 = 1.53 M L 0.225 L In one setup 34.8 g KNO 3 x 1 mol KNO 3 x 1 = 1.53 M 101.11g KNO 3 0.225 L Solution

11 11 Molarity Conversion Factors The units of molarity are used as conversion factors in calculations with solutions. Table 2.6 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

12 12 Molarity in Calculations How many grams of KCl are needed to prepare 125 mL of a 0.720 M KCl solution? STEP 1 Given 125 mL (0.125 L) of 0.720 M KCl Need Grams of KCl STEP 2 Plan L KCl mol KCl g KCl

13 13 Molarity in Calculations (cont.) STEP 3 Conversion factors 1 mol KCl = 74.55 g 1 mol KCl and 74.55 g KCl 74.55 g KCl 1 mol KCl 1 L KCl = 0.720 mol KCl 1 L and 0.720 mol KCl 0.720 mol KCl 1 L STEP 4 Calculate grams. 0.125 L x 0.720 mol KCl x 74.55 g KCl = 6.71 g KCl 1 L 1 mol KCl

14 14 How many grams of AlCl 3 are needed to prepare 125 mL of a 0.150 M solution? 1) 20.0 g AlCl 3 2) 16.7g AlCl 3 3) 2.50 g AlCl 3 Learning Check

15 15 Solution 3) 2.50 g AlCl 3 0.125 L x 0.150 mol x 133.3 g = 2.50 g AlCl 3 1 L 1 mol

16 16 How many milliliters of 2.00 M HNO 3 contain 24.0 g HNO 3 ? 1) 12.0 mL 2) 83.3 mL 3) 190. mL Learning Check

17 17 3) 190. mL 24.0 g HNO 3 x 1 mol HNO 3 x 1000 mL 63.02 g HNO 3 2.00 mol HNO 3 Molarity factor inverted = 190. mL HNO 3 Solution

18 18 Dilution In a dilution Water is added. Volume increases. Concentration decreases. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

19 19 Comparing Initial and Diluted Solutions In the initial and diluted solution The moles of solute are the same. The concentrations and volumes are related by the equation M 1 V 1 = M 2 V 2 initial diluted

20 20 Add GPS Guide to Calculating Dilution Quantities Basic 2e p 398

21 21 Dilution Calculations What is the molarity if 0.180 L of 0.600 M KOH is diluted to a final volume of 0.540 L? STEP 1 Prepare a table: M 1 = 0.600 MV 1 = 0.180 L M 2 = ?V 2 = 0.540 L STEP 2 Solve dilution equation for unknown. M 1 V 1 = M 2 V 2 M 1 V 1 / V 2 = M 2 STEP 3 Set up and enter values: M 2 = M 1 V 1 = (0.600 M)(0.180 L) = 0.200 M V 2 0.540 L

22 22 Learning Check What is the final volume if 15.0 mL of a 1.80 M KOH is diluted to give a 0.300 M solution? 1) 27.0 mL 2) 60.0 mL 3) 90.0 mL

23 23 Solution STEP 1 Prepare a table: M 1 = 1.80 MV 1 = 15.0 mL M 2 = 0.300MV 2 = ? STEP 2 Solve dilution equation for unknown. M 1 V 1 = M 2 V 2 V 2 = M 1 V 1 / M 2 STEP 3 Set up and enter values: V 2 = M 1 V 1 = (1.80 M)(15.0 mL) = 90.0 mL M 2 0.300 M


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