1. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 1 12.5 Molarity and Dilution Chapter 12 Solutions

2. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 2 Molarity (M) Molarity (M) is a concentration term for solutions the moles of solute in 1 L of solution moles of solute liter of solution

3. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 3 Preparing a 6.0 M Solution A 6.00 M NaOH solution is prepared by weighing out 60.0 g of NaOH (1.50 mol) and adding water to make 0.250 L of a 6.00 MNaOH solution

4. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 4 Calculating Molarity

5. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 5 What is the molarity of 0.500 L of a NaOH solution if it contains 6.00 g of NaOH? STEP 1 State the given and needed quantities. Given 6.00 g of NaOH in 0.500 L of solution Need molarity (M) STEP 2 Write a plan to calculate molarity. molarity (M) = moles of solute liters of solution grams of NaOH moles of NaOH molarity Example of Calculating Molarity

6. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 6 STEP 3 Write equalities and conversion factors needed. 1 mol of NaOH = 40.01 g of NaOH 1 mol NaOH and 40.01 g NaOH 40.01 g NaOH 1 mol NaOH STEP 4 Set up problem to calculate molarity. 6.00 g NaOH x 1 mol NaOH = 0.150 mol of NaOH 40.01 g NaOH 0.150 mol NaOH = 0.300 mol 0.500 L solution 1 L = 0.300 M NaOH solution Example of Calculating of Molarity (continued)

7. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 7 What is the molarity of 325 mL of a solution containing 46.8 g of NaHCO 3 ? A. 0.557 M NaHCO 3 solution B. 1.44 M NaHCO 3 solution C. 1.71 M NaHCO 3 solution Learning Check

8. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 8 STEP 1 State the given and needed quantities. Given 46.8 g of NaHCO 3 in 0.325 L of solution Need molarity (M) STEP 2 Write a plan to calculate molarity. molarity (M) = moles of solute liters of solution grams of NaHCO 3 moles of NaHCO 3 molarity Solution

9. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 9 STEP 3 Write equalities and conversion factors needed. 1 mol of NaHCO 3 = 84.01 g of NaHCO 3 1 mol NaHCO 3 and 40.01 g NaHCO 3 40.01 g NaOH 1 mol NaHCO 3 STEP 4 Set up problem to calculate molarity. 46.8 g NaHCO 3 x 1 mol NaHCO 3 = 0.557 mol of NaHCO 3 84.01 g NaHCO 3 0.557 mol NaHCO 3 = 1.71 M NaHCO 3 solution 0.325 L Solution (continued)

10. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 10 Molarity Conversion Factors The units of molarity are used as conversion factors in calculations with solutions.

11. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 11 Example of Using Molarity in Calculations How many grams of KCl are needed to prepare 0.125 L of a 0.720 M KCl solution? STEP 1 State the given and needed quantities. Given 0.125 L of a 0.720 M KCl solution Need grams of KCl STEP 2 Write a plan to calculate mass or volume. liters of KCl solution moles of KCl grams of KCl

12. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 12 Example of Using Molarity in Calculations (continued) STEP 3 Write equalities and conversion factors needed. 1 mol of KCl = 74.55 g of KCl 1 mol KCl and 74.55 g KCl 74.55 g KCl 1 mol KCl 1 L of KCl solution = 0.720 mol of KCl 1 L KCl solution and 0.720 mol KCl 0.720 mol KCl 1 L KCl solution

13. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 13 Example of Using Molarity in Calculations (continued) STEP 4 Set up problem to calculate mass or volume. 0.125 L x 0.720 mol KCl x 74.55 g KCl = 6.71 g of KCl 1 L 1 mol KCl

14. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 14 How many grams of AlCl 3 are needed to prepare 37.8 mL of a 0.150 M AlCl 3 solution? A. 0.00567 g of AlCl 3 B. 0.756 g of AlCl 3 C. 5.04 g of AlCl 3 Learning Check

15. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 15 Solution STEP 1 State the given and needed quantities. Given 37.8 mL of a 0.150 M AlCl 3 solution Need grams of AlCl 3 STEP 2 Write a plan to calculate mass or volume. milliliters of AlCl 3 solution liters of AlCl 3 solution moles of AlCl 3 grams of AlCl 3

16. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 16 Solution (continued) STEP 3 Write equalities and conversion factors needed. 1 mol of AlCl 3 = 133.33 g of AlCl 3 1 mol AlCl 3 and 133.33 g AlCl 3 133.33 g AlCl 3 1 mol AlCl 3 1000 mL of AlCl 3 solution = 1 L of AlCl 3 solution 1000 mL AlCl 3 solution and 1 L AlCl 3 solution 1 L AlCl 3 solution 1000 mL AlCl 3 solution 1 L of AlCl 3 solution = 0.150 mol of AlCl 3 1 L AlCl 3 solution and 0.150 mol AlCl 3 0.150 mol AlCl 3 1 L AlCl 3 solution

17. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 17 Solution (continued) STEP 4 Set up problem to calculate mass or volume. 37.8 mL x 1 L x 0.150 mol x 133.33 g 1000 mL 1 L 1 mol = 0.756 g of AlCl 3 (B)

18. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 18 How many milliliters of a 2.00 M HNO 3 solution contain 24.0 g of HNO 3 ? A. 12.0 mL of HNO 3 solution B. 83.3 mL of HNO 3 solution C. 190. mL of HNO 3 solution Learning Check

19. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 19 STEP 1 State the given and needed quantities. Given 24.0 g of HNO 3 ; 2.00 M HNO 3 solution Need milliliters of HNO 3 solution STEP 2 Write a plan to calculate mass or volume. g of solution moles of HNO 3 mL of HNO 3 Solution

20. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 20 STEP 3 Write equalities and conversion factors needed. 1 mol of HNO 3 = 63.02 g of HNO 3 1 mol HNO 3 and 63.02 g HNO 3 63.02 g HNO 3 1 mol HNO 3 1000 mL of HNO 3 = 2.00 mol of HNO 3 1000 mL HNO 3 and 2.00 mol HNO 3 2.00 mol HNO 3 1000 mL HNO 3 Solution (continued)

21. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 21 STEP 4 Set up problem to calculate mass or volume. 24.0 g HNO 3 x 1 mol HNO 3 x 1000 mL 63.02 g HNO 3 2.00 mol HNO 3 = 190. mL of HNO 3 solution (C) Solution (continued)

22. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 22 Dilution In a dilution, water is added volume increases concentration decreases

23. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 23 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

24. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 24 Calculating Dilution Quantities

25. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 25 Example of Dilution Calculations What is the final molarity of the solution when 0.180 L of 0.600 M KOH is diluted to 0.540 L? STEP 1 Prepare a table of the initial and diluted volumes and concentrations. Initial Solution Diluted Solution M 1 = 0.600 M M 2 = ? V 1 = 0.180 L V 2 = 0.540 L

26. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 26 Example of Dilution Calculations (continued) STEP 2 Solve the dilution expression for the unknown quantity. M 1 V 1 = M 2 V 2 V 2 V 2 M 2 = M 1 V 1 V 2 STEP 3 Set up the problem by placing known quantities in the dilution expression. M 2 = M 1 V 1 = (0.600 M)(0.180 L) = 0.200 M V 2 0.540 L

27. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 27 Learning Check What is the final volume, in milliliters, if 15.0 mL of a 1.80 M KOH solution is diluted to give a 0.300 M KOH solution? A. 27.0 mL of 0.300 M KOH solution B. 60.0 mL of 0.300 M KOH solution C. 90.0 mL of 0.300 M KOH solution

28. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 28 Solution STEP 1 Prepare a table of the initial and diluted volumes and concentrations. Initial Solution Diluted Solution M 1 = 1.80 M V 1 = 15.0 mL M 2 = 0.300M V 2 = ?

29. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 29 Solution (continued) STEP 2 Solve the dilution expression for the unknown quantity. M 1 V 1 = M 2 V 2 M 2 M 2 V 2 = M 1 V 1 M 2 STEP 3 Set up the problem by placing known quantities in the dilution expression. V 2 = M 1 V 1 = (1.80 M)(15.0 mL) M 2 0.300 M = 90.0 mL (C )