Chapter 12 Solutions

1. Chapter 12 Solutions 12.5 Molarity and Dilution

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. 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. Preparation of Solutions

5. Preparation of Solutions • A solution of a desired concentration can be prepared by diluting a small volume of a more-concentrated solution, a stock solution, with additional solvent. – Calculate the number of moles of solute desired in the final volume of the more-dilute solution and then calculate the volume of the stock solution that contains the amount of solute. – Diluting a given quantity of stock solution with solvent does not change the number of moles of solute present. – The relationship between the volume and concentration of the stock solution and the volume and concentration of the desired diluted solution is (Vs) (M s) = moles of solute = (Vd) (M d).

6. Preparation of Solutions

7. Calculating Molarity

8. Example of Calculating Molarity What is the molarity of 0.500 L of a NaOH solution if it contains 6.00 g of NaOH? STEP 1State 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

9. Example of Calculating of Molarity (continued) STEP 3Write 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

10. Learning Check What is the molarity of 325 mL of a solution containing 46.8 g of NaHCO3? A. 0.557 M NaHCO3 solution B. 1.44 M NaHCO3 solution C. 1.71 M NaHCO3 solution

11. Solution STEP 1State the given and needed quantities. Given 46.8 g of NaHCO3 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 NaHCO3 moles of NaHCO3 molarity

12. Solution (continued) STEP 3Write equalities and conversion factors needed. 1 mol of NaHCO3 = 84.01 g of NaHCO3 1 mol NaHCO3 and 84.01 g NaHCO3 84.01 g NaOH 1 mol NaHCO3 STEP 4 Set up problem to calculate molarity. 46.8 g NaHCO3 x 1 mol NaHCO3 = 0.557 mol of NaHCO3 84.01 g NaHCO3 0.557 mol NaHCO3 = 1.71 M NaHCO3 solution 0.325 L

13. Molarity Conversion Factors The units of molarity are used as conversion factors in calculations with solutions.

14. 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 1State 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

15. Example of Using Molarity in Calculations (continued) STEP 3Write 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

16. 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

17. Learning Check How many grams of AlCl3 are needed to prepare 37.8 mL of a 0.150 M AlCl3 solution? A. 0.00567 g of AlCl3 B. 0.756 g of AlCl3 C. 5.04 g of AlCl3

18. Solution STEP 1State the given and needed quantities. Given 37.8 mL of a 0.150 M AlCl3 solution Need grams of AlCl3 STEP 2 Write a plan to calculate mass or volume. milliliters of AlCl3 solution liters of AlCl3 solution moles of AlCl3 grams of AlCl3

19. Solution (continued) STEP 3Write equalities and conversion factors needed. 1 mol of AlCl3 = 133.33 g of AlCl3 1 mol AlCl3 and 133.33 g AlCl3 133.33 g AlCl3 1 mol AlCl3 1000 mL of AlCl3 solution = 1 L of AlCl3 solution 1000 mL AlCl3 solution and 1 L AlCl3 solution 1 L AlCl3 solution 1000 mL AlCl3 solution 1 L of AlCl3 solution = 0.150 mol of AlCl3 1 L AlCl3 solution and 0.150 mol AlCl3 0.150 mol AlCl3 1 L AlCl3 solution

20. 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 AlCl3 (B)

21. Learning Check How many milliliters of a 2.00 M HNO3 solution contain 24.0 g of HNO3? A. 12.0 mL of HNO3 solution B. 83.3 mL of HNO3 solution C. 190. mL of HNO3 solution

22. Solution STEP 1State the given and needed quantities. Given 24.0 g of HNO3; 2.00 M HNO3 solution Need milliliters of HNO3 solution STEP 2 Write a plan to calculate mass or volume. g of solution moles of HNO3 mL of HNO3

23. Solution (continued) STEP 3Write equalities and conversion factors needed. 1 mol of HNO3 = 63.02 g of HNO3 1 mol HNO3 and 63.02 g HNO3 63.02 g HNO3 1 mol HNO3 1000 mL of HNO3 = 2.00 mol of HNO3 1000 mL HNO3 and 2.00 mol HNO3 2.00 mol HNO3 1000 mL HNO3

24. Solution (continued) STEP 4 Set up problem to calculate mass or volume. 24.0 g HNO3 x 1 mol HNO3 x 1000 mL 63.02 g HNO3 2.00 mol HNO3 = 190. mL of HNO3 solution(C)

25. Dilution In a dilution, • water is added • volume increases • concentration decreases

26. NO3- Na+ Moles = 1.0 Volume = 1.0 L Molarity = 1.0 M NaNO3 solution

27. NO3- Na+ Moles = 1.0 Volume = 2.0 L Molarity = 0.50 M • Solution volume is doubled • Moles of solute remain the same • Solution concentration is halved

28. 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 M1V1 = M2V2 initial diluted

29. Calculating Dilution Quantities

30. 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 1Prepare a table of the initial and diluted volumes and concentrations. Initial Solution Diluted Solution M1 = 0.600 M M2= ? V1 = 0.180 L V2 = 0.540 L

31. Example of Dilution Calculations (continued) STEP 2 Solve the dilution expression for the unknown quantity. M1V1 = M2V2 M1V1 = M2V2 V2 V2 M2 = M1V1 V2 STEP 3 Set up the problem by placing known quantities in the dilution expression. M2 =M1V1 = (0.600 M)(0.180 L) = 0.200 M V2 0.540 L

32. 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

33. Solution STEP 1Prepare a table of the initial and diluted volumes and concentrations. Initial Solution Diluted Solution M1= 1.80 M V1 = 15.0 mL M2= 0.300 M V2 = ?

34. Solution (continued) STEP 2 Solve the dilution expression for the unknown quantity. M1V1 = M2V2 M1V1 = M2V2 M2 M2 V2= M1V1 M2 STEP 3 Set up the problem by placing known quantities in the dilution expression. V2 =M1V1 = (1.80 M)(15.0 mL) M2 0.300 M = 90.0 mL (C )