1. Chemistry: Atoms First Julia Burdge & Jason Overby Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 9.5 Homework: 61, 63, 65, 67, 71, 73, 75, 77, 79 and 81 Kent L. McCorkle Cosumnes River College Sacramento, CA

2. 4.1 A solution is a homogenous mixture of 2 or more substances The solute is(are) the substance(s) present in the smaller amount(s) The solvent is the substance present in the larger amount SolutionSolventSolute Soft drink (l) Air (g) Soft Solder (s) H2OH2O N2N2 Pb Sugar, CO 2 O 2, Ar, CH 4 Sn

3. Solutions solute Solvent When the solvent is water the solution is said to be aqueous

4. General Properties of Aqueous Solutions A solution is a homogenous mixture of two or more substances. The substance present in the largest amount (moles) is referred to as the solvent. The other substances present are called the solutes. A substance that dissolves in a particular solvent is said to be soluble in that solvent. 9.1

5. Concentration of Solutions Molarity (M), or molar concentration, is defined as the number of moles of solute per liter of solution. Other common rearrangements: 9.5

6. Worked Example 9.8 Strategy Convert the mass of glucose given to moles, and use the equations for interconversions of M, liters, and moles to calculate the answers. moles of glucose = For an aqueous solution of glucose (C 6 H 12 O 6 ), determine (a) the molarity of 2.00 L of a solution that contains 50.0 g of glucose, (b) the volume of this solution that would contain 0.250 mole of glucose, and (c) the number of moles of glucose in 0.500 L of this solution. Solution (a)molarity = = 0.139 M (b)volume = = 1.80 L (c)moles of C 6 H 12 O 6 in 0.500 L = 0.500 L×0.139 M = 0.0695 mol 50.0 g 180.2 g/mol 0.277 mol C 6 H 12 O 6 2.00 L solution 0.250 mol C 6 H 12 O 6 0.139 M solution = 0.277 mol Think About It Check to see that the magnitude of your answers are logical. For example, the mass given in the problem corresponds to 0.277 mole of solute. If you are asked, as in part (b), for the volume that contains a number of moles smaller than 0.277, make sure your answer is smaller than the original volume.

7. 4.5

8. Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution. Dilution Add Solvent Moles of solute before dilution (i) Moles of solute after dilution (f) = MiViMiVi MfVfMfVf = 4.5

9. How would you prepare 60.0 mL of 0.2 M HNO 3 from a stock solution of 4.00 M HNO 3 ? M i V i = M f V f M i = 4.00 M f = 0.200V f = 0.06 L V i = ? L V i = MfVfMfVf MiMi = 0.200 x 0.06 4.00 = 0.003 L = 3 mL 3 mL of acid + 57 mL of water= 60 mL of solution

10. Another Dilution Problem If 32 mL stock solution of 6.5 M H 2 SO 4 is diluted to a volume of 500 mL What would be the resulting concentration? M 1 *V 1 = M 2 *V 2 (6.5M) * (32 mL) = M 2 * (500.0 mL) M 2 = 500 mL 6.5 M * 32 mL M 2 = 0.42 M

11. Concentration of Solutions Dilution is the process of preparing a less concentrated solution from a more concentrated one. moles of solute before dilution = moles of solute after dilution

12. Concentration of Solutions In an experiment, a student needs 250.0 mL of a 0.100 M CuCl 2 solution. A stock solution of 2.00 M CuCl 2 is available. How much of the stock solution is needed? Solution:Use the relationship that moles of solute before dilution = moles of solute after dilution. (2.00 M CuCl 2 )(L c ) = (0.100 M CuCl 2 )(0.2500 L) L c = 0.0125 L or 12.5 mL To make the solution: 1)Pipet 12.5 mL of stock solution into a 250.0 mL volumetric flask. 2) Carefully dilute to the calibration mark. M c × L c = M d × L d

13. Concentration of Solutions Because most volumes measured in the laboratory are in milliliters rather than liters, it is worth pointing out that the equation can be written as M c × mL c = M d × mL d

14. Worked Example 9.9 Strategy M c = 12.0 M, M d = 0.125 M, mL d = 250.0 mL What volume of 12.0 M HCl, a common laboratory stock solution, must be used to prepare 150.0 mL of 0.125 M HCl? Solution 12.0 M × mL c = 0.125 M × 250.0 mL mL c = 0.125 M × 250.0 mL 12.0 M = 2.60 mL Think About It Plug the answer into Equation 9.4, and make sure that the product of concentration and volume on both sides of the equation give the same result.

15. Worked Example 9.10 Strategy (a) M d = Starting with a 2.0 M stock solution of hydrochloric acid, four standard solutions (1 to 4) are prepared by sequential diluting 10.00 mL of each solution to 250.00 mL. Determine (a) the concentrations of all four standard solutions and (b) the number of moles of HCl in each solution. Solution (a) M d1 = M d2 = M d3 = M d4 = 2.00 M × 10.00 mL 250.00 mL M c × mL c mL d ; (b) mol = M×L, 250.00 mL = 2.500×10 -1 L = 8.00×10 -2 M 8.00×10 -2 M × 10.00 mL 250.00 mL = 3.20×10 -3 M 3.20×10 -3 M × 10.00 mL 250.00 mL = 1.28×10 -4 M 1.28×10 -4 M × 10.00 mL 250.00 mL = 5.12×10 -6 M

16. Worked Example 9.10 (cont.) Think About It Serial dilution is one of the fundamental practices of homeopathy. Some remedies undergo so many serial dilutions that very few (if any) molecules of the original substance still exist in the final preparation. Solution (b) mol 1 = 8.00×10 -2 M × 2.500×10 -1 L = 2.00×10 -2 mol mol 2 = 3.20×10 -3 M × 2.500×10 -1 L = 8.00×10 -4 mol mol 3 = 1.28×10 -4 M × 2.500×10 -1 L = 3.20×10 -5 mol mol 4 = 5.12×10 -6 M × 2.500×10 -1 L = 1.28×10 -6 mol

17. Worked Example 9.11 Strategy Use the concentration given in each case and the stoichiometry indicated in the corresponding chemical formula to determine the concentration of the specified ion or compound. Using square-bracket notation, express the concentration of (a) chloride ion in a solution that is 1.02 M in AlCl 3, (b) nitrate ion in a solution that is 0.451 M in Ca(NO 3 ) 2, and (c) Na 2 CO 3 in a solution in which [Na + ] = 0.124 M. Solution (a) There are 3 moles of Cl - ion for every 1 mole of AlCl 3, AlCl 3 (s) → Al 3+ (aq) + 3Cl - (aq) so the concentration of Cl- will be three times the concentration of AlCl 3. [Cl - ] = [AlCl 3 ] × = × = = 3.06 M 3 mol Cl - 1 mol AlCl 3 1.02 mol AlCl 3 L 3 mol Cl - 1 mol AlCl 3 3.06 mol Cl - L