1. Concentrations of Solutions Amounts and Volumes

2. Objectives When you complete this presentation, you will be able to o Distinguish between solute, solvent, and solution. o Define the concentration of a solution in terms of molarity and percent composition. o Calculate the molarity of a solution given the volume of the solution and number of mols of a solute

3. Introduction The concentration of a solution is the measure of the amount of a solute is in a solution. A dilute solution has a small amount of solute. A concentrated solution has a large amount of solute. These terms are too qualitative - we need a quantitative measurement system.

4. Molarity We can use the number of mols of a solute, n, in a volume, V, of solution to give us a quantitative measurement. We call this Molarity (M). Molarity = M = number mols of solute volume of solution (L) n V

5. Molarity M = We can rearrange the formula to solve for mols or volume. n = M×V V = n V n M

6. Molarity Example 1: Intravenous (IV) saline solutions are often administered to patients in the hospital. One saline solution contains 0.90 g NaCl in exactly 100 mL of solution. What is the molarity of the solution? V = 100 mL = 0.100 L m NaCl = 0.90 g M NaCl = 58.5 g/mol n = m NaCl M NaCl = 0.90 g 58.5 g/mol = 0.015 mol M = n V = 0.015 mol 0.100 L = 0.15 M First, we write down the known values. This value comes from the atomic mass of Na (23.0 g/mol) plus the atomic mass of Cl (35.5 g/mol). Next, we need to calculate the number of mols of NaCl in solution. Now, we have enough information to calculate the molarity. We substitute our known values … … and do the calculation.

7. Molarity Find the molarity of each of the following solutions. 1.0.200 mols of NaOH in a solution of 500 mL 2.1.25 mols of CuCl 2 in a solution of 3.40 L 3.0.0352 mols of KCl in a solution of 25.0 mL 4.14.0 mols of Li 2 CO 3 in a solution of 5.25 L 0.400 M 0.368 M 1.41 M 2.67 M

8. Molarity Example 2: What is the mass of CuCl 2, M = 134 g/mol, in 425 mL of a 2.50 M solution of copper(II) chloride? V = 425 mL = 0.425 L M = 2.50 M M CuCl2 = 134 g/mol m = M n = (134 g/mol)(1.06 mol) = 142 g n = MV= (2.50 M)(0.425 L)= 1.06 mol First, we write down the known values. Next, we need to calculate the number of mols of CuCl 2 in solution. Now, we have enough information to calculate the mass of CuCl 2. We substitute in our known values … … and then do the calculation. We substitute in our known values … … and then do the calculation.

9. Molarity Find the number of mols of solute in each of the following solutions. 1.4.25 L of 0.250 M CuSO 4 2.12.5 mL of 6.50 M HCl 3.125 mL of 0.0450 M Na 3 PO 4 4.25.0 mL of 2.37 M AgNO 3 1.06 mol CuSO 4 0.0815 mol HCl 0.00563 mol Na 3 PO 4 0.0593 mol AgNO 3

10. Making Dilutions If we take a solution and add solvent to it, we are diluting the solution. o We have kept the same amount of solute in the solution and only changed the volume. The number of mols before dilution is equal to the number of mols after dilution. n 1 = n 2

11. Making Dilutions n 1 = n 2 If we expand this by using n = M×V, we get: o M 1 ×V 1 = M 2 ×V 2 o This is called the “dilution formula.” Dilutions are used with “stock” solutions. o We dilute the stock solutions to the concentration we desire.

12. Making Dilutions Example 3: How many milliliters of aqueous 2.00 M MgSO 4 solution must be diluted with water to prepare 100 mL of aqueous 0.400 M MgSO 4 ? M 1 = 2.00 M V 1 = ? mL g M 2 = 0.400 M V 2 = 100 mL = (0.400 M)(100 mL) 2.00 M M 1 V 1 = M 2 V 2 V 1 = M2V2M2V2 M1M1 = 20.0 mL First, we write down the known values. Next, we write down our dilution equation. Then, we rearrange to solve for V 1. We substitute our known values … … and do the calculation.

13. Making Dilutions How much stock solution is needed to prepare: 1.5.00 L of 0.250 M HCl from 12.0 M HCl? 2.500 mL of 0.100 M NaOH from 6.00 M NaOH? 3.125 mL of 0.0250 M HNO 3 from 13.5 M HNO 3 ? 4.25.0 mL of 0.0500 M AgNO 3 from 0.400 M AgNO 3 ? 0.104 L = 104 mL 0.00833 L = 8.33 mL 0.000231 L = 0.231 mL 0.00313 L = 3.13 mL

14. Percent Solutions The concentration of a solution in percent can be expressed in two ways: o as the ratio of the volume of the solute to the volume of the solution, %(v/v). %(v/v) = (V solute /V solution )(100%) o as the ratio of the mass of the solute to the mass of the solution, %(m/m). %(m/m) = (m solute /m solution )(100%)

15. Percent Solutions Example 4: What is the percent by volume of isopropanol in the final solution when 85 mL of isopropanol is diluted to a volume of 250 mL with water? V isopropanol = 85 mL V solution = 250 mL %(v/v) = (V isopropanol /V solution )(100%) %(v/v) = (85 mL/250 mL)(100%) = 34%(v/v)

16. Percent Solutions What is the %(v/v) of 1.25.0 ml of ethyl alcohol diluted to 100 mL? 2.12.5 mL of methyl alcohol diluted to 2.50 L? 3.32.0 mL of molasses diluted to 425 mL? 4.15.6 mL of ethyl alcohol diluted to 100. mL? 25.0%(v/v) 0.500%(v/v) 7.53%(v/V) 15.6%(v/V)

17. Summary The concentration of a solution is the measure of the amount of a solute is in a solution. We can use the number of mols of a solute in a liter of solution to give us a quantitative measurement called molarity, M. If we take a solution and add solvent to it, we are diluting the solution. n 1 = n 2 ➔ M 1 V 1 = M 2 V 2

18. Summary The concentration of a solution in percent can be expressed in two ways: as the ratio of the volume of the solute to the volume of the solution, %(v/v). o %(v/v) = (V solute /V solution )(100%) as the ratio of the mass of the solute to the mass of the solution, %(m/m). o %(m/m) = (m solute /m solution )(100%)