Concentrations of Solutions

1. Concentrations of Solutions Amounts and Volumes

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

3. Introduction • The concentrationof a solution is the measure of the amount of a soluteis in a solution. • A dilute solution has a smallamount of solute. • A concentrated solution has a largeamount 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 n volume of solution (L) V

5. Molarity M = • We can rearrange the formula to solve for mols or volume. n = M×V V = n n V 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 • mNaCl= 0.90 g • MNaCl = 58.5 g/mol 0.90 g 0.015 mol mNaCl n n = = M = = = 0.015 mol 0.100 L 58.5 g/mol MNaCl V This value comes from the atomic mass of Na (23.0 g/mol) plus the atomic mass of Cl (35.5 g/mol). First, we write down the known values. Next, we need to calculate the number of mols of NaCl in solution. Now, we have enough information to calculate the molarity. = 0.15 M We substitute our known values … … and do the calculation. We substitute our known values … … and do the calculation.

7. Molarity Find the molarity of each of the following solutions. • 0.200 mols of NaOH in a solution of 500 mL • 1.25 mols of CuCl2 in a solution of 3.40 L • 0.0352 mols of KCl in a solution of 25.0 mL • 14.0 mols of Li2CO3 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 CuCl2, 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 • MCuCl2= 134 g/mol Next, we need to calculate the number of mols of CuCl2 in solution. Now, we have enough information to calculate the mass of CuCl2. First, we write down the known values. n = MV = (2.50 M)(0.425 L) = 1.06 mol m = Mn = 142 g = (134 g/mol)(1.06 mol) We substitute in our known values … We substitute in our known values … … and then do the calculation. … and then do the calculation.

9. Molarity Find the number of mols of solute in each of the following solutions. • 4.25 L of 0.250 M CuSO4 • 12.5 mL of 6.50 M HCl • 125 mL of 0.0450 M Na3PO4 • 25.0 mL of 2.37 M AgNO3 1.06 mol CuSO4 0.0815 molHCl 0.00563 mol Na3PO4 0.0593 mol AgNO3

10. Making Dilutions • If we take a solution and add solvent to it, we are dilutingthe solution. • We have kept the same amountof solute in the solution and only changed the volume. • The number of mols before dilution is equal to the number of mols after dilution. n1= n2

11. Making Dilutions • n1 = n2 • If we expand this by using n = M×V, we get: • M1×V1 = M2×V2 • This is called the “dilutionformula.” • Dilutions are used with “stock” solutions. • We dilute the stock solutions to the concentration we desire.

12. Making Dilutions Example 3: How many milliliters of aqueous 2.00 M MgSO4 solution must be diluted with water to prepare 100 mL of aqueous 0.400 M MgSO4? • M1= 2.00 M • V1= ? mL g • M2= 0.400 M • V2 = 100 mL … and do the calculation. Then, we rearrange to solve for V1. We substitute our known values … Next, we write down our dilution equation. First, we write down the known values. (0.400 M)(100 mL) M2V2 ⟹ V1 = = 2.00 M M1 = 20.0 mL M1V1 = M2V2

13. Making Dilutions How much stock solution is needed to prepare: • 5.00 L of 0.250 M HCl from 12.0 M HCl? • 500 mL of 0.100 M NaOH from 6.00 M NaOH? • 125 mL of 0.0250 M HNO3 from 13.5 M HNO3? • 25.0 mL of 0.0500 M AgNO3 from 0.400 M AgNO3? 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: • as the ratio of the volumeof the soluteto the volumeof the solution, %(v/v). • %(v/v) = (Vsolute/Vsolution)(100%) • as the ratio of the massof the soluteto the massof the solution, %(m/m). • %(m/m) = (msolute/msolution)(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? Visopropanol = 85 mL Vsolution = 250 mL %(v/v) = (Visopropanol/Vsolution)(100%) %(v/v) = (85 mL/250 mL)(100%) = 34%(v/v)

16. Percent Solutions What is the %(v/v) of • 25.0 ml of ethyl alcohol diluted to 100 mL? • 12.5 mL of methyl alcohol diluted to 2.50 L? • 32.0 mL of molasses diluted to 425 mL? • 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. • n1 = n2 ➔ M1V1 = M2V2

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). • %(v/v) = (Vsolute/Vsolution)(100%) • as the ratio of the mass of the solute to the mass of the solution, %(m/m). • %(m/m) = (msolute/msolution)(100%)