Concentrations of Solutions Amounts and Volumes

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

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.

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

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

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 = L m NaCl = 0.90 g M NaCl = 58.5 g/mol n = m NaCl M NaCl = 0.90 g 58.5 g/mol = mol M = n V = mol 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.

Molarity Find the molarity of each of the following solutions mols of NaOH in a solution of 500 mL mols of CuCl 2 in a solution of 3.40 L mols of KCl in a solution of 25.0 mL mols of Li 2 CO 3 in a solution of 5.25 L M M 1.41 M 2.67 M

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 = 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.

Molarity Find the number of mols of solute in each of the following solutions L of M CuSO mL of 6.50 M HCl mL of M Na 3 PO mL of 2.37 M AgNO mol CuSO mol HCl mol Na 3 PO mol AgNO 3

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

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.

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 M MgSO 4 ? M 1 = 2.00 M V 1 = ? mL g M 2 = 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.

Making Dilutions How much stock solution is needed to prepare: L of M HCl from 12.0 M HCl? mL of M NaOH from 6.00 M NaOH? mL of M HNO 3 from 13.5 M HNO 3 ? mL of M AgNO 3 from M AgNO 3 ? L = 104 mL L = 8.33 mL L = mL L = 3.13 mL

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%)

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)

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

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

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%)