Concentration of Solutions Molarity (M), or molar concentration, is defined as the number of moles of solute per liter of solution.

Worked Example 9.8 For an aqueous solution of glucose (C6H12O6), 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. 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 = 50.0 g 180.2 g/mol = 0.277 mol Solution molarity = = 0.139 M volume = = 1.80 L moles of C6H12O6 in 0.500 L = 0.500 L×0.139 M = 0.0695 mol 0.277 mol C6H12O6 2.00 L solution 0.250 mol C6H12O6 0.139 M solution

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moles of solute before dilution = moles of solute after dilution 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

moles of solute before = moles of solute after Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution. Dilution Add Solvent Moles of solute before dilution (i) after dilution (f) = moles of solute before = moles of solute after n=CcVc n=CdVd CcVc = CdVd MiVi MfVf = 4.5

(2.00 M CuCl2)(Vc) = (0.100 M CuCl2)(0.2500 L) Concentration of Solutions In an experiment, a student needs 250.0 mL of a 0.100 M CuCl2 solution. A stock solution of 2.00 M CuCl2 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 CuCl2)(Vc) = (0.100 M CuCl2)(0.2500 L) Vc = 0.0125 L or 12.5 mL To make the solution: Pipet 12.5 mL of stock solution into a 250.0 mL volumetric flask. Carefully dilute to the calibration mark. Cc × Vc = Cd × Vd

How would you prepare 60.0 mL of 0.2 M HNO3 from a stock solution of 4.00 M HNO3? CcVc = CdVd Cc = 4.00M Cd = 0.200M Vd = 0.06 L Vc = ? L Vc = CdVd Cc = 0.200M x 0.06L 4.00M = 0.003 L = 3 mL 3 mL of acid + 57 mL of water = 60 mL of solution

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 also be used with mL as the volume as long as both volume measurements are the SAME unit.

Another Dilution Problem If 32 mL stock solution of 6.5 M H2SO4 is diluted to a volume of 500 mL What would be the resulting concentration? CcVc = CdVd (6.5M) (32 mL) = Cd (500 mL) 6.5 M * 32 mL Cd = 500 mL Cd = 0.42 M

Worked Example 9.9 What volume of 12.0 M HCl, a common laboratory stock solution, must be used to prepare 250.0 mL of 0.125 M HCl? Strategy Cc = 12.0 M, Cd = 0.125 M, Vd = 250.0 mL Solution 12.0 M × Vc = 0.125 M × 250.0 mL Vc = 0.125 M × 250.0 mL 12.0 M = 2.60 mL

AlCl3(s) → Al3+(aq) + 3Cl-(aq) Worked Example 9.11 Using square-bracket notation, express the concentration of (a) chloride ion in a solution that is 1.02 M in AlCl3, (b) nitrate ion in a solution that is 0.451 M in Ca(NO3)2, and (c) Na2CO3 in a solution in which [Na+] = 0.124 M. 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. Solution (a) There are 3 moles of Cl- ion for every 1 mole of AlCl3, AlCl3(s) → Al3+(aq) + 3Cl-(aq) so the concentration of Cl- will be three times the concentration of AlCl3. [Cl-] = [AlCl3] × = × = = 3.06 M 3 mol Cl- 1 mol AlCl3 1.02 mol AlCl3 L 3 mol Cl- 1 mol AlCl3 3.06 mol Cl- L

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