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2 3 Molarity and Dilutions

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2 3 Molarity and Dilutions Warm up • Discuss at your table: What does it mean if a solution is concentrated? What does it mean if a solution is dilute? Bonus: Try to use some new vocabulary from last week! Lab Debrief – use this data! • Class Averages: 20.4 45.2 62.9 74.6 Excel demo: graphing • Title • Units • Label axes • Intervals • Plot X: temperature Y: solubility Solubility as a function of temperature… Solutions Solvent solute When the solvent is water the solution is said to be aqueous We need a way to quantify the Concentration of Solutions Molarity (M), or molar concentration, is defined as the number of moles of solute per liter of solution. moles solute molarity = liters solution Molarity is a unit, but also can be used as a conversion factor Example 1: Molarity formula For an aqueous solution of glucose (C6H12O6), determine the molarity of a 2.00 L of a solution that contains 50.0 g of glucose Example 2: Molarity formula For an aqueous solution of glucose (C6H12O6), determine the volume of this solution that would contain 0.250 mole of glucose. Example 3: Molarity formula For an aqueous solution of glucose (C6H12O6) determine the number of moles of glucose in 0.500 L of this solution. 4.5 Example 1: Making a molar solution Explain how to make 1 liter of 0.125 M KBr solution Example 2: Making a molar solution Explain how to make 500 milliliters of 0.375 M MgCl2 solution Practice • In class: • HW: Complete lab Warm up • Discuss at your table: On a hot summer day you are sipping lemonade but it is far too sweet! What should you do if you want to dilute your beverage? Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution. Dilution Add Solvent Moles of solute Moles of solute before dilution (i) = after dilution (f) MiVi = MfVf 4.5 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 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. M1 V1 = M2 V2 How would you prepare 60.0 mL of 0.20 M HNO3 from a stock solution of 4.00 M HNO3? M1V1 = M2V2 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? M1V1 = M2V2 Example 1: Dilution from stock solution 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? Practice • Problems • HW: Prelab Worked Example 9.10 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. M × mL c c -1 Strategy (a) Md = ; (b) mol = M×L, 250.00 mL = 2.500×10 L mLd Solution 2.00 M × 10.00 mL (a) M = -2 d1 250.00 mL = 8.00×10 M 8.00×10-2 M × 10.00 mL M = = 3.20×10-3 M d2 250.00 mL 3.20×10-3 M × 10.00 mL M = = 1.28×10-4 M d3 250.00 mL 1.28×10-4 M × 10.00 mL M = = 5.12×10-6 M d4 250.00 mL Worked Example 9.10 (cont.) Solution -2 -1 -2 (b) mol1 = 8.0010 M 2.50010 L = 2.0010 mol -3 -1 -4 mol2 = 3.2010 M 2.50010 L = 8.0010 mol -4 -1 -5 mol3 = 1.2810 M 2.50010 L = 3.2010 mol -6 -1 -6 mol4 = 5.1210 M 2.50010 L = 1.2810 mol 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. 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, 3+ - AlCl3(s) → Al (aq) + 3Cl (aq) so the concentration of Cl- will be three times the concentration of AlCl3. - - 3 mol Cl [Cl ] = [AlCl3] × 1 mol AlCl3 1.02 mol AlCl 3 mol Cl- = 3 × L 1 mol AlCl3 3.06 mol Cl- = = 3.06 M L.
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