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Thursday, March 6, 2008 Discussion of Molarity Lab Results Introduce Section 15.2b -- Dilutions Homework: Pg. 555, #31a-d, 32, 33, 34

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Concentrated Solution To save time and space, usually buy solutions that are concentrated- stock solutions Add solvent to get the molarity you want Dilution- process of adding more solvent to a solution

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Dilutions Only water is added to the solutions Figure out how much water needs to be added Moles of solute after dilution = moles of solute before dilution

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What happens? Moles of solute stays the same Volume of water increases So Molarity decreases

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Example Prepare 500. mL of 1.00M acetic acetic acid, HC 2 H 3 O 2 from a 17.5 M stock solution of acetic acid. What volume of stock solution is required? 1 st - find the number of moles of acetic acid needed in the final solution V dilute x M dilute = moles solute

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Example (continued) Only source of acetic acid is the stock solution So moles of solute in dilute solution must equal moles in concentrated solution This is always true!

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Example (continued) Now we need to find the volume of 17.5 M acetic acid that contains 0.500 mol of HC 2 H 3 O 2 This is the unknown volume V

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Volume x molarity = moles Solving for V gives V = 0.0286 L or 28.6 mL of solution To make 500 mL of 1.00M acetic acid solution, we take 28.6 mL of 17.5 M acetic acid and dilute it to a total volume of 500 mL

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Because moles of solute remain the same before and after dilution, M 1 x V 1 = moles of solute = M 2 x V 2 1 is initial conditions, 2 is final conditions

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Preparing a dilution

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Example What volume of 18 M H 2 SO 4 would be needed to prepare 1.5 L of 0.10 M H 2 SO 4 ? V x 18 M = 1.5 L x 0.10 M V= 8.3 mL

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Common Stock Solutions Stock Solutions Sulfuric acid 18 M Nitric acid16 M HCl12 M

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Practice Problem Exercise 15.8 What volume of 12M HCl must be taken to prepare 0.75L of 0.25M HCl? 16 mL

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