Chapter 11A

# Chapter 11A - 1 Chemistry 1411 Dr Lyon Reactions in Aqueous...

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1 Chemistry 1411 --- Dr Lyon Reactions in Aqueous Solutions II: Calculations NOTE: This is Part 1 of 2 parts on chapter 11. 11-1 Calculations Involving Molarity Review: 1. 2. 1 1 M V If you know the percent mass and density of a solution , you can determine the molarity of a solution: E.g. A 82.5 % by mass solution of hydrobromic acid has a density of 1.63 g / mL. What is the molarity of the hydrobromic acid solution? 3 82.5 1.63 )% 100. 82.5 1.63 1 1 16.6 0 100. 80.91 soln KOH gHB r rso ln mo le sHB r aM a s s D M g Hbr soln mL HBr soln L HBr soln g HBr g HBr soln mol HBr mL HBr soln M g HBr soln mL HBr soln g HBr L H M Br soln == =    If you want to make a dilution of the more concentrated HBr solution to a less concentrated HBr solution, you can use the following relationship: 11 2 2 2 : . . MV Where the original concentrated HBr solution some volume of the original concentrated HBr solution that you will add to water to make a less concentrated HBr solution the diluted concentration of HBr solut ntha M io = = = = () 2 12 ( ). . t you want the result of adding some volume of the more concentrated solution to some volume of water the TOTAL VOLUME of the diluted solution V V V =+

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2 You CANNOT use this dilution relationship to solve stoichiometric solution problems (that is, neutralization questions)!!!!! E.g. You want to neutralize a 135 mL sample of 0.855 M calcium hydroxide. However, the stock solution (16.6 M HBr) is way too concentrated to use for this reaction. Instead, you would like to use a much less concentrated solution of HBr . The only source of HBr you have is the stock solution, therefore, you must perform a dilution of the more concentrated, 16.6 M HBr, solution to make the less concentrated HBr solution. Procedure: You are going to take 25.0 mL of the 16.6M HBr solution and add it to 150 mL of water. Now you must determine the new concentration of HBr solution: M 1 V 1 = M 2 V 2 M 1 = 16.6 M (the original concentration) (16.6 M) (.025 L) = M 2 (0.175 L) V 1 = 25.0 mL (1.0 L / 1000 mL) = 0.025 L (the volume of the 2.37 M = M 2 original 16.6 HBr solution that you use to add to the water.) M 2 = ???? V 2 = 175 mL (1.0 L / 1000 mL) = 0.175 mL (the volume of original HBr solution plus the volume of the water added—remember the two volumes are added together) The final procedure is to neutralize the calcium hydroxide solution. You want to know how many mL of the 2.37 M HBr solution is required to neutralize the entire sample of 0.855 M calcium hydroxide. 1. ) 2. Procedure: You must write a balanced equation for the neutralization reaction: 22 2 2( ) ( ) 2( ) ( HBr aq Ca OH H O l CaBr aq +→+ The balanced equation provides the stoichiometry of the reaction: that is; that 2 moles of HBr are needed to neutralize 1 mole of Ca(OH) 2 . Determine the numbers of moles of calcium hydroxide there are in the entire volume of the 0.855 M calcium hydroxide solution. 2 2 2 2 0.855 ( ) ' 0.135 ( ) ' 0.115 ( ) ' 1.0 ( ) ' moles Ca OH sol n L Ca OH sol n moles of Ca OH in the sol n L Ca OH sol n  =   3. Determine the number of moles of the HBr that are needed to neutralize 0.115 moles of Ca(OH) 2 .
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Chapter 11A - 1 Chemistry 1411 Dr Lyon Reactions in Aqueous...

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