Slide 1:1 Molarity Molarity is a term used to express concentration. The units of molarity are moles per liter (It is abbreviated as a capital M) When working problems, it is a good idea to change M into its units.


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Slide 3:3 A solution is prepared by dissolving 3.73 grams of AlCl3 in water to form 200.0 mL solution. A 10.0 mL portion of the solution is then used to prepare 100.0 mL of solution. Determine the molarity of the final solution. What type of problem(s) is this? Molarity followed by dilution. Solutions


Slide 4:4 A solution is prepared by dissolving 3.73 grams of AlCl3 in water to form 200.0 mL solution. A 10.0 mL portion of the solution is then used to prepare 100.0 mL of solution. Determine the molarity of the final solution. 1st: = mol L 3.73 g 200.0 x 10-3 L 0.140 2nd: M1V1 = M2V2 (0.140 M)(10.0 mL) = (? M)(100.0 mL) 0.0140 M = M2 molar mass of AlCl3 dilution formula final concentration Solutions


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Slide 6:6 50.0 mL of 6.0 M H2SO4 (battery acid) were spilled and solid NaHCO3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO3 must be used? H2SO4(aq) + 2NaHCO3 ? 2H2O(l) + Na2SO4(aq) + 2CO2(g) Solution Stoichiometry


Slide 7:7 50.0 mL 6.0 M ? g Look! A conversion factor! 50.0 mL of 6.0 M H2SO4 (battery acid) were spilled and solid NaHCO3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO3 must be used? H2SO4(aq) + 2NaHCO3 ? 2H2O(l) + Na2SO4(aq) + 2CO2(g) Solution Stoichiometry = Our Goal


Slide 8:8 50.0 mL 6.0 M ? g 50.0 mL of 6.0 M H2SO4 (battery acid) were spilled and solid NaHCO3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO3 must be used? H2SO4(aq) + 2NaHCO3 ? 2H2O(l) + Na2SO4(aq) + 2CO2(g) Solution Stoichiometry = Our Goal = g NaHCO3 H2SO4 50.0 mL 1 mol H2SO4 NaHCO3 2 mol NaHCO3 84.0 g mol NaHCO3 50.4


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:10 Solution Stoichiometry: Determine how many mL of 0.102 M NaOH solution are needed to neutralize 35.0 mL of 0.125 M H2SO4 solution. First write a balanced Equation. ____NaOH + ____H2SO4 ? ____H2O + ____Na2SO4 2 1 2 1


Slide 11:11 Solution Stoichiometry: Determine how many mL of 0.102 M NaOH solution is needed to neutralize 35.0 mL of 0.125 M H2SO4 solution. Now, let’s get organized. Place numerical Information and accompanying UNITS below each compound. ____NaOH + ____H2SO4 ? ____H2O + ____Na2SO4 2 1 2 1 0.102 M ? mL 35.0 mL Since 1 L = 1000 mL, we can use this to save on the number of conversions Our Goal


Slide 12:12 Determine how many mL of 0.102 M NaOH solution is needed to neutralize 35.0 mL of 0.125 M H2SO4 solution. Now let’s get to work converting. ____NaOH + ____H2SO4 ? ____H2O + ____Na2SO4 2 1 2 1 0.102 M ? mL 35.0 mL = mL NaOH H2SO4 35.0 mL H2SO4 0.125 mol 1000 mL H2SO4 NaOH 2 mol 1 mol H2SO4 1000 mL NaOH 0.102 mol NaOH 85.8 Units Match Solution Stoichiometry: shortcut


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Slide 14:14 What volume of 0.40 M HCl solution is needed to completely neutralize 47.1 mL of 0.75 M Ba(OH)2? 1st write out a balanced chemical equation Solution Stoichiometry


Slide 15:15 What volume of 0.40 M HCl solution is needed to completely neutralize 47.1 mL of 0.75 M Ba(OH)2? 2HCl(aq) + Ba(OH)2(aq) ? 2H2O(l) + BaCl2 0.40 M 47.1 mL 0.75 M ? mL = mL HCl Ba(OH)2 47.1 mL 1 mol Ba(OH)2 HCl 2 mol 0.40 mol HCl HCl 1000 mL 176 Units match Solution Stoichiometry


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Slide 18:18 Solution Stochiometry Problem: A chemist performed a titration to standardize a barium hydroxide solution. If it took 23.28 mL of 0.135 M hydrochloric acid to neutralize 25.00 mL of the barium hydroxide solution, what was the concentration of the barium hydroxide solution in moles per liter (M)? First write a balanced chemical reaction. ____HCl(aq) + ____Ba(OH)2(aq) ? ____H2O(l) + ____BaCl2(aq) 2 1 2 1 23.28 mL 0.135 mol L 25.00 mL ? mol L


Slide 19:19 Solution Stochiometry Problem: A chemist performed a titration to standardize a barium hydroxide solution. If it took 23.28 mL of 0.135 M hydrochloric acid to neutralize 25.00 mL of the barium hydroxide solution, what was the concentration of the barium hydroxide solution in moles per liter (M)? ____HCl(aq) + ____Ba(OH)2(aq) ? ____H2O(l) + ____BaCl2(aq) 2 1 2 1 23.28 mL 0.135 mol L 25.00 mL ? mol L = mol Ba(OH)2 L Ba(OH)2 25.00 x 10-3 L Ba(OH)2 Units Already Match on Bottom! 0.0629 Units match on top!


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Slide 21:21 48.0 mL of Ca(OH)2 solution was titrated with 19.2 mL of 0.385 M HNO3. Determine the molarity of the Ca(OH)2 solution. We must first write a balanced equation. Solution Stochiometry Problem:


Slide 22:22 48.0 mL of Ca(OH)2 solution was titrated with 19.2 mL of 0.385 M HNO3. Determine the molarity of the Ca(OH)2 solution. Ca(OH)2(aq) + HNO3(aq) ? H2O(l) + Ca(NO3)2(aq) 2 2 48.0 mL 19.2 mL 0.385 M = mol(Ca(OH)2) L (Ca(OH)2) 19.2 mL HNO3 48.0 x 10-3L ? M units match! 0.0770 Solution Stochiometry Problem:


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