 Solution Stoichiometry Edward A. Mottel Department of Chemistry Rose-Hulman Institute of Technology.

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Solution Stoichiometry Edward A. Mottel Department of Chemistry Rose-Hulman Institute of Technology

6/30/2015 Concentration Units Reading Assignment: Zumdahl: Chapter 4.1-4.3, 5 Solutions can be described by several different concentration units. Density (  ) Mole Fraction (  ) Molality (m) Molarity (M) Parts per Million (ppm)

6/30/2015 Applications Density (mass per volume) rough measure of concentration battery acid anti-freeze Mole Fraction (moles / total moles) colligative property measurements gas mixtures, vapor pressure, distillation

6/30/2015 Applications Molality (moles of solute / kilogram of solvent) colligative property measurements change in freezing point, boiling point vapor pressure, osmotic pressure Molarity (moles of solute / liter of solution) bench top laboratory applications titration

6/30/2015 Applications Analytical applications Parts per Million (ppm) trace analysis, impurities Parts per Billion (ppb) environmental Parts per Trillion (ppt)

6/30/2015 Preparation of Solutions From pure compounds By the mixture of a solid and a solution By the mixture of two solutions Note: masses are always additive volumes involving concentrated solutions may not be additive

6/30/2015 Solution Preparation Prepare 500. mL of 3.0 M sulfuric acid from sulfur trioxide gas and water. Outline the process Solve the problem H2OH2OSO 3

6/30/2015 Prepare 500. mL of 3.0 M sulfuric acid from sulfur trioxide gas and water. Write a balanced equation Define terms What is molarity? Moles of H 2 SO 4 needed Moles of SO 3 needed Mass of SO 3 needed Procedure needed to make solution Remember volumes may change when materials are mixed together

Prepare 500. mL of 3.0 M sulfuric acid from sulfur trioxide gas and water. Dissolve 120. grams of sulfur trioxide in enough water to give 500. mL of solution. H2OH2OSO 3

6/30/2015 Solution Preparation Prepare 275 mL of 0.450 M aqueous sulfuric acid solution from 18.0 M sulfuric acid and water. Outline the process Solve the problem H2OH2OH 2 SO 4

6/30/2015 Prepare 275 mL of 0.450 M aqueous sulfuric acid solution from 18.0 M sulfuric acid and water. Write a balanced equation Define terms What is molarity? Moles of sulfuric acid in product Process needed to make solution Remember volumes may change when materials are mixed together Moles of sulfuric acid in reactant

6/30/2015 Prepare 275 mL of 0.450 M aqueous sulfuric acid solution from 18.0 M sulfuric acid and water. Dissolve 6.87 mL of 18.0 M sulfuric acid in some water and then dilute to 275 mL of total solution volume. H2OH2OH 2 SO 4

6/30/2015 Solution Preparation Prepare 400. mL of 2.50 M sulfuric acid from 1.00 M sulfuric acid and 6.00 M sulfuric acid. Assume volumes are additive. Outline the process Solve the problem H 2 SO 4

6/30/2015 Prepare 400. mL of 2.50 M sulfuric acid from 1.00 M sulfuric acid and 6.00 M sulfuric acid. Write a balanced equation Define terms What is molarity? Moles of sulfuric acid in product Moles of sulfuric acid in reactants Volumes of reactants needed Process needed to make solution Volumes assumed to be additive Need this assumption or more information

6/30/2015 Prepare 400. mL of 2.50 M sulfuric acid from 1.00 M sulfuric acid and 6.00 M sulfuric acid. Write a balanced equation Define terms What is molarity? Moles of sulfuric acid in product Moles of sulfuric acid in reactants Volumes of reactants needed Volumes assumed to be additive Need this assumption or more information What information would be needed if this isn’t assumed? Volumes of reactants needed

Prepare 400. mL of 2.50 M sulfuric acid from 1.00 M sulfuric acid and 6.00 M sulfuric acid. H 2 SO 4 x mL of 1.00 M H 2 SO 4 y mL of 6.00 M H 2 SO 4 x + y = 400 mL (x · 1.00 M) + (y · 6.00 M) = 400 mL · 2.50 M y = 120. mL

6/30/2015 Neutralization Balanced Equation Moles react with moles moles of acid = moles of base M acid x V acid M base x V base

6/30/2015

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