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Solutions Notes

Solutions Notes. Words to Know. Solution – homogenous mixture Solvent – substance present in the largest amount Solutes – substance present in the smallest amount Aqueous solution – solutions with water as the solvent Concentration – the amount of solute in a given volume of solution

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Solutions Notes

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  1. Solutions Notes

  2. Words to Know • Solution – homogenous mixture • Solvent – substance present in the largest amount • Solutes – substance present in the smallest amount • Aqueous solution – solutions with water as the solvent • Concentration – the amount of solute in a given volume of solution • Concentrated – large amount of solute dissolved in solvent • Dilute – small amount of solute dissolved in solvent

  3. Saturated – a solution that contains as much solute as will dissolve at that temperature • Unsaturated – a solution that hasn’t reached that limit of solute that will dissolve • Supersaturated - a solution that contains more solute than should dissolve at that temperature

  4. Effect of Temperature on Solubility • Increasing the temperature of a solution, increases the amount of solute that can be dissolved • Decreasing the temperature of a solution, causes the solute to recrystallize

  5. Effect of Pressure on Solubility • Pressure has a major effect on the solubility of gas-liquid systems • An increase in pressure increases the solubility of a gas in the liquid

  6. “Like dissolves like” – a solvent usually dissolves solutes that have polarities similar to itself Generally, “like dissolves like.” Polar molecules dissolve other polar molecules and ionic compounds. Nonpolar molecules dissolve other nonpolar molecules. Alcohols, which have characteristics of both, tend to dissolve in both types of solvents, but will not dissolve ionic solids.

  7. Colligative properties Colligative properties - the physical changes that result from adding solute to a solvent. Colligative Properties depend on how many solute particles are present as well as the solvent amount, but they do NOT depend on the type of solute particles. • Boiling Point Elevation • Freezing Point Depression • Osmotic Pressure • Vapor Pressure Lowering

  8. Mass % = 1.00 g x 100 % 101.0 g Solution Composition - Mass Percent Mass percent – describes a solution’s composition expresses the mass of solute present in a given mass of solution Mass Percent = mass of solute x 100% mass of solution* * mass of solution = mass of solute + mass of solvent Example – A solution is prepared by mixing 1.00g of C2H5OH, with 100.0g of H2O. Calculate the mass percent of ethanol. Mass Percent = mass of solute x 100% mass of solution Given mass of solute = 1.00 g mass of solution = 100.0 g + 1.00 g = 101.0 g Mass % = 0.990 %

  9. Solution Composition – Molarity Molarity – measure of concentration - number of moles of solute per volume of solution in liters Molarity = moles of solute = mol = M L of solution L Example – Calculate the molarity of a solution prepared by dissolving 11.5 g NaOH in enough water to make 1.50L solution. 1 1 0.192 M mol NaOH 11.5 g NaOH x ___________ x __________ = 40.0 g NaOH 1.50 L NaOH

  10. 1.50 L 0.100 M Ex: Calculate the mass of solid AgCl formed when 1.50L of a 0.100M AgNO3 solution is reacted with excess NaCl. NaCl + AgNO3 AgCl + NaNO3 ? g 0.100 mol AgNO3 143.2 1 mol AgCl g AgCl 1.50 L AgNO3 x ______________ x ____________ = x ____________ 21.5 g L AgNO3 1 1 mol AgNO3 1 mol AgCl

  11. Example – How many moles of Ag+ ions are present in 25mL of a 0.75M Ag2SO4 solution? Ag2SO4 2 Ag+1 + SO4-2 1 L Ag2SO4 0.75 mol Ag2SO4 2 mol Ag+1 25 mL Ag2SO4 x _____________ x ______________ x ______________ = 1000 1 L Ag2SO4 1 mol Ag2SO4 mL Ag2SO4 0.038 mol Ag+1

  12. Standard Solution • Standard Solution – a solution whose concentration is accurately known Example – A chemist needs 1.0 L of a 0.200M K2Cr2O7 solution. How much solid K2Cr2O7 must be weighed out to make this solution? 1.0 L K2Cr2O7 0.200 mol K2Cr2O7 294.2 g K2Cr2O7 x ________________ x ______________ = 1 L K2Cr2O7 1 mol K2Cr2O7 59 g K2Cr2O7

  13. _____ V1 = M2V2 V1 = (0.10 M)(1.5 L) ____________ M1 16 M Dilution Dilution – process of adding more solvent to a solution Moles of solute before dilution = Moles of solute after dilution M1V1 = M2V2 Example: What volume of 16M H2SO4 must be used to prepare 1.5L of a 0.10M H2SO4 solution? Given V1 = ? M1 = 16 M V2 = 1.5 L M2 = 0.10 M V1 = 0.0094 L

  14. V2 = (1.00 M)(500.0 mL) _______________ 17.5 M _____ V2 = M1V1 M2 Example: Prepare 500.0mL of 1.00 M HC2H3O2 from a 17.5 M stock solution. What volume of the stock solution is required? Given V1 = 500.0 mL M1 = 1.00 M M2 = 17.5 M V2 = ? V2 = 28.6 mL

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