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Concentrations of Solutions

Concentrations of Solutions. The properties and behavior of solutions often depend not only on the type of solute but also on the concentration of the solute. Concentration : the amount of solute dissolved in a given quantity of solvent or solution many different concentration units

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Concentrations of Solutions

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  1. Concentrations of Solutions • The properties and behavior of solutions often depend not only on the type of solute but also on the concentrationof the solute. • Concentration:the amount of solute dissolved in a given quantity of solvent or solution • many different concentration units • (%, ppm, g/L, etc) • often expressed as molarity

  2. Concentrations of Solutions • Molarity (M):the number of moles of solute per liter of solution • Molarity = moles of solute volume of solution in liters • 6.0 M HCl • 6.0 moles of HCl per liter of solution. • 9.0 M HCl • 9.0 moles of HCl per liter of solution.

  3. Concentration of Solutions • To calculate the molarity of a solution: • Divide amount of solute by volume of solution • convert amount of solute to moles • convert volume to liters • Units of molarity are mol/L (M)

  4. M = 225 g glucose x 1000 mL x 1 mol 825 mL soln 1 L 180.2 g M = 1.51 M glucose Concentration of Solutions Example: What is the molarity of a solution prepared by dissolving 225 g of glucose (C6H12O6) in enough water to make 825 mL of solution?

  5. Concentration of SolutionsInterconverting Molarity, Moles, and Volume • Molarity can be used as a conversion factor. • The definition of molarity contains 3 quantities: Molarity = moles of solute volume of solution in liters • If you know two of these quantities, you can find the third.

  6. Use molarity as a conversion factor Concentration of Solutions Interconverting Molarity, Moles, and Volume Example: How many moles of HCl are present in 2.5 L of 0.10 M HCl? Given: 2.5 L of soln 0.10M HCl Find:mol HCl Mol HCl = 2.5 L soln x 0.10 mol HCl 1 L of soln = 0.25 mol HCl

  7. Use M as a conversion factor Concentration of Solutions Interconverting Molarity, Moles, and Volume Example: What volume of a 0.10 M NaOH solution is needed to provide 0.50 mol of NaOH? Given: 0.50 mol NaOH 0.10 M NaOH Find: vol soln Vol soln = 0.50 mol NaOH x 1 L soln 0.10 mol NaOH = 5.0 L solution

  8. Concentration of Solutions Interconverting Molarity, Moles, and Volume Example: How many grams of NaNO3 are needed to prepare 250.0 mL of 1.00 M NaNO3? Given: 250.0 mL soln 1.00 M NaNO3 Find: g NaNO3 Conversion factors: Molarity, molar mass Strategy: mL  L  mol  grams

  9. Concentration of Solutions Interconverting Molarity, Moles, and Volume g NaNO3 = 250.0 mL soln x 1 L x 1.00 mol 1000 mL 1 L soln x 85.0 g NaNO3 1 mol = 21.25 g NaNO3 = 21.3 g NaNO3

  10. Concentration of Solutions Dilution Calculations • Many laboratory chemicals such as acids are purchased as concentrated solutions (stock solutions). • 12 M HCl • 12 M H2SO4 • More dilute solutions are prepared by taking a certain quantity of the stock solution and diluting it with water.

  11. Concentration of Solutions Dilution Calculations • A given volume of a stock solution contains a specific number of moles of solute. • 25 mL of 6.0 M HCl contains 0.15 mol HCl • If 25 mL of 6.0 M HCl is diluted with 25 mL of water, the number of moles of HCl present does not change. • Still contains 0.15 mol HCl

  12. Concentration of Solutions Dilution Calculations Moles solute = moles solute before dilution after dilution • Although the number of moles of solute does not change, the volume of solution does change. • The concentration of the solution will change since Molarity = mol solute Vol soln

  13. Concentration of Solutions Dilution Calculations • When a solution is diluted, the concentration of the new solution can be found using: Mi x Vi = Mf x Vf where Mi = initial concentration (mol/L) Vi = initial volume Mf = final concentration (mol/L) Vf = final volume

  14. Concentration of Solutions Dilution Calculations Example: What is the concentration of a solution prepared by diluting 25.0 mL of 6.00 M HCl to a total volume of 50.0 mL? Given: Vi = 25.0 mL Mi = 6.00 M Vf = 50.0 mL Find: Mf Vi x Mi = Vf x Mf

  15. Concentration of Solutions Dilution Calculations Vi x Mi = Vf x Mf 25.0 mL x 6.00 M = 50.0 mL x Mf Mf = 25.0 mL x 6.00 M = 3.00 M 50.0 mL Note:Vi and Vf do not have to be in liters, but they must be in the same units.

  16. Concentration of Solutions Dilution Calculations Example: What volume of 0.50 M NaNO3 would be needed to prepare 100.0 mL of 0.10 M NaNO3 by dilution? Given: Mi = 0.50 M Vf = 100.0 mL Mf = 0.10 M Find: Vi Vi x Mi = Vf x Mf

  17. Concentration of Solutions Dilution Calculations Vi x Mi = Vf x Mf Vi x 0.50 M = 100.0 mL x 0.10 M Vi = 100.0 mL x 0.10 M = 2.0 x 101 mL 0.50 M

  18. Concentration of Solutions Solution Stoichiometry • Recall that reactions occur on a mole to mole basis. • For pure reactants, we measure reactants usingmass • For reactants that are added to a reaction as aqueous solutions, we measure the reactants using volume of solution.

  19. Concentration of Solutions Solution Stoichiometry grams A Molar mass moles A Molarity A Vol Soln A Molar ratio Molar mass grams B moles B Molarity B Vol Soln B

  20. Concentration of Solutions Solution Stoichiometry Example: If 25.0 mL of 2.5 M NaOH are needed to neutralize (i.e. react completely with) a solution of H3PO4, how many moles of H3PO4 were present in the solution? Given: 25.0 mL 2.5 M NaOH Find: moles of H3PO4 3NaOH (aq) + H3PO4 (aq)  Na3PO4 (aq) + 3H2O(l)

  21. Concentration of Solutions Solution Stoichiometry Molarity A moles A Vol Soln A Molar ratio Molarity B moles B Vol Soln B

  22. Concentration of Solutions Solution Stoichiometry Molarity NaOH moles NaOH Vol NaOH Soln Molar ratio moles H3PO4

  23. Concentration of Solutions Solution Stoichiometry Strategy: mL NaOH soln L NaOH soln mol H3PO4 mol NaOH

  24. Concentration of Solutions Solution Stoichiometry Mol H3PO4 = 25.0 mL x 1 L 1000 mL x 2.50 mol NaOH x 1 mol H3PO4 1 L 3 mol NaOH = 0.0208 mol H3PO4

  25. Concentration of Solutions Solution Stoichiometry • Solution stoichiometry can be used to determine the concentration of aqueous solutions used in reactions. • Concentration of an acid can be determined using a process called titration. • reacting a known volume of the acid with a known volume of a standard base solution (i.e. a base whose concentration is known)

  26. Concentration of Solutions Solution Stoichiometry Example: If 35.50 mL of 2.5 M NaOH are needed to neutralize 50.0 mL of an H3PO4 solution, what is the concentration (molarity) of the H3PO4 solution? Given: 35.50 mL 2.5 M NaOH 50.0 mL of H3PO4 sol’n Find: molarity (mol/L) H3PO4 3NaOH (aq) + H3PO4 (aq)  Na3PO4 (aq) + 3H2O(l)

  27. Concentration of Solutions Solution Stoichiometry Strategy: M = moles L • To find the concentration of H3PO4 soln, we need both # moles and volume of H3PO4. • Since volume is given, we can simply find moles and plug into the equation for M.

  28. Concentration of Solutions Solution Stoichiometry Molarity NaOH moles NaOH Vol NaOH Soln Molar ratio moles H3PO4

  29. Concentration of Solutions Solution Stoichiometry Mol H3PO4 = 35.5 mL x 1 L 1000 mL x 2.50 mol NaOH x 1 mol H3PO4 1 L 3 mol NaOH = 0.0296 mol H3PO4 We’re not done….we need molarity.

  30. Concentration of Solutions Solution Stoichiometry Molarity = moles L = 0.0296 mol H3PO4 x 1000 mL 50.0 mL L = 0.592 M H3PO4

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