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Standard 5.1

Standard 5.1. Solutions. Concentration. All solutions consist of two parts. The solute refers to the substance being dissolved and the solvent refers to the substance the solute is dissolved in (almost always water). Concentration.

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Standard 5.1

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  1. Standard 5.1 Solutions

  2. Concentration • All solutions consist of two parts. The solute refers to the substance being dissolved and the solvent refers to the substance the solute is dissolved in (almost always water).

  3. Concentration • All solutions consist of two parts. The solute refers to the substance being dissolved and the solvent refers to the substance the solute is dissolved in (almost always water). • Several units can be used to describe the concentration of a solution. We will use molarity most often. The molarity of a solution represents the moles of solute per liter of solution and can be represented by the equation: Molarity(M) = moles of solute (mol) / Liters of solution (L)

  4. Concentration • A 0.75 M solution of sodium chloride contains 0.75 moles of NaCl for every liter of solution and is referred to as a 0.75 molar solution.

  5. Concentration • A 0.75 M solution of sodium chloride contains 0.75 moles of NaCl for every liter of solution and is referred to as a 0.75 molar solution. • Typical molarity problems will give you 2 measurements (out of molarity, moles, or liters) and ask you to solve for the 3rd.

  6. Some things to note: • Grams are not a part of the molarity equation. If a gram amount is given, it must first be converted to moles before you can plug it into the equation.

  7. Some things to note: • Grams are not a part of the molarity equation. If a gram amount is given, it must first be converted to moles before you can plug it into the equation. • Volumes must be in liters. Any amounts in mL must be divided by 1000 before using it in the equation (Example: 15.5 mL is equal to 0.0155 L.)

  8. Some things to note: • Grams are not a part of the molarity equation. If a gram amount is given, it must first be converted to moles before you can plug it into the equation. • Volumes must be in liters. Any amounts in mL must be divided by 1000 before using it in the equation (Example: 15.5 mL is equal to 0.0155 L.) • To solve the equation more easily, you can make a fraction out of the molarity by adding a denominator of 1. Then you can cross multiply and divide.

  9. Determine the molarity of a solution containing 20.0 g of Cu(NO3)2 in a volume of 350 mL.

  10. Determine the molarity of a solution containing 20.0 g of Cu(NO3)2 in a volume of 350 mL. • convert grams to moles 20.0 g / (63.55 + 28.01 + 96.00) = 0.107 mol

  11. Determine the molarity of a solution containing 20.0 g of Cu(NO3)2 in a volume of 350 mL. • convert grams to moles 20.0 g / (63.55 + 28.01 + 96.00) = 0.107 mol • convert mL to L 350 mL = 0.350 L

  12. Determine the molarity of a solution containing 20.0 g of Cu(NO3)2 in a volume of 350 mL. • convert grams to moles 20.0 g / (63.55 + 28.01 + 96.00) = 0.107 mol • convert mL to L 350 mL = 0.350 L • solve for M M = 0.107 mol / 0.350 L = 0.305 M Cu(NO3)2

  13. How many mL of a 1.50 M NaCl solution can be made by dissolving 30.0 grams of NaCl in water?

  14. How many mL of a 1.50 M NaCl solution can be made by dissolving 30.0 grams of NaCl in water? • convert grams to moles 30.0 g / (22.99 + 35.45) = 0.513 moles

  15. How many mL of a 1.50 M NaCl solution can be made by dissolving 30.0 grams of NaCl in water? • convert grams to moles 30.0 g / (22.99 + 35.45) = 0.513 moles • solve for L 1.50 M / 1 = 0.513 mol / L L = 0.342 L

  16. How many mL of a 1.50 M NaCl solution can be made by dissolving 30.0 grams of NaCl in water? • convert grams to moles 30.0 g / (22.99 + 35.45) = 0.513 moles • solve for L 1.50 M / 1 = 0.513 mol / L L = 0.342 L • convert L to mL 0.342 L = 342 mL

  17. Concentration (continued) • Commercial products often use percent by mass as a measure of concentration. Percent by mass can be calculated by dividing the grams of solute by the grams of the entire solution.

  18. Concentration (continued) • Commercial products often use percent by mass as a measure of concentration. Percent by mass can be calculated by dividing the grams of solute by the grams of the entire solution. % by mass = (mass of solute (g) / mass of solution (g)) x 100

  19. Concentration (continued) • Commercial products often use percent by mass as a measure of concentration. Percent by mass can be calculated by dividing the grams of solute by the grams of the entire solution. % by mass = (mass of solute (g) / mass of solution (g)) x 100 • The mass of the solution is equal to the mass of the solute plus the mass of the solvent.

  20. Concentration (continued) • Commercial products often use percent by mass as a measure of concentration. Percent by mass can be calculated by dividing the grams of solute by the grams of the entire solution. % by mass = (mass of solute (g) / mass of solution (g)) x 100 • The mass of the solution is equal to the mass of the solute plus the mass of the solvent. • If the volume of water (solvent) is given in mL, the mass in grams will have the same value since 1 mL of water has a mass of exactly 1 gram.

  21. Determine the percent by mass of a solution made by dissolving 50.0 grams of solid KBr in 250 mL of water.

  22. Determine the percent by mass of a solution made by dissolving 50.0 grams of solid KBr in 250 mL of water. • The solute is KBr, the solution is the KBr and water together.

  23. Determine the percent by mass of a solution made by dissolving 50.0 grams of solid KBr in 250 mL of water. • The solute is KBr, the solution is the KBr and water together. % by mass = (50.0 g / (50.0 + 250)) x 100 = 16.7%

  24. Determine the percent by mass of a solution made by dissolving 50.0 grams of solid KBr in 250 mL of water. • The solute is KBr, the solution is the KBr and water together. % by mass = (50.0 g / (50.0 + 250)) x 100 = 16.7% Complete problems on p.481 and 488 along with the first 2 worksheets (Solution Vocab. and Solution Problems).

  25. Titrations • A titration is an experiment involving two solutions (one has to have a concentration that is known) in order to find the exact concentration of a solution in which the concentration is unknown.

  26. Titrations • A titration is an experiment involving two solutions (one has to have a concentration that is known) in order to find the exact concentration of a solution in which the concentration is unknown. • The solution in which the concentration is known (called thetitrant) is carefully dispensed from a buret into the solution of unknown concentration (called the titrate) until a specially selected indicator causes the solution to change colors.

  27. Solving titration problems

  28. Solving titration problems • Determine the moles of titrant using the molarity equation.

  29. Solving titration problems • Determine the moles of titrant using the molarity equation. • Determine the moles of titrate in the sample by multiplying by B / A from the balanced equation.

  30. Solving titration problems • Determine the moles of titrant using the molarity equation. • Determine the moles of titrate in the sample by multiplying by B / A from the balanced equation. • Determine the concentration of titrate using the molarity equation.

  31. A solution of Ca(OH)2 with an unknown concentration and a volume of 15.0 mL is titrated with a 0.500 M solution of HCl according to the reaction below.Ca(OH)2 + 2 HCl 2 H2O + CaCl2Determine the molarity of the Ca(OH)2 if 11.78 mL of HCl are required to reach the endpoint.

  32. A solution of Ca(OH)2 with an unknown concentration and a volume of 15.0 mL is titrated with a 0.500 M solution of HCl according to the reaction below.Ca(OH)2 + 2 HCl 2 H2O + CaCl2Determine the molarity of the Ca(OH)2 if 11.78 mL of HCl are required to reach the endpoint. • 0.500 M = mol / 0.01178 L mol = .00589 mol HCl

  33. A solution of Ca(OH)2 with an unknown concentration and a volume of 15.0 mL is titrated with a 0.500 M solution of HCl according to the reaction below.Ca(OH)2 + 2 HCl 2 H2O + CaCl2Determine the molarity of the Ca(OH)2 if 11.78 mL of HCl are required to reach the endpoint. • 0.500 M = mol / 0.01178 L mol = .00589 mol HCl 2. .00589 x ½ = .002945 mol Ca(OH)2

  34. A solution of Ca(OH)2 with an unknown concentration and a volume of 15.0 mL is titrated with a 0.500 M solution of HCl according to the reaction below.Ca(OH)2 + 2 HCl 2 H2O + CaCl2Determine the molarity of the Ca(OH)2 if 11.78 mL of HCl are required to reach the endpoint. • 0.500 M = mol / 0.01178 L mol = .00589 mol HCl • .00589 x ½ = .002945 mol Ca(OH)2 • M = .002945 mol / .0150 L = .196 M Ca(OH)2

  35. A solution of Ca(OH)2 with an unknown concentration and a volume of 15.0 mL is titrated with a 0.500 M solution of HCl according to the reaction below.Ca(OH)2 + 2 HCl 2 H2O + CaCl2Determine the molarity of the Ca(OH)2 if 11.78 mL of HCl are required to reach the endpoint. • 0.500 M = mol / 0.01178 L mol = .00589 mol HCl • .00589 x ½ = .002945 mol Ca(OH)2 • M = .002945 mol / .0150 L = .196 M Ca(OH)2 Complete the titration lab and titration worksheet.

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