Solutions Chapter 12. Vocabulary Solution: a homogeneous mixture of two or more substances in a...
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Transcript of Solutions Chapter 12. Vocabulary Solution: a homogeneous mixture of two or more substances in a...
![Page 1: Solutions Chapter 12. Vocabulary Solution: a homogeneous mixture of two or more substances in a single phase Solvent: the dissolving medium in a solution.](https://reader036.fdocuments.net/reader036/viewer/2022062407/56649d825503460f94a67f78/html5/thumbnails/1.jpg)
Solutions
Chapter 12
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Vocabulary
• Solution: a homogeneous mixture of two or more substances in a single phase
• Solvent: the dissolving medium in a solution
• Solute: the substance dissolved in a solution
• Solution can be liquid or solid or gaseous
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liquid-liquid
solid-liquid
solid-solid
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gas-liquid
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Factors Affecting the Rate of Dissolution
• Surface area: greater surface area speeds up the rate
• Agitation: shaking or stirring speeds up the rate
• Temperature: higher temperature usually favor a higher rate, except in the case of most gases, then a lower temperature speeds up the rate
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Like Dissolves Like
• Polar solutes dissolve in polar solvents• Nonpolar solutes dissolve in nonpolar
solvents• Molecules can have a polar end and
nonpolar end (especially if they are large)• This is why soap will dissolve grease and
then be washed away in water
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Concentration of Solutions
• Concentration: a measure of the amount of solute in a given amount of solvent or solution
• Molarity: the number of moles of solute in one liter of solution
• Molarity (M) = amount of solute (mol) volume of solution (L)
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Molarity
• One mole of NaOH has a mass of 39.998 g.• If this mass is dissolved in enough water to
make exactly 1.00 L of solution, the solution is a 1 M solution.
• If 20.0 g of NaOH is dissolved in enough water to make 1.00 L of solution, what is the molarity? 0.500 M NaOH
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• Notice that 1 mol of solute added to 1 L of solvent does not usually make 1 M solution
• Adding the solute to 1 L will cause the volume to increase
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The preparation of a 0.0128 M solution of CuSO4 5H2O
Start by calculating the mass needed. Convert the mol to mass by multiplying by the molar mass. This mass is 3.20 g.
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The resulting solution has. 0.0128 mol of solute dissolved in 1.000 L of solution, which is a 0.0128 M concentration
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Problem
• You have 3.50 L of solution that contains 90.0 g of NaCl. What is the molarity of the solution?
• Given : solute mass of 90.0 g NaCl and solution volume of 3.50 L
• First change g to mol• 90.0 g NaCl = 1.54 mol NaCl
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Continued
• Divide mol of solute by liters of solution• 1.54 mol NaCl 3.50 L solution• 0.440 M NaCl
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Problem
• You have 0.8 L of a 0.5 M HCl solution. How many moles of HCl does this solution contain?
• M = mol/L so mol = M L• 0.5 M 0.8 L = 0.4 mol HCl• How many grams is this? (use 3 sig figs)• 14.6 g HCl
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Problem
• If you have 61.9 mL of a 2M solution of HCl, how many moles do you have?
• Convert 61.9 mL to liters by dividing by 1000.
• 0.0619 L• mol = M L• mol = 0.1238 = 0.124 mol HCl
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Dilutions
• Concentrated solutions are purchased in standard molarities called stock solutions
• You can prepare a less concentrated solution by diluting the stock solution with solvent (usually water)
• The number of moles of solute does not change when you dilute
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Making Dilutions
• The molarity times the volume of the stock solution equals the molarity times the volume of the dilution
• M1V1 = M2V2
• Where M1V1 is the molarity and volume of the stock solution and M2V2 is the molarity and volume of the dilution
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Problem
• What volume, in mL of 2.00 M calcium chloride stock solution would you use to make 0.50 L of 0.300 M calcium chloride solution?
• Write down what you are given• M1 = 2.00 M CaCl2 V1 = ?
• M2 = 0.300 M V2 = 0.50 L
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Continued• (2.00 M)(x) = (0.300)(0.50)• 2.00x = 0.15• x = 0.075 L• Since the answer has to be in mL:• (0.075)(1000)• 75 mL of the stock solution must be
measured and then water added to make 0.50 L
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Problems• What volume of a 3.00 M KI stock
solution would you use to make 0.300 L of a 1.25 M KI solution?
• How many mL of a 5.0 M H2SO4 stock solution would you need to prepare 100.0 mL of a 0.25 M H2SO4?
• If you dilute 20.0 mL of a 3.5 M solution to make 100.0 mL of solution, what is the molarity of the dilute solution?
0.125 L or 125 mL
5.0 mL
0.70 M