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NOTES: 16.1-16.2 – Solutions and Concentration

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Solutions – a REVIEW: ● SOLUTION – a homogeneous mixture of pure substances ● the SOLVENT is the medium in which the SOLUTES are dissolved. (The solvent is usually the most abundant substance.) ● Example: Solution: Salt Water Solute: Salt Solvent: Water

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The process of dissolution is favored by: ● A decrease in the energy of the system (exothermic) ● An increase in the disorder of the system (entropy)

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Liquids Dissolving in Liquids ● Liquids that are soluble in one another (“mix”) are MISCIBLE. -“LIKE dissolves LIKE” ● POLAR liquids are generally soluble in other POLAR liquids. ● NONPOLAR liquids are generally soluble in other NONPOLAR liquids.

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Liquids Dissolving in Liquids ● Liquids that are insoluble in one another (do not mix) are IMMISCIBLE. ● Example: oil and water

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Factors affecting RATE of dissolution: ● Surface area / particle size Greater surface area, faster it dissolves ● Temperature Most solids dissolve faster @ higher temps ● Agitation Stirring / shaking will speed up rate of dissolution

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SATURATION: ● Unsaturated solution – is able to dissolve more solute ● Saturated solution – has dissolved the maximum amount of solute ● Supersaturated solution – has dissolved excess solute (at a higher temperature); (solid crystals generally form when this solution is cooled)

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SOLUBILITY ● SOLUBILITY = the AMOUNT of solute that will dissolve in a given amount of solvent **Key difference between this and the RATE of dissolving!!

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Factors Affecting SOLUBILITY: ● The nature of the solute and solvent: different substances have different solubilities ● Temperature: many solids substances become more soluble as the temp of a solvent increases; however, gases are less soluble in liquids at higher temps. ● Pressure: Only affects the solubility of gases. As pressure increases, the solubility of gases increases.

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SOLUBILITY CURVES: ● shows how the solubility of a particular substance in a particular solvent changes as temperature changes.

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HENRY’S LAW: ● at a given temperature the solubility (S) of a gas is directly proportional to the pressure (P) or S 1 /P 1 = S 2 /P 2

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Henry’s Law EXAMPLE: If the solubility of a gas in water is 0.77 g/L at 3.5 atm of pressure, what is its solubility (in g/L) at 1.0 atm of pressure? (the temp. is held constant at 25°C)

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Henry’s Law EXAMPLE: If the solubility of a gas in water is 0.77 g/L at 3.5 atm of pressure, what is its solubility (in g/L) at 1.0 atm of pressure? (the temp. is held constant at 25°C) S 1 /P 1 = S 2 /P 2 0.77 g/L=S 2 3.5 atm1.0 atm S 2 = 0.22 g/L

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Concentration of Solution ● CONCENTRATION refers to the amount of solute dissolved in a solution. ● a DILUTE solution is one that contains a low concentration of solute ● a CONCENTRATED solution contains a high concentration of solute

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MOLARITY

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MOLARITY example #1: What is the molarity of a 2.0 L solution containing 4.0 mol NaCl?

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MOLARITY example #1: What is the molarity of a 2.0 L solution containing 4.0 mol NaCl? MOLARITY = mol solute / L solution = 4.0 mol NaCl / 2.0 L = 2.0 mol/L or 2.0 M NaCl

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MOLARITY example #2: A salt solution contains 0.90 g of NaCl in 100.0 mL of solution. What is the molarity of the solution?

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MOLARITY example #2: A salt solution contains 0.90 g of NaCl in 100.0 mL of solution. What is the molarity of the solution? mol NaCl = 0.90 g x 1 mol = 0.0154 mol 58.5 g

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MOLARITY example #2: A salt solution contains 0.90 g of NaCl in 100.0 mL of solution. What is the molarity of the solution? MOLARITY = 0.0154 mol / 0.100 L = 0.154 mol / L or 0.154 M NaCl

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MOLARITY example #3: How many moles of solute are present in 1.5 L of 0.24 M Na 2 SO 4 ?

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MOLARITY example #3: How many moles of solute are present in 1.5 L of 0.24 M Na 2 SO 4 ? 1.5 L x 0.24 mol= 0.36 mol Na 2 SO 4 1 L

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MOLARITY BY DILUTION ● When you dilute a solution, you can use this equation:

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DILUTION example #1: How many liters of 5.0 M CuSO 4 would be needed to prepare 0.1 L of 0.5 M CuSO 4 ?

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DILUTION example #1: How many liters of 5.0 M CuSO 4 would be needed to prepare 0.1 L of 0.5 M CuSO 4 ? (5.0 M) (V 1 ) = (0.5 M) (0.1 L) V 1 =0.01 L or 10. mL

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DILUTION example #2: How many milliliters of a stock solution of 2.00 M MgSO 4 would you need to prepare 100.0 mL of 0.400 M MgSO 4 ?

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DILUTION example #2: How many milliliters of a stock solution of 2.00 M MgSO 4 would you need to prepare 100.0 mL of 0.400 M MgSO 4 ? (2.00 M) (V 1 ) = (0.400 M) (100.0 mL) V 1 =20.0 mL

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Example #3: Describe how you would prepare 2.50 L of 0.665 M Na 2 SO 4 solution starting with: a) solid Na 2 SO 4 b) 5.00 M Na 2 SO 4 Dissolve 236 g of Na 2 SO 4 in enough water to create 2.50 L of solution.

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Example #3: Describe how you would prepare 2.50 L of 0.665 M Na 2 SO 4 solution starting with: a) solid Na 2 SO 4 b) 5.00 M Na 2 SO 4 Add 0.333 L of Na 2 SO 4 to 2.17 L of water.

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PERCENT SOLUTIONS: ● If both the solute and the solvent are liquids, the concentration of the solute can be expressed as a PERCENT BY VOLUME: % (v/v) = volume of solute x 100% solution volume

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Percent by Vol. example #1 What is the percent volume of acetone in water if 25 mL of acetone is added to 75 mL of water?

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Percent by Vol. example #1 What is the percent volume of acetone in water if 25 mL of acetone is added to 75 mL of water? % v/v=solute vol. / solution vol. =25 mL / (75 + 25 mL) =25 mL / 100 mL =0.25 x 100% =25% acetone (v/v)

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Percent by Vol. example #2: What is the percent by volume of ethanol (C 2 H 5 OH) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water?

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Percent by Vol. example #2: What is the percent by volume of ethanol (C 2 H 5 OH) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water? % (v/v) =85 mL / 250 mL =34% ethanol (v/v)

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MASS PERCENT: ● If a solid solute is dissolved into a liquid, the concentration of the solute can be expressed as a PERCENT BY MASS: % (m/v) = mass of solute (g) x 100% solution volume (mL)

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Mass % example #1: What is the % (m/v) of a 100 mL aqueous solution containing 10 g of NaCl?

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Mass % example #1: What is the % (m/v) of a 100 mL aqueous solution containing 10 g of NaCl? % (m/v)=mass solute / vol. solution =10 g NaCl / 100 mL =0.10 x 100% =10% NaCl (m/v)

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Mass % example #2: What is the percent of NaCl in a solution made by dissolving 24 g of NaCl in water to a final volume of 174 mL?

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Mass % example #2: What is the percent of NaCl in a solution made by dissolving 24 g of NaCl in water to a final volume of 174 mL? % (m/v)=mass solute / vol. solution =24 g NaCl / 174 mL =0.138 x 100% =13.8% NaCl (m/v)

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