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Unit X - Solutions
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Chapter Goals 1. Understand the process of dissolving and why certain substances dissolve in water. 2. Understand the qualitative descriptions of the concentration of a solution. 3. Understand the factors that affect the rate at which a solid dissolves. 4. Understand and use the concentration terms: mass percent and molarity. 5. Learn to calculate the concentration of a solution made by dilution.
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6. Understand the strategy for solving problems in solution reactions. 7. Understand the effect of a solute on solution properties.
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Introduction Most of the important chemistry that keeps plants, animals, and humans functioning occurs in aqueous solutions. Examples: A solution is a homogeneous mixture Can be a solid, liquid, or a gas The substance present in the largest amount is the solvent Doing the dissolving All other substances are the solutes. The substance being dissolved
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Aqueous solutions are solutions with water as the solvent.
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Solubility Objectives: To understand the process of dissolving & To learn why certain substances dissolve in water. Question: What causes substances to dissolve?
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Remember – Like dissolves like Substances with similar polarities dissolve Example: H 2 O (H +1 ; O -2 ) & NaCl (Na +1 ;Cl -1 ) Both are polar, thus one dissolves in another Substances with different polarities do not have the ability to dissolve Example: H 2 O & Oil
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In order for a substance to dissolve in water, a “hole” must be made in the water structure
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Solution Composition Objective: To learn the qualitative terms associated with the concentration of a solution. Saturated: When a solution contains as much solute as will dissolve at that temperature Unsaturated: A solution in which more solute can be can be dissolved than is dissolved already Supersaturated: A solution contains more dissolved solid than a saturated solution
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A solution is a mixture The amounts of substances can vary Concentrated: Where a relatively large amount of solute is dissolved Dilute: A relatively small amount is solute is dissolved Examples and Applications…
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Factors Affecting the Rate of Dissolution Objective: To understand the factors that affect the rate at which a solute dissolves. There are 3 factors that affect the rate of dissolution: 1. Surface Area 2. Stirring 3. Temperature Examples and Applications…
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Solution Composition: Mass Percent Objective: To understand the concentration term mass percent and learn how to calculate it. Water is the “Universal Solvent” Similar to percent composition & forming a Pie Graph
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Example 1: A solution is prepared by mixing 2.50 g of CaCl 2 with 50.0 g of water. Calculate the mass % of CaCl 2 in the solution. Example 2: A Concentrated HCl solution contains 37.2% by mass HCl. What mass of HCl is contained in 35.5 g of concentrated HCl? Example 3: What mass of H 2 O must be added to 425 g of formaldehyde to prepare a 40.0% by mass solution?
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Solution Composition: Molarity Objectives: To understand molarity & To use molarity to calculate the number of moles of solute present. Sometimes it is more convenient to measure the volume of a solution, rather than it mass Chemists use the term concentration to describe this The amount of solute in a given volume of solution
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The most commonly used expression of concentration is molarity (M) The number of moles of solute per volume of solution
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Example 4: Calculate the molarity of a solution prepared by dissolving 15.6 g of solid KBr in enough water to make 1.25 L of solution. Example 5: How many moles of Na + ions are present in 42.0 mL of 0.350 M NaCl solution? How many grams of Na + ions would this represent?
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Dilution Objective: To learn to calculate the concentration of a solution made by diluting a stock solution. To save time and space in the science lab, solutions are often purchased in concentrated form Called stock solutions These solutions will be diluted How much water needs to be added?
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Remember, only water will be added! Moles solute before = moles solute after Know these 2 formulas:
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Example 6: What volume of 19 M NaOH must be used to prepare 1.0 L of a.15 M NaOH solution? Example 7: What volume of water is needed to prepare 500.0 mL of a.250 M Ca(NO 3 ) 2 solution from a 5.00 M Ca(NO 3 ) 2 solution?
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Stoichiometry of Solution Reactions Objective: To understand the strategy for solving stoichiometric problems for solution reactions. Formulas:
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Example 8: Calculate the mass of solid sodium sulfate that must be added to 250.0 mL of a.200 M solution of barium nitrate to precipitate all of the ions to form barium sulfate. Also calculate the mass of barium sulfate formed. What are the ?’s: 1. g Na 2 SO 4 2. g BaSO 4 Have: Ba(NO 3 ) 2 =.200 M =.250 L
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Balanced Equation: Moles of Ba(NO 3 ) 2 =.05 mol Find the following: 1..05 mol Ba(NO 3 ) 2 → g Na 2 SO 4 2..05 mol Ba(NO 3 ) 2 → g BaSO 4 1.. 05 (1 / 1) 142.04 = 7.10 g Na 2 SO 4 2.. 05 (1 / 1) 233.39 = 11.7 g BaSO 4
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