Lecture 4.2 – The Dissolving Process and the Rate of Dissolving
Chapter 15 Solutions. 1.To understand the process of dissolving 2.To learn why certain substances...
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Transcript of Chapter 15 Solutions. 1.To understand the process of dissolving 2.To learn why certain substances...
![Page 1: Chapter 15 Solutions. 1.To understand the process of dissolving 2.To learn why certain substances dissolve in water 3.To learn qualitative terms describing.](https://reader035.fdocuments.in/reader035/viewer/2022062222/5697bff11a28abf838cbb6df/html5/thumbnails/1.jpg)
Chapter 15
Solutions
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1. To understand the process of dissolving2. To learn why certain substances dissolve in water3. To learn qualitative terms describing the concentration of
a solution 4. To understand the factors that affect the rate at which a
solid dissolves
Objectives Section 1 – Homogenous and Heterogeneous Solutions
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Forming solutions
• Solution = homogeneous mixture
• Can be solid, liquid, or gas
• Solute =
• Solvent = – Aqueous solution
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Solubility
• Solubility of ionic substances– Strong attractive forces that hold ionic crystal
together are overcome by the strong attraction between the ionic crystal and the water molecule
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– When ionic compounds dissolve they break into individual ions
• NaCl• CaCl2
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• Solubility of polar substances– Polar molecules can form hydrogen bonds with
water and dissolve– Like dissolves like
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• Substances insoluble in water– Nonpolar molecules will not dissolve in water– Water-water hydrogen bonds keep the water from
mixing with the nonpolar molecules
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• How substances dissolve– A “hole” must be made in the water structure for
each solute particle. – The lost water-water interactions must be
replaced by water-solute interactions. – “like dissolves like”
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Solution Composition
• Saturated – solution contains as much solute as will dissolve at that temperature– If more solute is added it will not dissolve
• Unsaturated – solution that has not reached the limit of solute that will dissolve in it. – If more solute is added it will dissolve
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• Supersaturated - occurs when a solution is saturated at an elevated temperature and then allowed to cool but all of the solid remains dissolved • Contains more dissolved solute than a saturated solution• Very unstable
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Solution Composition: An Introduction
• A supersaturated solution is clear before a seed crystal is added.
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Solution Composition: An Introduction
• Crystals begin to form in the solution immediately after the addition of a seed crystal.
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Solution Composition: An Introduction
• Excess solute crystallizes rapidly.
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• Solution concentration = the amount of solute in a given amount of solution
• Qualitative measurements of solution concentration– Concentrated – lots of solute– Dilute – not a lot of solute
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Solution Composition: An Introduction
Which solution is more concentrated?
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Solution Composition: An Introduction
Which solution is more concentrated?
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Factors affecting the rate of dissolving
• Surface area – dissolving occurs at surface of substance being dissolved
• Stirring – removes newly dissolved particles from the solute surface and continuously exposes the surface to fresh solvent
• Temperature - molecules moving more rapidly, more interaction between solvent and solute
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1. To understand mass percent and how to calculate it 2. To understand and use molarity 3. To learn to calculate the concentration of a solution
made by diluting a stock solution
Objectives Section 2 – Concentration of Solutions
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Solution Composition: Mass Percent
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• A solution is prepared by mixing 1.00 g of ethanol with 100.0 g of water. Calculate the mass percent of ethanol in this solution
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• Cow’s milk typically contains 4.5% by mass of lactose, calculate the mass of lactose in 175 g of milk.
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Solution Composition: Molarity
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• What is the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in 1500 mL of solution?
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• Give the concentration of the ions in a 0.50 M solution of Co(NO3)2
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• How man moles of Ag+ ions are present in 25 mL of a 0.75 M AgNO3 solution?
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Solution Composition: Molarity • Standard solution - a solution whose concentration is
accurately known • To make a standard solution
– Weigh out a sample of solute.
– Transfer to a volumetric flask.
– Add enough solvent to mark on flask.
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Molarity
• To make a 0.5-molar (0.5M) solution, first add 0.5 mol of solute to a 1-L volumetric flask half filled with distilled water.
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Molarity
• Swirl the flask carefully to dissolve the solute.
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Molarity
• Fill the flask with water exactly to the 1-L mark.
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• A chemist needs 1.00 L of 0.200 M K2Cr2O7 solution. How much solid K2Cr2O7 (molar mass = 294.2 g/mol) must be weighed out to make this solution?
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Dilution
• Water can be added to an aqueous solution to dilute the solution to a lower concentration.
• Only water is added in the dilution – the amount of solute is the same in both the original and final solution.
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• Diluting a solution – Transfer a measured amount of original solution to
a flask containing some water. – Add water to the flask to the mark (with swirling)
and mix by inverting the flask.
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Making Dilutions
• The total number of moles of solute remains unchanged upon dilution, so you can write this equation.
• M1 and V1 are the molarity and volume of the initial solution, and M2 and V2 are the molarity and volume of the diluted solution.
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• If we want to prepare 500. mL of 1.00 M acetic acid from a 17.5 M stock solution, what volume of the stock solution is required?
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1. To learn to solve stoichiometric problems involving solution reactions
2. To do calculations involving acid-base reactions3. To learn about normality and equivalent weight 4. To use normality in stoichiometric calculations 5. To understand the effect of a solute on solution properties
Objectives Section 3 – Properties of Solutions
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Stoichiometry of Solution Reactions
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• Calculate th mass of solid NaCl that must be added to 1.50 L of a 0.100 M AgNO3 solution to precipitate all of the Ag+ ions in the form of AgCl. Calculate the mass of AgCl formed.
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• When Ba(NO3)2 and K2CrO4 react in aqueous solution the yellow solid BaCrO4 is formed. Calculate the mass of BaCrO4 that forms when 3.50 x 10-3 mol of solid Ba(NO3)2 is dissolved in 265 mL of 0.0100 M K2CrO4 solution.
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Neutralization Reactions • An acid-base reaction is called a neutralization reaction. • Steps to solve these problems are the same as before.
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• What volume of a 0.100 M HCl solution is needed to neutralize 25.0 mL of a 0.350 M NaOH solution?
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Normality
• Unit of concentration– One equivalent of acid – amount of acid that furnishes 1
mol of H+ ions – One equivalent of base – amount of base that furnishes 1
mol of OH ions – Equivalent weight – mass in grams of 1 equivalent of acid
or base
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Normality
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Normality
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• Phosphoric acid, H3PO4 can furnish three H+ ions per molecule. Calculate the equivalent weight of H3PO4
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Normality
• To find number of equivalents
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• A solution of sulfuric acid contains 86g of H2SO4 per liter of solution. Calculate the normality of the solution.
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Colligative properties
• a solution property that depends on the number of solute particles present
• The presence of solute “particles” causes the liquid range to become wider. – Boiling point increases – Freezing point decreases
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• Vapor pressure lowering – – Vapor pressure:
– Adding a nonvolatile solute to a solution lowers the solvent’s vapor pressure
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• Boiling point elevation –– Boiling point
– Because vapor pressure is lowered, boiling point increases
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• Freezing point depression – solute particles interfere with attractive forces holding solvent particles together