ch. 13: solutions dr. namphol sinkaset chem 152: introduction to general chemistry
TRANSCRIPT
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Ch. 13: SolutionsCh. 13: Solutions
Dr. Namphol Sinkaset
Chem 152: Introduction to General Chemistry
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I. Chapter OutlineI. Chapter Outline
I. Introduction
II. Concentration
III. Preparation of a Solution
IV. Dilution
V. Solution Stoichiometry
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I. IntroductionI. Introduction
• Solution chemistry is the most well studied – why?
• Although solutions tend to be liquid based, the general definition allows for other types of solutions.
• solution: homogeneous mixture of two or more substances
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I. Types of SolutionsI. Types of Solutions
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I. Solution ComponentsI. Solution Components
• There are two parts of a solution. solute: substance present in smaller
amount solvent: substance present in larger
amount
• For stoichiometry, the important aspect of a solution is its concentration.
• concentration: amount of solute present in a certain volume of solution
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II. Solution ConcentrationII. Solution Concentration• Solutions can be either dilute or
concentrated.
• dilute: small amount of solute relative to amount of solvent
• concentrated: large amount of solute relative to amount of solvent
• There are several different ways to express solution concentration.
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II. Mass PercentII. Mass Percent
• Mass percent expresses the number of grams of solute per 100 g of solution.
%100solvent mass solute mass
solute mass percent mass
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II. Using Mass PercentII. Using Mass Percent
• If expressed as a fraction over 100, the mass percent can be used as a conversion factor.
• For a solution that is 23.2% ethanol by mass:
solution g 100.0
ethanol g 23.2
ethanol g 23.2
solution g 100.0OR
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II. Sample ProblemII. Sample Problem
• Calculate the mass percent of a solution containing 15.5 g of fructose and 249.6 g of water.
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II. Sample ProblemII. Sample Problem
• Ocean water contains 3.5% sodium chloride by mass. How much sodium chloride does a 2.00 L sample of ocean water contain? Note that ocean water has a density of 1.027 g/mL.
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II. MolarityII. Molarity
• The most common concentration unit is molarity, which is moles solute per L of solution.
In a solution, solute is evenly dispersed in the solvent!!
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II. Using MolarityII. Using Molarity
• Molarity can be used as a conversion factor between moles of solute and liters of solution.
• For a 0.500 M NaCl solution:
solution L 1
NaCl mole 0.500
NaCl mole 0.500
solution L 1OR
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II. Sample ProblemII. Sample Problem
• Calculate the molarity of a solution formed when 24.2 g NaCl is dissolved in 124.1 mL of water.
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II. Sample ProblemII. Sample Problem
• How many grams of Na2HPO4 are needed to make 1.50 L of a 0.500 M Na2HPO4 solution?
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II. Concentration of IonsII. Concentration of Ions
• When soluble ionic compounds are added to water, they dissociate.
• Soluble ionics dissociate because the solvent-solute attraction is greater than the solute-solute attraction.
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II. The Dissolving ProcessII. The Dissolving Process
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II. NaCl in SolutionII. NaCl in Solution
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II. Ion ConcentrationsII. Ion Concentrations• Sometimes, we need to know the
concentration of an individual ion.
• When calculating, we must account for the ratio seen in the formula of the ionic compound.
• e.g. MgCl2 has two anions for every one cation; anion will be twice as concentrated.
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II. Sample ProblemII. Sample Problem
• Calculate the concentration of the ions when 19.6 g of iron(III) sulfate is dissolved in enough water to make 200.0 mL of solution.
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III. Solution CreationIII. Solution Creation• The last sample problem is an example
of a calculation needed in order to create a certain volume of solution of a certain concentration.
• This type of calculation is very common in any research lab.
• To make the solution, special glassware and a specific procedure must be used.
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III. Creating 1.00 M NaClIII. Creating 1.00 M NaCl
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IV. DilutionIV. Dilution
• Less concentrated solutions can be made from more concentrated solutions in a process called dilution.
• The more concentrated solution is known as a stock solution.
• To perform a dilution, you need to know how much of the stock solution to use.
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IV. Dilution EquationIV. Dilution Equation
• The following dilution equation makes it easy to calculate how much stock solution is needed.
2211 VMVM
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IV. Sample ProblemIV. Sample Problem
• How many mL of a 2.0 M NaCl solution are needed to make 250.0 mL of a 0.50 M NaCl solution?
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V. Solution StoichiometryV. Solution Stoichiometry
• Since molarity is a ratio between moles of solute and volume of solution, it can be used in stoichiometric calculations.
• The key is to remember that molarity breaks down into units of mole/L!
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V. Sample ProblemV. Sample Problem
• How many mL of 0.10 M HCl reacts with 0.10 g Al(OH)3 according to the reaction below?
Al(OH)3(s) + 3HCl(aq) AlCl3(aq) + 3H2O(l)
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V. Sample ProblemV. Sample Problem
• How much PbCl2 forms when 267 mL 1.50 M lead(II) acetate reacts with 125 mL 3.40 M sodium chloride according to the reaction below?
Pb(CH3COO)2(aq) + 2NaCl(aq) PbCl2(s) + 2NaCH3COO(aq)