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Ch. 13: Solutions

Ch. 13: Solutions. Dr. Namphol Sinkaset Chem 152: Introduction to General Chemistry. I. Chapter Outline. Introduction Concentration Preparation of a Solution Dilution Solution Stoichiometry. I. Introduction. Solution chemistry is the most well studied – why?

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Ch. 13: Solutions

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  1. Ch. 13: Solutions Dr. Namphol Sinkaset Chem 152: Introduction to General Chemistry

  2. I. Chapter Outline • Introduction • Concentration • Preparation of a Solution • Dilution • Solution Stoichiometry

  3. I. 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

  4. I. Types of Solutions

  5. I. 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

  6. II. 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.

  7. II. Mass Percent • Mass percent expresses the number of grams of solute per 100 g of solution.

  8. II. 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: OR

  9. II. Sample Problem • Calculate the mass percent of a solution containing 15.5 g of fructose and 249.6 g of water.

  10. II. 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.

  11. II. 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!!

  12. II. 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: OR

  13. II. Sample Problem • Calculate the molarity of a solution formed when 24.2 g NaCl is dissolved in 124.1 mL of water.

  14. II. Sample Problem • How many grams of Na2HPO4 are needed to make 1.50 L of a 0.500 M Na2HPO4 solution?

  15. II. 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.

  16. II. The Dissolving Process

  17. II. NaCl in Solution

  18. II. 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.

  19. II. 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.

  20. III. 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.

  21. III. Creating 1.00 M NaCl

  22. IV. 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.

  23. IV. Dilution Equation • The following dilution equation makes it easy to calculate how much stock solution is needed.

  24. IV. 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?

  25. V. 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!

  26. V. 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)

  27. V. 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)

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