Reactions in Aqueous Solution: Solubility, Dissociation, and Concentration

1. Chapter 4 Reactions in Aqueous Solution

2. Solutions • Solutions are homogeneous mixtures of two or more substances • Aqueous solutions have water as a solvent • Solutions have two components • Solute: this is the dissolved substance • Solvent: this is the substance in which the solute is dissolved • Solutions form when there is a sufficient attraction between the solute and the solvent molecules

3. Solutions: homogeneous mixtures • Common solutions with different types of solutes and solvents

4. Solubility • The solubility is the maximum amount of solute that can be dissolved in a specific amount of solvent • A solute that dissolves in a solvent is said to be soluble • A solute that does not dissolve in a solvent is said to be insoluble • When ionic compounds dissolve in water they usually dissociate into ions • However not all ionic compounds are soluble in water • We will use the following table to determine the solubility of ionic compounds

5. Solubility of ionic compounds • Determine if each of the following is soluble or insoluble in water • KOH • AgBr • CaCl2 • Pb(NO3)2 • PbSO4

6. Solubility of ionic compounds

7. Dissociation • Dissociation is the process by which ionic compounds dissolve in water into their anions and cations • Not all ionic compounds are soluble in water • Molecular compounds interact with water but do not dissociate • When compounds containing polyatomic ions dissociate, the polyatomic ion stays together as one ion

8. Dissociation: Examples • Potassium iodide dissociates in water into potassium cations and iodide anions • Copper (II) Sulfate dissociates in water into copper (II) cations and sulfate anions

9. Electrolytes • Electrolytes are compounds that dissociate in water to give ions and therefore conduct electricity • Strong electrolytes dissociate completely in water (conduct easily) • Weak electrolytes partially dissociate in water (some conduction) • Nonelectrolytes are compounds that dissolve in water but do not produced ions (do not conduct)

10. Solution concentration • The concentration is the amount of solute in a given amount of solvent • Dilute solutions have low amounts of solute per amount of solution (weak coffee) • Concentrated solutions have high amounts of solute per amount of solution (strong coffee) • The concentration of a solution is expressed as

11. Unit of concentration: Molarity • The molarity of a solution is defined as the moles (mols) of solute per volume (L) of solution • A 1.0 M solution of NaCl is defined as • To make a solution of a specified molarity, you usually put the solute into a flask and then add water to the desired volume

12. Unit of concentration: molarity Preparation of 1.0 M NaCl solution

13. Making a standard solution

14. Unit of concentration: molarity • Example: Calculate the molarity of a solution made by putting 15.5 g of NaCl into a beaker and adding water to make a 1.50 L solution of NaCl (MM NaCl = 58.44 g/mol)

15. Molarity as a Conversion Factor • The units of molarity are used as conversion factors in calculations • Examples • Express 3.5 M HCl as a conversion factor • Determine the mass of CaCl2 in 1.75 L of a 1.50 M solution

16. Solution dilution • Dilution is the process of adding more solvent (usually water), to increase the volumes and decrease the concentration of the solution

17. Solution dilution • The dilution formula is • Where are the molarity and volume of the initial concentrated solution and are the molarity and volume of the diluted solution • Concentrations and volumes can be written in most units as long as they are consistent • When diluting acids, always add the concentrated acid to the water (not the other way around)

18. Solution dilution: Example • How would you prepare 5.00 L of a 1.50 M KCl solution from a 12.0 M stock solution? • Use the dilution formula • We are given the molarity and volume of the dilute solution and the molarity of the concentrated solution, we just need to figure out the amount of the concentrated solution needed …. • Divide both sides by • = • Answer: take 0.625 L of the stock solution, place it in a 5.00 L volumetric flask, and add water to the 5 L mark

19. Solution Stoichiometry • A balanced chemical equation tells us the relationship between the moles of reactants and the products in a reaction • Molarity is the relationship between the moles of solute and the liters of solution • We can now calculate the moles in a reaction in a solution by knowing its molarity and volume • The general solution map for these kinds of calculations is as follows Volume A Moles A Moles B Volume B

20. Solution stoichiometry • How much 0.125 M NaOH solution is needed to completely neutralize 0.225 L of 0.175 M H2SO4? Volume Moles Moles Volume

21. Classification of Chemical reactions • The first method of classification is based on the process that occurs Chemical Reaction Oxidation reduction reactions Precipitation reactions Acid-base reactions Gas-evolution reactions Combustion reactions

22. Classification of Chemical reactions • The second method of classification is based on what the atoms do Chemical Reaction Double displacement synthesis decomposition Single displacement

23. Precipitation reactions • Precipitation reactions produce a solid • Often this occurs when the cations of one solution exchange with the anion of the other solution • If the ion exchange results in forming a compound insoluble in water, it will come out of solution as a precipitate

24. Precipitation reactions • When potassium iodide is mixed with lead (II) nitrate, a brilliant yellow precipitate of PbI2(s) forms • The key to predicting precipitation reactions is understanding that only insoluble compounds form precipitates • In this reaction when KI(aq) is mixed with Pb(NO3)2(aq) we look at the products of ion exchange to see if either is insoluble • KNO3 is soluble and written KNO3(aq) • PbI2 is insoluble and written PbI2 (s) – this is the yellow solid

25. Molecular Equations for precipitation reactions • Steps for writing equations for precipitation reactions • 1. Write out balanced formulas for the reactants including their phases • 2. Combine the cation of one reactant with the anion from the other (ion exchange or the cross rule) to obtain the products • 3. Balance the reaction by finding the stoichiometric coefficients for reactants and products • 4. Use the solubility rules to determine which products are insoluble. If all the products are soluble, write NR for No reaction on the right side of the arrow • 5. If one of the products is insoluble (s), a liquid (l), a gas (g), a weak acid or a weak base, write its formula as the product of the reaction

26. Equations for precipitation reactions • Example: write an equation for the precipitation reaction that occurs (if any) when sodium carbonate (Na2CO3 (aq)) is reacted with copper (II) chloride (CuCl2(aq)) • Steps (1+2) : use the Cross rule to construct the products and (step 3) balance the reaction • Steps (4+5) Use the solubility rules to determine if any of the products are insoluble and label accordingly 2 Soluble insoluble

27. Molecular & ionic Equations • The molecular equation is a chemical equation showing the complete neutral formulas for every compound for example • The total ionic equation is a chemical equation showing all the species are they are present in solution • Only dissociate aqueous compounds, not solids, liquids, gases, weak acids or weak bases • In the net ionic equation we cancel ions common to both sides of the total ionic equation, these are called spectator ions and write out what is left

28. Acid-base reactions • These are reactions that form water upon mixing of an acid and a base • The acid is the compound that produces H+(aq) when dissociating in aqueous solution • The base is the compound that produces OH-(aq) when dissociating in aqueous solution • These are also known as neutralization reactions • The net ionic equation for many acid-base reactions is (as long as the salt is soluble)

29. Acid-base reactions • Some common acids and bases

30. Acid-base reactions • Write out the molecular, complete ionic and net ionic equations for

31. Acid-base titration • This is a laboratory technique used to determine the concentration of an unknown acid or base • It uses a base (or acid) to neutralize a measured volumes of an acid (or base) • It requires a few drops of an indicator such as phenolphthalein to identify the end point • Equivalence point is the point where the moles of base (or acid) added via the burette = moles of the acid (or base) in the measured volume • Indicator is the dye whose color depends on the acidity or basicity of the solution • End point is the point at which the indicator’s color changes

32. Acid-base titration Example: • A base is in the burette and acid is in the flask with some phenolphthalein • As the base from the burette is added the OH- reacts with the H+ to make water. Still excess acid is present so the solution stays colorless • At the endpoint, just enough base has been added to neutralize all of the acid. At this point the indicator changes color

33. Acid-base titration • Example: When the number of moles of NaOH added = number of moles of HC2H3O2 (acetic acid) initially the indicator (phenolphthalein) changes to pink • Phenolphthalein is colorless in acid and pink in base

34. Acid-base titration • Example: The titration of 7.49 mL of HCl solution of unknown concentration requires 13.54 mL of 0.287 M KOH solution to reach the end point. What is the concentration of the HCl solution? • The neutralization reaction is • At the end point mole KOH = mole HCl

35. Gas Evolution reactions • Gas evolution reactions are reactions in one of the products is a gas • Some reactions form a gas directly as a result of ion exchange • Other reactions form a gas by the decomposition of one of the ion-exchange products into a gas and water

36. Gas evolution reactions • Compounds that decompose into water and a gas