Unit 4: Aqueous Reactions and Solution Chemistry

1. Unit 4: Aqueous Reactions and Solution Chemistry By: Ms. Buroker

2. General Properties of Aqueous Solutions A solution is made of a solute and solvent. *Solute: Dissolved Material *Solvent: Dissolving Material (Usually found in the greatest amount)

3. Aqueous solutions • Dissolved in water. •Water is a good solvent because the molecules are polar. 1.)The oxygen atoms have a partial negative charge. 2.)The hydrogen atoms have a partial positive charge. 3.)The angle is 105º.

4. Solvation • This is known as HYDRATION when the solvent is water. •The process of breaking the ions of salts apart. •Ions have charges and attract the opposite charges on the water molecules.

5. Solubility • How much of a substance will dissolve in a given amount of water. • Varies greatly, but if they do dissolve the ions are separated, and they can move around. • The ionic solid dissociates into its component ions as it dissolves. • Water can also dissolve non-ionic compounds if they have polar bonds.

6. General Properties of Aqueous Solutions Electrolytes and Nonelectrolytes Aqueous solution Aqueous solution containing ions containing no ions

7. Electrolytes Strong Electrolytes: Compounds which dissolve almost completely in solution (exist as ions). Weak Electrolytes: Compounds which do not dissolve or ionize in solution … molecular compounds mostly.

8. Acidic Solutions • Acids-form H+ions when dissolved. • Strong acids disassociate completely. • You Must Memorize These!! • H2SO4, HNO3, HCl, HBr, HI, HClO4, HCLO3 • Weak acids do NOT dissociate completely. • Write the dissociation equation for Hydrochloric acid Acetic Acid HClH+ + Cl- HC2H3O2 ↔ H+ + C2H3O2-

9. Common Reactions Synthesis A + B  AB Decomposition AB  A + B Single Replacement A + BX  AX + B or A + XB  XA + B Double Replacement AX + BY  AY + BX Don’t forget the metal reactivity series and the non-metal reactivity series!!

10. Precipitation Reactions • When aqueous solutions of ionic compounds are poured together a solid forms. • A solid that forms from mixed solutions is a precipitate Watch this video for examples of precipitation reactions

11. Precipitation reaction • Called METATHESIS–Greek word for “to transpose.” • We can predict the products • Can only be certain by experimenting • The anion and cation switch partners • AgNO3(aq) + KCl2(aq)  • Zn(NO3)2(aq) + BaCr2O7(aq) • CdCl2(aq) + Na2S(aq) 

12. Precipitations Reactions • Only happen if one of the products is insoluble, otherwise all the ions stay in solution-nothing has happened. • Three things drive a double replacement reaction … the formation of a gas, a solid, or a liquid.

13. Solubility Rules Molecular Equation Complete Ionic Equation Net Ionic Equation *Minus Spectator Ions

14. Acid- Base Reactions Acids Substances that are able to ionize in aqueous solutions to form a hydrogen ion and thereby increase the concentration of H+(aq) ions. Bases Substances that accept (react with H+) ions. Substances that increase the OH- concentration when added to water.

15. Strong and Weak Acids and Bases Acids and bases which completely ionize in solution are considered strong acids and strong bases (strong electrolytes)! The opposite is true concerning those which weakly ionize … they are weak acids and weak bases.

16. Strong and Weak Acids and Bases Strong Acids Hydrochloric, HCl Hydrobromic, HBr Hydroiodic, HI Chloric, HClO3 Perchloric, HClO4 Nitric, HNO3 Sulfuric, H2SO4 (Most acids are weak) Strong Bases Group 1A metal hydroxides Heavy group 2A metal hydroxides (Ca, Sr, Ba) (most common weak base, NH3)

17. Let’s Practice! Classify each of the following dissolved substances as a strong electrolyte, weak electrolyte, or nonelectrolyte: CaCl2, HNO3, C2H5OH (ethanol), HCHO2 (formic acid), KOH Answer: CaCl2= strong , HNO3= strong, C2H5OH (ethanol)= non, HCHO2= weak (formic acid), KOH= strong

18. Neutralization Reactions Reactions which take place between an acid and a metal hydroxide; they produce water and a salt. Write a balanced complete chemical equation for the reaction between aqueous solution of acetic acid and barium hydroxide. Write the ionic equation for this reaction.

19. Acid-Base Reactions With Gas Formation The sulfide ion, the carbonate ion, and the bicarbonate ion are bases which react with acids to form gases that have low solubility in water. H2S forms when: HCl(aq) + Na2S(aq) H2S(g) + NaCl (aq) Carbonates & Bicarbonates react to form CO2(g)

20. CO32- and HCO3- react with acids first to give carbonic acid, then carbonic acid decomposes into water and CO2(g). HCl(aq) + NaHCO3(aq) NaCl(aq) + H2CO 3(aq) Then… H2CO3(aq)  H2O(l) + CO2(g) Actually …. HCl(aq) + NaHCO3(aq) NaCl(aq) + H2O(l) + CO2(g)

21. Oxidation- Reduction Reactions When an atom or molecule has become more positively charged … that is, it has lost electrons … we say that it has been oxidized. Oxidation is Loss When an atom or molecule has become more negatively charged … that is, it has gained electrons … we say that it has been reduced. Reduction is Gain OIL RIG

22. Example of a Non-Redox • Let’s look at a precipitation reaction: AgCl + Na2SO4 Ag2SO4 + NaCl • Notice that the charges of each of the ions are the same on both sides: Ag+1 Ag+1 Na+1 Na+1 SO4-2 SO4-2 Cl-1 Cl-1

23. Example of a Redox • Look at this synthesis reaction: Fe + O2 Fe2O3 • What happens to the charge on each of the elements? Fe0 Fe+3 O20 O-2

24. Which reactions qualify as redox? • Redox reactions: reactions in which the charge of elements change from one side to the other • The element that is Oxidized loses electrons, which means a more positive charge (the iron in the previous example) • The element that is Reduced gains electrons, which means a more negative charge (the oxygen in the previous reaction)

25. Corrosion at the terminal of a battery, caused by attack of the metal by sulfuric acid from the battery. When a metal undergoes corrosion, it loses electrons and forms cations. For example: Ca(s) + 2H+(aq)  Ca2+ + H 2(g)

26. Oxidation Numbers Oxidation Numbers are basically the hypothetical charge we assign to atoms in a reaction to keep track of the electrons. Rules: 1.) For an atom in its elemental form, the oxidation number is zero. 2.) For any monatomic ion, the oxidation number equals the charge on the ion. 3.)Nonmetals usually have negative oxidation numbers, although they can sometimes be positive. (The oxidation number of hydrogen is +1 when bonded to nonmetals and -1 when bonded to metals. 4.) The sum of the oxidation numbers of all atoms in a neutral compound is zero. The sum of the oxidation numbers in a polyatomic ion equals the charge on the ion.

27. Examples of Determining Oxidation Numbers • What are the oxidation numbers of each element in KNO3? • O = -2 • K = +1 • Since the oxidation numbers of a neutral compound must equal 0: • (+1) + (N) +(3*-2) = 0 • N = +5

28. Examples of Determining Oxidation Numbers (continued) • What are the oxidation numbers of each element in (NH4)+1? • H = +1 • Since the oxidation numbers of a polyatomic ion must equal its charge: • (N) + (4*+1) = +1 • N = -3

29. Do these on your own: Determine the oxidation numbers for each element in the following chemicals: • H2CO3 • K3PO4 • Al(NO3)3

30. Answers • H2CO3 • H=+1 • O=-2 • C=+4 • K3PO4 • K=+1 • P=+5 • O=-2 • Al(NO3)3 • Al=+3 • N=+5 • O=-2

31. Some Examples Assign Oxidation Numbers to the elements in the following equations: Fe(s) + Ni(NO3)2(aq)  Fe(NO3)2(aq) + Ni (s) Write the balanced molecular and net ionic equation for the reaction of aluminum with hydrobromic acid. Which element is oxidized in the reaction?

32. A couple more terms… From the previous example, we know: Al + H2O  Al2O3 + H2 Al0Al+3 H+1H0 • The aluminum is oxidized • The hydrogen is reduced • H2O is called the oxidizing agent (the e- acceptor) • Al is called the reducing agent (the e- donor)

33. Balancing Redox Reactions Now that we know what they are, how do we balance them? Will it be the same as before? Of course not. Are we excited to learn more techniques that will improve our Chemistry knowledge? Of course we are.

34. Where do redox reactions occur? • Redox reactions typically take place in solutions • Because of this redox reactions are most often written as net ionic equations • If the reaction takes place in only water, you can balance by inspection (just like normal) • However, if the reactions involve acidic or basic solutions balancing is different…

35. Why would a redox reaction take place in an acid or base? • First, let’s define: • Acid: H+ ion donator when dissolved in water • Base: H+ ion acceptor when dissolved in water • When these types of solutions are present, it creates an opportunity for charges to get messed with • Extra H+ ions will attract electrons • Removing H+ ions creates leftover OH- ions in the water

36. Writing Half Reactions • In any case, when balancing a redox reaction it will be handy to write the reaction as two half reactions • One showing only the oxidation • One showing only the reduction • We will treat the two half reactions separately, and then combine them at the end

37. Writing Half Reactions: Examples • Write the half reactions for this redox: • CeCl4 + Sn(NO3)2 SnCl4 + Ce(NO3)3 • The half reactions would look like this: • Ce+4 Ce+3 (reduction) • Sn+2 Sn+4 (oxidation) • Notice the cancellation of the spectator ions, nitrate and chloride (they are net ionic equations)

38. To balance a redox reaction in an acidic solution: • Write the reaction as two half reactions • For each half reaction: • Balance the elements except H and O • Balance the oxygen by adding water • Balance the hydrogen by adding H+ ions • Balance the charge by adding electrons (e-) • Multiply the half reactions to equalize the number of electrons in each • Combine the half reactions and cancel anything identical (which should always include the electrons)

39. To balance a redox reaction in an acidic solution: Example Balance the following redox reaction that occurs in an acidic solution: MnO4-1 + Fe+2 Fe+3 + Mn+2 • Write the reaction as two half reactions Oxidation: Fe+2 Fe+3 Reduction: MnO4-1 Mn+2

40. To balance a redox reaction in an acidic solution: Example Fe+2 Fe+3MnO4-1 Mn+2 • For each half reaction: • Balance the elements except H and O Not needed in these half reactions • Balance the oxygen by adding water MnO4-1 Mn+2 + 4H2O • Balance the hydrogen by adding H+ ions 8H+ + MnO4-1 Mn+2 + 4H2O • Balance the charge by adding electrons (e-) Fe+2 Fe+3+ e- 5 e-+ 8H+ + MnO4-1 Mn+2 + 4H2O

41. To balance a redox reaction in an acidic solution: Example Fe+2 Fe+3 + e- 5 e- + 8H+ + MnO4-1 Mn+2 + 4H2O • Multiply the half reactions to equalize the number of electrons in each 5(Fe+2 Fe+3 + e- ) 5 e- + 8H+ + MnO4-1 Mn+2 + 4H2O

42. To balance a redox reaction in an acidic solution: Example 5(Fe+2 Fe+3 + e- ) 5 e- + 8H+ + MnO4-1 Mn+2 + 4H2O • Combine the half reactions and cancel anything identical (which should always include the electrons) 5 e- + 8H+ + MnO4-1 + 5Fe+2 Mn+2 + 4H2O + 5Fe+3 + 5e- 8H+ + MnO4-1 + 5Fe+2 Mn+2 + 4H2O + 5Fe+3

43. To balance a redox reaction in an basic solution: Follow the steps for an acidic solution, then: • To both sides of the reaction, add OH- ions to equal the number of H+ ions remaining • Form water on the side containing both H+ and OH-, and then cancel water from both sides if possible

44. To balance a redox reaction in an basic solution: Example Balance this redox reaction that takes place in a basic solution: Ag + CN-1 + O2 Ag(CN)2-1 + H2O Oxidation: Ag + CN-1 Ag(CN)2-1 Reduction: O2 H2O

45. To balance a redox reaction in an basic solution: Example Ag + CN-1 Ag(CN)2-1 O2 H2O Ag + 2CN-1 Ag(CN)2-1 O2 H2O Ag + 2CN-1 Ag(CN)2-1 O22H2O Ag + 2CN-1 Ag(CN)2-1 4H+ + O2 2H2O

46. To balance a redox reaction in an basic solution: Example Ag + 2CN-1 Ag(CN)2-1 4H+ + O2 2H2O Ag + 2CN-1 Ag(CN)2-1 + e-4e-+ 4H+ + O2 2H2O 4(Ag + 2CN-1 Ag(CN)2-1 + e-) 4e- + 4H+ + O2 2H2O 4e- + 4H+ + O2 + 4Ag + 8CN-1 4Ag(CN)2-1 + 4e-+ 2H2O 4H+ + O2 + 4Ag + 8CN-14Ag(CN)2-1 + 2H2O

47. To balance a redox reaction in an basic solution: Example 4H+ + O2 + 4Ag + 8CN-1 4Ag(CN)2-1 + 2H2O To both sides of the reaction, add OH- ions to equal the number of H+ ions remaining 4OH- + 4H+ + O2 + 4Ag + 8CN-1 4Ag(CN)2-1 + 2H2O + 4OH-

48. To balance a redox reaction in an basic solution: Example 4OH- + 4H+ + O2 + 4Ag + 8CN-1 4Ag(CN)2-1 + 2H2O + 4OH- Form water on the side containing both H+ and OH-, and then cancel water from both sides if possible 4OH- + 4H+ + O2 + 4Ag + 8CN-1 4Ag(CN)2-1 + 2H2O + 4OH- 4H2O + O2 + 4Ag + 8CN-1 4Ag(CN)2-1 + 2H2O + 4OH- 2H2O + O2 + 4Ag + 8CN-1 4Ag(CN)2-1 + 4OH- 2H2O + O2 + 4Ag + 8CN-1 4Ag(CN)2-1 + 4OH-

49. Concentration of Solutions Molarity = moles of solute Liters of solution Molality = moles of solute Kilograms of solvent Mole Fraction = moles of part Moles in total solution M1V1 = M2V2

50. When expressing the concentration of electrolytes, it’s important to remember that ionic compounds break apart completely in solution. 1.0M NaCl solution … 1.0M Na+ and 1.0M Cl- ions in solution