Net Ionics Guide

1. Net Ionics Guide

2. The Basics • A net ionic equation is a chemical equation in simplest form, allowing you to seethe essential parts of the equation. • Net ionics deal with compounds and molecules in a solution (that is, mixed with water). • Compounds that have (aq) after it means that the compound is aqueous and is able to be dissolved in water. • Aqueous compounds are split into their ions.

3. Example Problems • Double-replacement): • 2KNO3(aq) + Na2CO3(aq) ---> K2CO3(s) + 2NaNO3(aq) • Precipitate: • Mg(NO3)2(aq) + Na2CO3(aq) --->2NaNO3(aq) +MgCO3(s) • Acid-Base: • HCl(aq) + NaOH(aq) ---> NaCl(aq) + H2O(l ) • Oxidation-Reduction (Redox): • Cu2+ (aq) + Zn(s) ---> Cu(s) + Zn2+(aq)

4. Things to keep in mind • Review and remember your basic ions and charges, solubility rules, common Redox formations, and complex ion formations. • Remember that all elements must be accounted for when you finish, if you miss something (besides spectators) you need to put in.

5. Double-Replacement • Net Ionic Double-Replacement reactions involve the same basic rules of DR reactions, involving rules for net ionics. • First, you must identify which compounds are aqueous, and which are solids, liquids and gases, this is where the basic rules you learned come in handy.

6. Pb(NO3)2(aq) + 2KI(aq) ---> 2KNO3(aq) + PbI2(s) • The substance states have been identified in the parenthesis • Pb2+(aq) + 2NO3-(aq) + 2K+(aq) + 2I-(aq) ---> 2K+(aq) + 2NO3-(aq) + PbI2(s) • The aqueous substances dissolve into their ions • Pb2+(aq) + 2I-(aq) ---> PbI2(s) • The spectator ions or ions that are exactly the same on both sides of the equation and do not change state in the reaction, are removed from the equation, leaving only the ions that change state. You now have a net ionic equation.

7. Simple, right?

8. Now you try. The compounds have already been identified. Indicate which ions dissolve. The ladies will applause correct selections, and will shoot you if you are wrong. • AgNO3(aq) + KI(aq) ---> KNO3(aq) + AgI(s) If you heard three applause's, click here!

9. What happened?! • Have you been paying attention?! SOLID compounds (i.e. a rock, steel) DO NOT dissolve in water and thus DO NOT split into it’s ionic parts! Got it? Try again.

10. Spectator Ions • Now its time to see which ions are the exact same on both sides and do not change states. Click on the buttons with spectator ions on them. • Ag+(aq) + NO3- (aq) + K+(aq) + I-(aq) ---> K+(aq) + NO3-(aq) + AgI(s) NO3-and K+ AgI

11. Uh… no • This ion has changed states in the equation from aqueous to solid. It is therefore not a spectator ion. Pay attention! Try, try again!

12. Good! • You correctly identified all the spectator ions! • Remember that sometimes they may appear the same on both sides, however, sometimes it may be an ion on one side and a normal atom or a different charge ion on the other side, and is not a spectator!

13. Advanced work • Relax! Its not as bad as it sounds. I meant that there are certain situations when cool stuff happens and you need to know what they are.

14. Precipitation Reactions • The problem you just did was called a precipitation reaction. This just means that one of the products formed is a solid and does not dissolve in water. It precipitates or falls out of solution and sinks to the bottom of the water mixture.

15. Gas forming reactions • Essentially the same as a precipitate reaction, except instead of a solid being formed, a gas is formed in the reaction and floats up and out of the water mixture. An example would be a reaction with Hydrogen gas bubbling in a mixture.

16. Acid-Base reactions • In this type of reaction an acid (characterized by being an ion bonded with the H+ ion) and a base (characterized by being bonded with the hydroxide ion OH-) combine together and form water and a salt. • NaOH + HCl ---> NaCl (a salt) + H2O • Basically most will be double-replacement.

17. Much Harder Reactions • Or as they are more commonly called: Oxidation-Reduction or redox reactions. • Redox reactions are reactions where electron movement is the most important factor. A redox reaction will have something lose electrons and something gain electrons. • Redox reactions can be most types of reaction.

18. Redox (continued) • If an element or ion loses one or more electrons, it is said to be oxidized. If one gains one or more electrons it is said to be reduced. • A nifty way to remember this is with the phrase “LEO (a lion) says GER!” Where LEO stands for “Loses Electrons: Oxidized,” and GER stands for “Gains Electrons: Reduced.”

19. More Redox... • When you find yourself in a redox reaction (a reaction where, for example, Fe2+ changes to Fe3+) you usually need to find the oxidation numbers of each element.

20. Oxidation number example • Say you have CrO4 (Chromate), the oxidation number of each oxygen is 2-, and four of them would be a total of 8-. Chromate’s ionic charge is 2- so to equal 2-, chromium must have an oxidation number of +6. • This means that there are 6 valence electrons missing from chromium in order for it to bond with oxygen.

21. Example (cont.) • So you know that, in this case, chromium is +6, however, on the product side, chromium has separated from oxygen and is now in its ion form, Cr2+. How has this happened? • A redox reaction has occurred. The Chromium has gained 4 electrons and has been reduced. Where did it get them from?

22. Redox Example • I hope you pay attention, because you’re gonna do this in a few minutes! • The answer is the chromium received the electrons from whatever chromate had reacted with. The other component was oxidized. • In every redox reaction there is ALWAYS at least one thing oxidized and one thing reduced.

23. Reaction Example • Say, theoretically, the following reaction occurs: • Cr042+ + C (graphite) ---> Cr2+ + CO2 (g) • What type of redox reaction is this (think back to what you’ve learned) ? • Precipitate Gas forming • Acid-Base Double Replacement

24. Incorrect • I’m sorry, you are wrong. Think harder!

25. Correct • You see? Not hard! • Now notice that carbon starts as a 0 charge but acquires an oxidation number of 4+. This is where chromium’s electrons came from! The four electrons left carbon so that it could form CO2 and went to chromium when carbon forced chromium off of it’s oxygen.

26. Cr042+ + C (graphite) ---> Cr2+ + CO2 (g) • C4+ + 4e- • Cr+6 + 4e-

27. Cr042+ + C (graphite) ---> Cr2+ + CO2 (g) • 4 O + C4+ 2CO2

28. Explanation • Note that the problem would produce 2 moles of carbon dioxide, and that the initial problem is unbalanced. In net ionics, balancing is only important when predicting certain products, and in this case, there was no need.

29. Redox #2 • When there is more to consider in a reaction, a half-reaction is commonly used to to help you better understand what is going on. • Note the following equation:

30. Redox #2 (cont.) • The reaction is set up as a net ionic. Click on the button that depicts spectator ions. • C2O4 and MnO4 Mn+2 and CO2 • K+ and Na+ C2 and O2

31. Wrong. • If you’re still having problems, I suggest you start over and re-read the section on spectator ions. • Try again Back to Spectator Basics

32. Right. • Good Work. Now we are left with the equation:

33. Oxidation numbers • Now you must find the individual charge of each element within the molecules on both sides of the equation. • MnO4- has an overall charge of -1, and each oxygen is worth -2. What must the charge of Manganese be? • +7 -9 • +3 +1

34. No. • x + (-8) = -1 What would the x value be?

35. Yes. • Your getting the hang of this Tex. Now the hard part. • The carbon on the oxalate ion must have a charge of+6 (+3 each). Each carbon loses an electron to obtain the +4 charge that it must have in order to form carbon dioxide. Those electrons go to the manganese, and at the end, manganese changes from +7 to +2. Now to show you the half reaction.

36. Half Reaction • When you use half reactions, all you do is split apart the oxidizing part from the reduction part and denote them as separate reactions, a half reaction, as each only shows half of what goes on. • Mn+7 + 5e- ---> Mn+2 • 2C+3 ---> C+4 +1e-

37. Balancing Half Reactions • Mn+7 + 5e- ---> Mn+2 +7 + (-5) = +2 • 2C+3 ---> C+4 +2e- 2(+3) = +4 + (-2) • As you can see in carbon’s half reaction, the number of electrons don’t balance out. If you add up the charges on each side of the equation and take the sum, a balanced equation would equal zero. It should look like this: • 2C+3 ---> 2C+4 +2e- • The next step is to put the two equations together.

38. Round ‘em up! • Mn+7+ 2C+3 + 5e- ---> Mn+2 + 2C+4 +2e- • Add up the charges, and you see this: • +8 = +8 • When the charges line up, that means you’ve done the problem correctly. Sometimes however, the charges will not be the same and you must balance. • Balancing half reactions follows the exact same rules as normal balancing, taking into account the electrons in the problem, and treating them as molecules.

39. THE END • Be sure to beg Mr. Kirk to give you a practice net ionic problem, as well as a redox problem. Use this presentation as a guide, and you’ll do fine.