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Redox titrations | Chemical reactions | AP Chemistry | Khan Academy

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  1. - [Voiceover] We've already seen how to do an acid-base titration.
  2. Now let's look at a redox titration. Let's say we have a solution
  3. containing iron two plus cations. We don't know the concentration
  4. of the iron two plus cations, but we can figure out the concentration
  5. by doing a redox titration. Let's say we have 10 milliliters of our solution,
  6. and let's say it's an acidic solution. You could have some sulfuric acid in there.
  7. In solution, we have iron two plus cations and a source of protons from our acid.
  8. To our iron two plus solution, we're going to add some potassium permanganate.
  9. In here, we're going to have some potassium permanganate,
  10. KMnO4. Let's say the concentration
  11. of our potassium permanganate is .02 molar.
  12. That's the concentration that we're starting with. Potassium permanganate is, of course,
  13. the source of permanganate anions, because this would be K plus
  14. and MnO4 minus. Down here, we have a source of permanganate anions.
  15. We're going to drip in the potassium permanganate solution.
  16. When we do that, we're going to get a redox reaction.
  17. Here is the balanced redox reaction. If you're unsure
  18. about how to balance a redox reaction, make sure to watch the video
  19. on balancing redox reactions in acid. Let's look at some oxidation states really quickly
  20. so we can see that this is a redox reaction. For oxygen, it would be negative two.
  21. We have four oxygens, so negative two times four
  22. is negative eight. Our total has to add up to equal negative one.
  23. For manganese, we must have a plus seven, because plus seven and negative eight
  24. give us negative one. Manganese has an oxidation state of plus seven.
  25. Over here, for our products, we're going to make Mn two plus.
  26. Manganese two plus cation in solution, so the oxidation state is plus two.
  27. Manganese is going from an oxidation state of plus seven
  28. to plus two. That's a decrease or a reduction
  29. in the oxidation state. Therefore, manganese is being reduced
  30. in our redox reaction. Let's look at iron two plus.
  31. We have iron two plus as one of our reactants here.
  32. That means the oxidation state is plus two. For our products,
  33. we're making iron three plus, so an oxidation state of plus three.
  34. Iron is going from plus two to plus three.
  35. That's an increase in the oxidation state. Iron two plus is being oxidized
  36. in our redox reaction. As we drip in our potassium permanganate,
  37. we're forming our products over here. These ions are colorless in solution.
  38. As the permanganate reacts, this purple color disappears
  39. and we should have colorless, we should have a colorless solution.
  40. Let's say we've added a lot of our permanganate. Everything is colorless.
  41. But then we add one more drop, and a light purple color persists.
  42. Everything was clear, but then we add one drop of permanganate
  43. and then we get this light purple color. This indicates the endpoint of the titration.
  44. The reason why this is the endpoint is because our products are colorless.
  45. So if we get some purple color, that must mean we have some unreacted,
  46. a tiny excess of unreacted permanganate ions in our solution.
  47. That means we've completely reacted all the iron two plus
  48. that we originally had present. So we stop our titration at this point.
  49. We've reached the endpoint. We've used a certain volume
  50. of our potassium permanganate solution. Let's say we finished down here.
  51. If we started approximately there, we can see that we've used
  52. a certain volume of our solution. Let's say it took 20 milliliters.
  53. We used up 20 milliliters of our potassium permanganate solution
  54. to completely titrate our iron two plus. Our goal was to find the concentration of iron two plus.
  55. If we're going to find the concentration of iron two plus,
  56. we could figure out how many moles of permanganate were necessary
  57. to completely react with our iron two plus. We could figure out moles
  58. from molarity and volume. Let's get some more room down here.
  59. We know that molarity is equal to moles over liters.
  60. The molarity of permanganate is .02. We have .02
  61. for the concentration of permanganate ions. Moles is what we're solving for.
  62. It took us 20 milliliters for our titration, which we move our decimal place one, two, three,
  63. so we get .02 liters. So solve for moles.
  64. .02 times .02 is equal to .0004.
  65. So we have .0004. This is how many moles of permanganate
  66. were needed to completely react with all of the iron two plus
  67. that we originally had in our solution. It took .0004 moles of permanganate
  68. to completely react with our iron. All right. Next, we need to figure out
  69. how many moles of iron two plus that we originally started with.
  70. To do that, we need to use our balance redox reaction.
  71. We're going to look at the coefficients, because the coefficients tell us mole ratios.
  72. The coefficient in front of permanganate is a one. The coefficient in front of iron two plus
  73. is a five. If we're doing a mole ratio of permanganate
  74. to iron two plus, permanganate would be a one
  75. and iron would be a five. So we set up a proportion here.
  76. One over five is equal to ... Well, we need to keep permanganate
  77. in the numerator here. How many moles of permanganate were necessary
  78. to react with the iron two plus? That was .0004.
  79. So we have .0004 moles of permanganate. X would represent how many moles of iron two plus
  80. we originally started with. We could cross-multiply here to solve for x.
  81. Five times .0004 is equal to .002.
  82. X is equal to .002. X represents the moles of iron two plus
  83. that we originally had present. We're almost done,
  84. because our goal was to find the concentration of iron two plus cations.
  85. Now we have moles and we know the original volume,
  86. which was 10 milliliters. To solve for the concentration
  87. of iron two plus, we just take how many moles of iron two plus we have,
  88. which is .002, so we have .002 moles of iron two plus.
  89. We started with a total volume of 10 milliliters,
  90. which is equal to .01 liters. So we have .01 liters here.
  91. And .002 divided by .01 is equal to .2.
  92. So this is equal to .2 molar. That was the original concentration
  93. of iron two plus ions in solution. You could have used
  94. the MV is equal to MV equation and modified it, because our ratio isn't one to one here.
  95. That's another way to do it. But I prefer to actually sit down
  96. and do these calculations and think about exactly what's happening.

Zum Nachlesen

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