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Chapter 21, question 6, asks us to determine how many moles of silver chloroporipoprecipricitate out, given that there are four complexes with the same oxidation state, one mole each in water with h2o and cl minus ligands and cations.
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These complexes are all octahedral.
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And we have to figure out, the first thing that we actually have to figure, out is the four complexes that are going to be in solution.
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So to determine this, first we need to determine what oxidation state these four complex are going to be in.
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And the easiest way to figure this out, so they're all chromium complexes.
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And the first complex that we can think about is two, is actually this complex right here.
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So it has three ligands that are water.
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Three ligands that are cl minus, and chromium has a 3 plus oxidation state with an overall charge of this complex being 0.
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So we have to keep this chromium at the 3 plus oxidation state.
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The actual overall charge of these complexes doesn't matter for this question.
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So the second one, the second complex that we can think about, we can actually replace one of the cl minuses with a water ligand...