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Ch.4 - Reactions in Aqueous Solution

Chapter 4, Problem 48b

Determine the oxidation number of sulfur in each of the following substances: (b) sulfurous acid, H2SO3

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Hello. Everyone in this video we're going to calculate the oxidation number of sulfur and H. Two S. 203. So some things to notice as first we have a neutral compound. Whenever we have a neutral compound this means that the sum of all the oxidation numbers will equal to zero. So I'll just say sigma equals to zero. Sigma just means some. Alright, next thing that I noticed is that we have a hydrogen that's bonded with non medals in that case then our hydrogen will have an oxidation number of plus one. Alright, another thing that I noticed is that well we have an oxygen and that's going to always have an oxidation state of -2. So we'll put that the oxygen or yeah the oxygen number. So our station number of the oxygen Adam is going to be -2. Alright now I'm gonna go ahead and rewrite our compound. So we have h. two s. 203. Alright so we have our unknown S. So we'll just put that S. Two X. I have a. Two there because we don't know or have a. Two there because we have two atoms of our sulfur and X. For the S. The sulfur. Alright then we have our Hira gens and that's going to be we have two atoms and each will have a plus one charge. So that's equal to two. And for our oxygen's we have three oxygen atoms And each has a oxygenation number of -2. So three times 2 is -6. All right now putting this or laying this out more mathematically correctly that would be two Plus our native six plus our two X. And because we have a neutral compound, all these oxidation numbers will equal to zero. Now simplifying this a little bit further, we'll get to x minus four equals to zero. I'm gonna go ahead and add both sides by four To get this. And now isolating our X. We divide each side by two, Giving us x equals to two. Let's not also forget that X we said is going to equal to R. S. And so from this calculation, we get that the oxidation number of us for this compound is equal to positive too. And this is going to be our final answer for this problem. Thank you all so much for watching.