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How to employ The 1615 0037; Expires 123109?

hello this lesson is entitled polyprotic.acids.polyprotic acids like sulfuric acid.h2so4 than one ionizable h so we see two.equations that relate to equilibrium.constants the first one has one H and.there's a certain equilibrium constant.that relates to that and then the second.equation shows the second H popping off.and there's a different equilibrium.constant that applies for that case if.you look at the magnitudes of the KA.values if you just look at the power of.10 the first one in this case is 10 to.the negative second and the second one.is 10 to the negative 8 that's a million.times different which is quite a bit and.the reason that the second ka value is.so much smaller is because it's harder.to remove additional protons kind of.like the bully at lunch in grade school.it was tatertot day and the bully might.come by and grab one of your tater tots.and that first tater tot was fairly easy.for him to steal because you weren't.expecting it well the second tater tot.would be much harder because now you're.on to him so not that it can't be done.but it's a lot tougher.in the same way losing one proton is one.thing losing a second proton is much.more difficult now here's the nice thing.about polyprotic acids usually ka 2 is.at least a thousand times smaller than.ka one in such cases.one can calculate the hydrogen ion.concentration and the pH based only on.ka one that is ignore ka two and pretend.you have a monoprotic acid let's show.you an example find the pH of a point.zero zero three seven molar carbonic.acid solution carbonic acid has two.ionizable protons so you can see that we.have two ka values so here's the first H.popping off and let's write an ice.expression where we're gonna combine the.initial the change and then we're gonna.write just the at equilibrium line point.zero zero three seven which is the.initial then that amount will go down.some amount X and with the hydrogen ion.and this one here on the right is called.the bicarbonate ion start both of them.start at zero and then go up by some.amount X so substituting the values in.to the equilibrium constant expression.there's our ka one value there we have.products x times X and there we have the.reactant value if we use the quadratic.formula to solve that X comes out to be.three point nine seven times 10 to the.negative fifth molar and then of course.we can find the pH by taking the.negative logarithm of that if instead we.use the shortcut which we talked about.in an earlier lesson which says that X.is very small so in the denominator.point zero zero three seven minus.something very very small is essentially.zero point zero zero three seven that's.a lot easier to solve than using the.quadratic formula and if you do solve.that you'll get this value.3.99 times 10 to the negative 5th which.you can see is very very close to the.value you would get from the quadratic.formula in fact to two places past the.decimal the pH is exactly the same so.this is where the pH stands after we've.lost one proton we're going to go.through on the next slide what.additional effect takes place when we.lose the second proton so after what.went on on the previous slide the.hydrogen ion and the bicarbonate ion.concentrations after the first.ionization are right here 3.9 7 times 10.to the negative fifth I'll go ahead and.use the value the actual value that you.get from the quadratic formula and not.the shortcut value but it really doesn't.make any difference let's now show what.happens when for some of these a second.proton is removed now this one is a.little bit more of a complex expression.because we need to pick up from the.point at which we've already lost one.proton which means that the initial.concentration for this case for the.bicarbonate is 3.9 7 times 10 to the.negative fifth and that amount will go.down some bit the hydrogen ion.concentration initially is also 3.9 7.times 10 to the negative fifth that.amount will go up somewhat and we will.then produce for the first time some.carbonate ion so instead of X this time.I'm going to use Y we need to use ka to.here which is five point six times ten.to the negative 11th so this is much.smaller than the ka one value you can.see we've done products.in the numerator reactants in the.denominator you can see that this looks.like it's going to be a lot of work to.solve and it is we've simplified it here.and if you solve that for y feel free to.do it if you want or you could just.trust me.the additional concentration of H that.will be produced due to some of the.second protons popping off is right here.five point six times ten to the negative.11th molar now when we assumed that only.one proton popped off.we got three point nine seven times 10.to the negative fifth molar.those are a million times different in.other words this number that we computed.fairly easily by assuming only one.proton coming off and in fact we could.use a shortcut for that too that's a.million times more than all of then what.we get if we do all this work to find.out how much additional H+ is popped off.in other words the pH is still going to.be four point four zero and nothing will.have changed so if you have a in this.case let's say diprotic acid that's weak.if you assume that just the first proton.has been ionized you're pretty much good.to go.thanks for joining me.

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How can I fill out Google's intern host matching form to optimize my chances of receiving a match?

I was selected for a summer internship 2016. I tried to be very open while filling the preference form: I choose many products as my favorite products and I said I'm open about the team I want to join. I even was very open in the location and start date to get host matching interviews (I negotiated the start date in the interview until both me and my host were happy.) You could ask your recruiter to review your form (there are very cool and could help you a lot since they have a bigger experience). Do a search on the potential team. Before the interviews, try to find smart question that you are Continue Reading

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While I did not study physics this is something that relates to my field as well. One thing to remember is the scope of the field which you are talking about. With physics it might seem narrower than History or Archaeology but I suspect that when you boil it down it isn’t. It would be impossible to cover everything in a subject even going all the way through to gaining a doctorate. The answer you got and posted up is very accurate and extremely good advice. What a lot of it boils down to in education (especially nowadays) is not so much teaching specific facts but teaching themes and how to find Continue Reading

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