WEBVTT chemistry/ap-chemistry/hovasapian
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Hello, and welcome back to Educator.com, and welcome back to AP Chemistry.
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Today, we are going to continue our coverage of the AP practice exam, and we are going to go over the free response questions.
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This is going be Part 1; so let's just jump right on in.
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OK, so the first question says you have this reaction, ammonia plus water goes to ammonium ion plus hydroxide ion; that is a standard hydrolysis reaction.
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And the question reads as following: In aqueous solution, ammonia reacts as represented above (which--the equation--I'm going to write that down in just a moment); in a .0180 Molar NH₃ at 25 degrees Celsius, the hydroxide ion concentration is 5.6x10^-4 Molar.
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In answering the following questions, assume that the temperature is constant at 25 degrees and that the volumes are additive (that is an important part).
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This is the first question, and of course, these free response questions come in several parts, and the first part of the free response, you are actually going to have 40 minutes to do it.
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There is going to be one question that you answer, and there are going to be a couple of questions that you actually choose from: you are going to choose to answer one or the other.
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Now, of course, in my coverage, I am going to answer both of them; so all of the questions are going to be answered; but know that you actually have a choice on the AP exam.
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OK, so part A says: Write the equilibrium constant expression for the reaction represented above.
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All right, well, before we do anything, let's go ahead and write the reaction and write down what we know.
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Now, of course, you are going to have this in front of you, but I want it to be on a page here, so that we can reference it if we need to.
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So, we have: ammonia, plus water (H₂O, which I am going to write as HOH, simply because I like doing it that way, because it is easier for me to remember that the H actually breaks apart, and the OH becomes the hydroxide ion); now, this is in equilibrium with NH₄^+, plus our hydroxide ion.
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That is a good way of thinking about it.
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Now, they actually tell me that the concentration of the NH₃ (the initial concentration of NH₃...oh, that is fine; let's just go ahead and write NH₃ here) is going to be 0.0180 Molar.
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Molarity, moles per liter: again, nowadays (over the past 20 years or so), we have been using a capital M to represent molarity; I have been using an m with a line over it, so that just means molarity--moles per liter--concentration.
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Now, another thing that they give us here is actually the hydroxide ion concentration; they didn't necessarily need to do that, but it does make the problem a little bit easier.
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The hydroxide ion concentration is 5.60x10^-4 moles per liter.
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OK, so part A says: Write the equilibrium constant expression for the reaction represented above.
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OK, so this is going to be NH₃; this is aqueous, by the way; this is liquid; the NH₄, ammonium, is aqueous, and this is also aqueous.
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So, when we write any equilibrium constant expression, of course, anything that is liquid or solid does not show up in the expression.
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And, because this is a hydrolysis reaction, hydrolysis...
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...which is actually a really, really bad name for it--you want to think of it as more of an association constant, as opposed to a dissociation constant like an acid.
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An acid like HCl breaks up into H^+ and Cl⁻; a base like ammonia associates, in the sense that it actually pulls a hydrogen off the water and becomes NH₄^+.
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That is where this hydrogen is coming from.
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So, we have K < font size="-6" > b < /font > is equal to (you don't need to write the K < font size="-6" > b < /font > part; you can just write K; it's not a problem--no points will be taken off for that) the concentration of ammonium (NH₄^+) times the concentration of hydroxide, divided by the concentration of ammonia at equilibrium; that is it.
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This is part A: they are just asking for the equilibrium expression.
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OK, part B says: Determine the pH of a .018 Molar NH₃ solution.
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In other words, what is the pH of this solution?
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Well, this is where it got a lot easier: they already gave us the hydroxide ion concentration, so we can find the pOH.
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Well, pH is just 14.0 minus the pOH; well, the pOH is equal to -log of the hydroxide ion concentration, which is -log of 5.60x10^-4, which is equal to 3.25.
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Therefore, the pH is equal to 14.0 minus 3.25, giving us a pH of 10.75--a basic solution.
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Now, this confirms the fact that we actually did this right: 10.75 is basic, sure enough; hydroxide ion is what is floating around in solution, so we definitely confirm the fact that it is a basic solution; so we are on the right track.
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Whether we did the arithmetic right--that is a different story (we did do it right, but at least you have some qualitative way of deciding whether you are actually in the right place.
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If you ended up with a pH of 4.2, something is wrong; that is an acidic solution; bases create basic solutions by releasing hydroxide ion into solution.
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OK, so now, let's go ahead and do part C.
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Part C says: Determine the value of the base ionization constant K < font size="-6" > B < /font > : so they want to know what the actual numerical value of that is.
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Well, we have the expression for it: NH₄^+ + OH⁻ over NH₃.
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OK, so let's go ahead, and we are going to have to use a little ICE chart here: so let me go ahead and rewrite the equation: NH₃ + HOH in equilibrium with (you know what, that bothers me; I like my left arrow on the bottom)...goes to...NH₄^+ + OH⁻.
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Initial, Change, Equilibrium; OK, before anything happens, I start with 0.0180 Molar (when we do this ICE chart, we have to use molarity).
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Now, when they say a .018 Molar solution, that means that is how much they actually put in the solution, before anything happens.
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We start off with...that actually doesn't matter, so it's not 0; that doesn't matter; there is no ammonium, and there is no hydroxide.
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Well, a certain amount of this reacts; that same amount shows up here and shows up here; and the reason is because this is a 1:1 ratio.
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Our equilibrium concentration is 0.0180-x; this doesn't matter; this is +x, and this is +x.
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Well, here is what is really, really nice: they already gave us x; we know what x is--the hydroxide ion concentration and the ammonium concentration is 5.6x10^-4.
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Again, it made the problem a lot easier; I didn't have to deal with any algebra here.
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I know that x is 5.40x10^-4 Molar, so I take 5.40x10^-4 Molar; I put that one there for this one; I take the other x; I put 5.4x10^-4 there...I'm sorry, 5.6; I copied that 4 wrong--my apologies.
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This is 5.6x10^-4 Molar--is the hydroxide and the ammonium concentration: I put those up in the numerator: now the NH₃ equilibrium is .0180-x.
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So, I just put .0180-5.6x10^-4 in the denominator.
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Let me actually write that out so that you see it: so K < font size="-6" > b < /font > equals 5.60x10^-4, times 5.60x10^-4 (I have a tendency to write them separately, even though they are the same number, simply to remind myself that they are two different species), over 0.0180-5.60x10^-4.
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And when I do this, I end up with 1.80x10^-5.
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This is my value for the base dissociation constant of ammonia at this temperature; that is it.
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OK, now let's do Part D--Part D says: Determine the percent ionization NH₃ in .0180 Molar NH₃; so, in this particular case, what is the percent ionization?
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Well, if you remember, the percent ionization is the amount...a measure of the amount that is dissociated, over the total amount that you started with.
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A fraction--that is all it is; it is the part over the whole, expressed as a percentage (times 100).
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As far as the AP exam is concerned, you don't even need to express it as a percentage--you just need to write the decimal answer; so let's go ahead and do that.
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Percent ionization equals amount dissociated, and we know what the amount dissociated is: it is the amount of hydroxide ion, because for every x that dissociates from NH₃, that much ammonium is produced; that much hydroxide is produced; it's a 1:1 ratio.
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The amount dissociated, over the initial amount of the ammonia, times 100: and that gives us a percentage.
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Well, the amount dissociated is 5.60x10^-4, correct?
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The initial amount was 0.0180 Molar, times 100: that gives us 3.11%; or you can express it as .031, .0311...that is it; it doesn't matter--the AP exam doesn't care if you express it as a decimal or as a percentage.
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That is it, nice and straightforward, for that one.
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OK, now Part E--this is where it starts to get a little bit interesting, but again, not altogether difficult; we just need to keep track of the stoichiometry.
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So, we have several things going on: acid reaction, base reaction..we have to understand the chemistry.
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If you understand the chemistry, you can reason out the math; not the other way around, because at this level, the chemistry--the problems--it is not like you have one type of problem where you do the same thing over and over and over again.
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You don't just attack it in one algorithmic fashion; you have to know what is going on to decide what is happening, to decide how you are going to manipulate the equation or this or that.
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There are several parts going on: this is a true puzzle--you have to put the pieces together.
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OK, it says: In an experiment, a 20-milliliter sample of a .0180 Molar NH₃ was placed in a flask and titrated to the equivalence point and beyond, using a .0120 Molar HCl.
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Now, I'm presuming you have the question in front of you, so I'm not going to go ahead and draw pictures here.
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They just took the ammonia, and they decided to add some .0120 Molar HCl to it to equivalence point.
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Equivalence means that all of the base was neutralized by acid in equal amounts: neutralization means...titration--neutralization equivalence means pH 7: you bring it to an equivalence point.
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Not necessarily pH 7, but make sure that all of the hydroxide has been equally used up by an equal amount of hydrogen ion.
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Not pH 7...that is not equivalence.
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pH 7 is equivalence for a strong acid/strong base; this is a weak base, titrated with a strong acid.
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OK, now, the first part says: Determine the volume of .0120 Molar HCl that was added to reach the equivalence point.
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OK, so they said...so we have a certain amount of base, ammonia; that ammonia--when we are adding hydrogen ion to it, it is going to end up forming ammonium and hydroxide ion.
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That hydroxide ion is going to be neutralized; so, let's go ahead and find out how much ammonia is in there, that we can actually convert completely to ammonium by adding H^+ (in other words, completely producing hydroxide ion).
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OK, we have: 20.0 milliliters, times .0180 Molar ammonia solution; well, .0180 moles per liter...I'm going to work in millimoles per milliliter.
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OK, as long as your prefix is the same, you can use the same on the top and bottom: millimoles per milliliter.
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If it is Joules per mole, you can do millijoules per millimole, as long as you are consistent.
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So, let's do millimole per milliliter; and what you end up getting is 0.36 milliliters of NH₃.
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Well, 0.36 milliliters of NH₃ implies 0.36 millimoles of H^+ to add to completely convert all of that.
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OK, that implies that we need 0.36 millimoles of...not H^+; this is OH⁻--millimoles of OH⁻ can form, because the equation is 1:1.
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That means, for this much OH⁻ to form, we need .036 millimoles of H^+ ion.
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Well, the H^+ ion is going to come from the hydrochloric acid; therefore, we have 0.36 millimoles of HCl, times the molarity of the HCl; milliliters, millimoles; this time, the millimoles is on the bottom, because we need to convert it that way: it's 0.0120 millimoles per milliliter, and what we end up with is 30.0 milliliters.
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So, let's review this one again: I have a certain amount of ammonia; I'm going to titrate it--in other words, I'm going to add hydrochloric acid to that solution in order to use up all of the hydroxide that is produced.
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Well, if I have 20 milliliters of a .018 millimole per milliliters, I have (I'm getting my units all mixed up here) 0.36 millimoles of ammonia.
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Well, .36 millimoles of ammonia has the capacity to produce .36 millimoles of hydroxide; .36 millimoles of hydroxide requires .36 millimoles of hydrogen ion.
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.36 millimoles of hydrogen ion comes from 30 milliliters of this solution, given the fact that its molarity is .0120.
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So, .0112 there you go--I think the 2 is a little bit more clear there.
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30 milliliters of the hydrochloric acid solution will bring it to equivalence.
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OK, so now, let's do part 2: part 2 says: Determine the pH of the solution in the flask after a total of 15 milliliters of .0120 Molar HCl was added.
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OK, so we know that we need 30 milliliters to bring it to equivalence; they are asking what the pH in the flask is after 15 milliliters, which is half equivalence.
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This is important; I am going to do this problem in a complete analytic way, the way that we have done it--the way that I have described it--using an ICE chart.
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However, afterwards I will tell you how to do it in the short way, by recognizing that it is half equivalence; basically, all you have to do is: you have to take the negative log of the K < font size="-6" > b < /font > , what we found before.
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That automatically...half equivalence means the pH equals the pK < font size="-6" > a < /font > , or in this case, the pK < font size="-6" > b < /font > .
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But let's just do it analytically, the way that we normally do; that way, we never get lost.
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OK, so let's see what we have: so again, here we have to do the stoichiometry first.
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In other words, we have to take care of the number of moles that are reacting and how many moles are left over, because you are dealing with volumes.
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Any time you add one volume to another, the volume changes; that means the molarity changes.
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We have to make sure to do the stoichiometry first, so we write: stoichiometry first.
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Now, they are saying we have 15.0 milliliters of the hydrochloric acid solution--so 15.0 milliliters, times 0.0120 millimoles per milliliter, gives us 0.18 millimoles of hydrogen ion.
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Well, remember, we started off with 0.36 millimoles of NH₃; if we add .18 millimoles of H^+ to it, that means we are going to use up half of this, .018 millimoles of NH₃ (right?--because for every mole of this, one mole of that is going to react).
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That is going to leave us with .18 millimoles of NH₃; OK.
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But now, we have added 15 milliliters to the initial 20 milliliters of ammonia solution; so our total volume is now 35 milliliters.
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So now, we have 0.18 millimoles of ammonia floating around in 20+15 milliliters; so that changes the molarity, so now, the molarity is 0.005142 moles per liter.
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Now, this is the molarity of the ammonia; this is the stoichiometry--this is before anything has had a chance to come to equilibrium.
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Remember, when we do these problems, we do the stoichiometry first; we act as though the reaction that is running--the titration reaction--the H^+ plus the OH⁻ formed from the NH₃--goes to completion.
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And then, we allow the system to come to equilibrium; so now that we know that we have this much molarity ammonia, now we have to allow it to come to equilibrium.
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OK, let's see: another thing we should probably realize here--it is probably a good idea to write the equation: NH₃ + H₂O (I know, I didn't write it as HOH this time; it is not that necessary here) + OH.
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This ammonia--in the process of being titrated that way, it not only produces this much OH⁻ that can be neutralized with the H^+, but in the process, ammonium is also produced.
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You should know that the ammonium concentration is also 0.005142 molarity--that is very, very important.
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In the process of this reacting, if this much reacted--if .18 millimoles of NH₃ reacted--that means .18 millimoles of ammonium (NH₄^+) was created.
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That is the whole idea: so now, I not only have .005142 Molar NH₃ floating around, I have .005142 Molar NH₄^+ floating around.
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Watch this very, very carefully.
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OK, now we will do the equilibrium.
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Now...actually, you know what, let me just write "equilibrium second"; so we did stoichiometry first; now we do the equilibrium part.
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Now, I'm going to rewrite the equation one more time: NH₃ + H₂O (so I can do my ICE chart) goes to NH₄^+ + OH⁻.
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My initial; my change; my equilibrium concentrations--I'm going to put that in the equilibrium expression, with the 1.80x10^-5 that I calculated up above.
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OK, so my initial concentration is 0.005142 for ammonia; water doesn't matter; the ammonium is the same thing (I'm going to do this in red).
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My initial concentration, before the system has come to equilibrium, is 0.005142; this is strictly by virtue of adding the acid.
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I added the acid; I used up half the ammonia; I created that much ammonium; the OH⁻ is taken care of with the OH^+.
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This is 0; now the system comes to equilibrium; a little bit of this dissociates; a little bit of that shows up; a little bit of this shows up; and now, we have 0.005142-x; I have 0.005142+x; and this is +x.
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Now, I plug this, this, and this into the K < font size="-6" > b < /font > expression.
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So, what I end up with is the following: the K < font size="-6" > b < /font > that we found before is 1.80x10^-5 (not a lot of dissociation--OK, that is what the K < font size="-6" > b < /font > tells me) equals 0.005142+x, times x (the ammonium concentration times the hydroxide concentration), over the ammonia concentration, 0.005142-x.
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I need to find x; that is my OH⁻; I'm going to take the -log of that to get my pOH; I'm going to subtract that from 14 to get my pH.
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Be very, very clear about what they want: they are asking for pH here.
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Now, notice: .005142+x; .005142-x; x...the dissociation is probably really, really small, so for all practical purposes, I can ignore that x, and I can just write this as 0.005142, times x, over 0.005142.
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Well, this and this just leave x; therefore, x (which is the OH⁻ concentration) is equal to 1.80x10^-5.
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I get a pOH equal to 4.75; and therefore, my pH is equal to 14.00-4.75, which is equal to 9.75.
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That is the pH of my solution after 15 milliliters have been added.
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Those 15 milliliters of H^+ that I added (of the HCl) is .18 millimoles of H^+; it reacts with .18 millimoles of the OH⁻, so that is taken care of, leaving .18 moles of NH₃.
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That NH₃ starts to dissociate a little, creating some more hydroxide ion; that hydroxide ion is why this solution is actually basic, at 9.75.
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Make sure you understand the chemistry: if this is confusing, go back to the lessons where we discussed weak acid-weak base chemistry; we did a lot of problems in this, and we were very, very systematic about what is going on.
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You need to understand what is happening.
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OK, so let's see what our final section says: this is part 3 of part E: so now, it says: Determine the pH of the solution in the flask after a total of 40 milliliters of the .0120 Molar HCl was added.
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OK, so now, we have...well, we know that 30.0 milliliters from the first part of section E was used to bring the solution to equivalence (to the equivalence point).
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So, they say they are adding 40 milliliters; so "equivalence point" means there is no more base to react with; so now, what you have is an extra 10 milliliters (40 milliliters minus the 30 milliliters to bring it to equivalence).
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So now, you have an extra 10 milliliters: so, 10.0 milliliters, times 0.0120 millimoles per milliliter (which is the molarity of the hydrochloric acid solution)...
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I hope you all understand what it is that is happening here: there was a certain amount of ammonia, .36 millimoles; to bring it to equivalence, we added 30 milliliters of this hydrochloric acid solution.
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Well, by adding this H^+, we converted all of the ammonia over to ammonium and OH⁻.
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OK, .36 millimoles produce .36 millimoles of OH⁻; that .36 millimoles of OH⁻ reacted with the 30 milliliters of the acid, .36 millimoles of hydroxide ion, to form water.
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That is it--that is all used up; there is no more ammonia in the solution, so now, any excess acid that I add is going to be pure acid: that is what is going on here.
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It is going to be pure acid.
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OK, so we end up with...let's see, where are we?...10 times .0120; we get 0.120 millimoles of HCl.
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Well, HCl is a strong acid; it completely dissociates into H^+ plus Cl⁻; in case you need to see that...that means .12 millimoles of HCl produces...that implies...0.120 millimoles of H^+.
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Well, we want the pH of the solution, so let's just take the negative log of...oh, no, no; sorry, sorry, we can't do that; look at that--I almost made the same mistake that I am telling you to be careful of.
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We have .12 millimoles of H^+; now we need to find the molarity, because we are doing concentration.
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So now, 0.120 millimoles of H^+...but the total volume...I have the initial 30 milliliters...no, I have the initial 20-milliliter sample (right? 20 milliliters), into which I have now added 40 milliliters; so now, my molarity is .120 millimole, divided by 60 milliliters.
00:29:43.000 --> 00:29:52.000
I end up with a molarity of .002 Molar; this time, I'll go ahead and use the capital M.
00:29:52.000 --> 00:30:05.000
So, .002 molarity of hydrogen ion, because, when we take the negative log of something, we are taking the negative log of the hydrogen ion concentration, not the number of moles...
00:30:05.000 --> 00:30:07.000
So now, we can go ahead and finish off the problem.
00:30:07.000 --> 00:30:27.000
The pH equals -log(0.02); OK, and we get 2.70 (is the final pH).
00:30:27.000 --> 00:30:34.000
I hope that made sense: there was nothing in here that was altogether that difficult, but you do definitely need to understand the chemistry.
00:30:34.000 --> 00:30:49.000
Of all the questions that I think can come up in the AP exam, these acid-base and these solubility-product questions, I personally think, are the most difficult, because there is a lot of chemistry going on.
00:30:49.000 --> 00:31:00.000
Thermodynamics is actually not that bad, but it's these equilibrium problems, where you really need to understand what is happening in order for you to put it together the way you need to.
00:31:00.000 --> 00:31:07.000
OK, so let's go ahead and move on to the next question: let's see what we have here.
00:31:07.000 --> 00:31:16.000
Question #2: OK, so again, you are going to be given a choice of answering either question #2 or #3; we are, of course, going to do both.
00:31:16.000 --> 00:31:28.000
Question #2: OK, answer the following questions regarding light and its interactions with molecules, atoms, and ions.
00:31:28.000 --> 00:31:51.000
Part A says: The longest wavelength of light with enough energy to break the Cl-Cl bond (so we have the chlorine bond in Cl₂ gas) is 495 nanometers.
00:31:51.000 --> 00:31:58.000
So again, the longest wavelength of light with enough energy to break the Cl-Cl bond is 495 nanometers.
00:31:58.000 --> 00:32:11.000
Let me make my nanometer a little bit clearer: nm; and in case you don't know, nano- is 10^-9, so this is 495x10^-9 meters.
00:32:11.000 --> 00:32:20.000
OK, 1 says: Calculate the frequency in inverse seconds (or hertz) of the light; OK.
00:32:20.000 --> 00:32:29.000
Well, we know that there is a relationship between frequency and wavelength, and that is: frequency equals the speed of light over the wavelength.
00:32:29.000 --> 00:32:46.000
We just put it in: the speed of light is going to be 3.0x10⁸ meters per second; and we are going to express 495 nanometers...now, we don't just put the nanometers; units need to match.
00:32:46.000 --> 00:33:06.000
When doing most problems in chemistry, you have to watch your units, especially when you are dealing with thermodynamics, and particularly when you are dealing with quantum mechanics--things like this--because you are going to get nanometers; you are going to get centimeters; you are going to get micrometers; you are going to get meters; so be very, very careful.
00:33:06.000 --> 00:33:14.000
495 nanometers...well, that is just the same as 495x10^-9 meters.
00:33:14.000 --> 00:33:30.000
We need meters to cancel with meters; and when we do that, we get 6.06x10^14 inverse seconds, or if you want to write hertz, that is not a problem.
00:33:30.000 --> 00:33:35.000
That is the frequency of the light, based on the wavelength, which was 495.
00:33:35.000 --> 00:33:39.000
OK, I'm going to go ahead and switch over to blue here.
00:33:39.000 --> 00:33:48.000
Now, part 2 says: Calculate the energy in Joules of a photon of the light.
00:33:48.000 --> 00:33:56.000
Well, the equation for energy...and by the way, these equations--you don't have to have them memorized; they are going to be on the AP exam for you.
00:33:56.000 --> 00:34:04.000
You are going to be given several pages of equations, and they are actually going to tell you what the equations are related to.
00:34:04.000 --> 00:34:15.000
You need to know how to use them, of course, but you don't have to memorize any of these; you will be given a periodic table, a table of reduction potentials, equations...pretty much everything that you need, as far as reference is concerned.
00:34:15.000 --> 00:34:22.000
The energy of a photon is equal to Planck's constant times the frequency.
00:34:22.000 --> 00:34:30.000
OK, so we just put it in--nice, easy, simple calculation; we just need to make sure that the units are right.
00:34:30.000 --> 00:34:48.000
Let's see--Planck's constant: 6.626x10^-34 Joule-seconds, times 6.06x10^14 inverse seconds.
00:34:48.000 --> 00:35:02.000
Second cancels with second, and you are left with a number of 4.02x10^-19 Joules...and yes, they wanted the answer in Joules, so we leave it in Joules.
00:35:02.000 --> 00:35:14.000
Definitely make sure you include the units here: in general, the units are not too big of a deal, but if the question actually specifies that they want it in a specific unit, you are actually going to lose points if you don't put the right unit in there.
00:35:14.000 --> 00:35:20.000
That is it: frequency equals c/λ; energy equals Planck's constant times the frequency.
00:35:20.000 --> 00:35:29.000
OK, now, part 3 says: Calculate the minimum energy, in kilojoules per mole, of the Cl-Cl bond.
00:35:29.000 --> 00:35:40.000
OK, calculate the minimum energy in kilojoules per mole of the Cl-Cl bond; it takes one photon to break one Cl-Cl bond.
00:35:40.000 --> 00:35:46.000
Well, if I have one mole of these, I just basically multiply by Avogadro's number.
00:35:46.000 --> 00:36:04.000
So, let's see what we have: I have 4.02x10^-19 Joules per photon; that is what I calculated.
00:36:04.000 --> 00:36:24.000
This e=Hν--that is the energy of one photon; so it's Joules per photon...times...I have 6.02x10^23 photons in a mole (right?--it's particles per mole; in this case, the particle we are talking about is a photon).
00:36:24.000 --> 00:36:34.000
Well, so now, photon cancels photon; I have Joules per mole, but they specifically said they want kilojoules per mole.
00:36:34.000 --> 00:36:44.000
Oh, I should probably let you know: I have a bit of a habit here--kilojoule is a small kJ; I have a habit of always writing it with a capital KJ.
00:36:44.000 --> 00:36:58.000
In general, it's not a problem, but those of you who ever work on certain computer software programs--if you enter a capital KJ, it will actually tell you that it is wrong, so just change it to a small kj; it's just one of those things.
00:36:58.000 --> 00:37:10.000
They want kilojoules per mole; well, one kilojoule is equal to 1,000 Joules; so now, we can cancel the Joule, and the final unit we are left with is kilojoules per mole; everything is good.
00:37:10.000 --> 00:37:23.000
When we multiply this times that and divide it by 1,000, we get the answer: 242 kilojoules per mole.
00:37:23.000 --> 00:37:39.000
That means, if I have a mole of chlorine gas, I need to hit it with 242 kilojoules of energy of that frequency in order to break every single bond in that gas sample.
00:37:39.000 --> 00:37:48.000
OK, now let's go to part 2 (or part B, actually--part B, which also has several parts).
00:37:48.000 --> 00:38:01.000
The first part of part B says: Well, now a certain line in the spectrum of atomic hydrogen is associated with electronic transition in the H atom from the sixth energy level to the second energy level.
00:38:01.000 --> 00:38:06.000
So, actually, let me go ahead and...before I start the first part...
00:38:06.000 --> 00:38:22.000
We have n=6 (the sixth energy level) going to n=2; this is a drop in energy--that means the electron is going from an excited state down to a lower state--it has been excited from 2 to 6; now it is dropping down from 2 to 6.
00:38:22.000 --> 00:38:28.000
It is releasing that energy, and it is causing a line to show up in the hydrogen spectrum.
00:38:28.000 --> 00:38:38.000
OK, so 1--it says: Indicate whether the H atom emits energy, or whether it absorbs energy, during the transition; justify your answer.
00:38:38.000 --> 00:38:51.000
Well, we just answered that question: energy is going from the sixth energy level...the electron is dropping down to the second energy level; that is a drop in energy.
00:38:51.000 --> 00:38:56.000
When it drops in energy, it actually emits that extra energy as light.
00:38:56.000 --> 00:39:08.000
So, the H atom emits: OK.
00:39:08.000 --> 00:39:31.000
In other words, higher to lower transition is emission; lower to higher is absorption.
00:39:31.000 --> 00:39:40.000
OK, now we will do part 2--part 2 says: Calculate the wavelength, in nanometers, of the radiation associated with the spectral line.
00:39:40.000 --> 00:39:51.000
OK, we want to find the wavelength: well, there is going to be an energy difference.
00:39:51.000 --> 00:39:58.000
That energy is going to be given off as light; that energy is going to be associated with a certain frequency; that frequency is going to be associated with a certain wavelength.
00:39:58.000 --> 00:40:03.000
That is what we need to do--basically just work our way backwards from energy to frequency to wavelength.
00:40:03.000 --> 00:40:17.000
That is it; so, the Δenergy equals the energy of the sixth level, minus the energy of the second level (right?).
00:40:17.000 --> 00:40:32.000
However, let's...you know what, Δ...initial...Δ, by definition, is final minus initial; so the final is actually the second level; the sixth is the first level, the initial.
00:40:32.000 --> 00:40:46.000
This is actually going to be a transition; the ΔE, mathematically, is going to be the energy of second level, minus the energy of the sixth level (final minus initial--just by definition, that is what Δ is).
00:40:46.000 --> 00:41:07.000
Let's see: the equation for the energy of a given level E < font size="-6" > n < /font > is equal to -2.178x10^-18, divided by n²; and that is in Joules.
00:41:07.000 --> 00:41:26.000
This is one of the equations that is actually given to you; for a particular energy level (for a particular primary quantum number--that is what n is--primary quantum number), the energy associated is equal to -2.178x10^-18 over n², that quantum number squared; and that is in Joules.
00:41:26.000 --> 00:41:48.000
The energy of the sixth level equals -2.178x10^-18, divided by 6 squared, which equals -6.05x10^-20 Joules.
00:41:48.000 --> 00:42:05.000
E₂ is equal to -2.178x10^-18, divided by 2 squared, equals -5.45x10^-19 Joules.
00:42:05.000 --> 00:42:15.000
Therefore, my ΔE is equal to 4.84x10^-19 Joules.
00:42:15.000 --> 00:42:35.000
Well, energy (or Δenergy that it releases---this is the amount of energy that is released)--that energy is equal to Planck's constant, h, times ν, which is equal to (well, we know what ν is; ν is c over λ) hc over λ.
00:42:35.000 --> 00:42:45.000
And, when we solve this equation for λ, we get a wavelength that is equal to Planck's constant, times the speed of light, divided by the energy.
00:42:45.000 --> 00:43:10.000
Well, that is it--we just put the numbers in: Planck's constant is 6.626x10^-34 Joule-seconds; the speed of light is 3.0x10⁸ meters per second; divided by energy, which is in Joules, which is 4.84x10^-19.
00:43:10.000 --> 00:43:16.000
You know what I am going to do?--I'm actually going to put the units in here, so that you see how the units work out.
00:43:16.000 --> 00:43:36.000
This is going to be...let me write: 6.626x10^-34 Joule-seconds; this second cancels this second; this Joule cancels this Joule; my final unit is going to be in meters; so I'm going to switch the page here.
00:43:36.000 --> 00:43:47.000
That is going to be 4.11x10^-7 meters; now, what do they want it in?
00:43:47.000 --> 00:44:16.000
They actually wanted it in nanometers, which is...so this meter is 4.11x10^-7 meters, times 10^-9 nanometers, which is equal to 4.11x10² nanometers, which is equal to 411 nanometers.
00:44:16.000 --> 00:44:24.000
There is your answer: we found the difference in the energy levels; we took that energy; we set it equal to Planck's constant times the frequency.
00:44:24.000 --> 00:44:29.000
Frequency is equal to c/λ again, we just have a couple of equations to work with when we are dealing with light.
00:44:29.000 --> 00:44:35.000
It's pretty much ν=c/λ and energy=h times ν.
00:44:35.000 --> 00:44:39.000
That is it, and we get 411 nanometers.
00:44:39.000 --> 00:44:43.000
It is going to emit energy of that wavelength.
00:44:43.000 --> 00:44:53.000
That is it; OK, now let's go ahead and finish off this section with part 3 of B.
00:44:53.000 --> 00:45:13.000
Account for the observation that the amount of energy associated with the same electronic transition (that is, 6 to 2) in the helium + ion is greater than that associated with the corresponding transition in the hydrogen atom.
00:45:13.000 --> 00:45:22.000
OK, what we did is: we just calculated the energy associated with a 6-to-2 transition of the hydrogen atom, 411 nanometers.
00:45:22.000 --> 00:45:41.000
They are saying that the same transition--the sixth primary level down to an atom in the helium +, so helium has 2 electrons, right?--1 of the electrons has been pulled away from it, leaving basically a helium atom (a helium atom has 1 electron).
00:45:41.000 --> 00:45:56.000
That electron makes a transition from level 6 to level 2; but, as it turns out, they are telling us that the transition from level 6 to level 2 has greater energy than the one that we did for the hydrogen atom.
00:45:56.000 --> 00:45:59.000
They are asking us to explain that: how can you explain that?
00:45:59.000 --> 00:46:10.000
Well, it's very, very simple: an energy transition...you need energy to go from level 2 to level 6--it takes a certain amount of energy to go up that way.
00:46:10.000 --> 00:46:14.000
So, when you drop back down, that extra energy is emitted.
00:46:14.000 --> 00:46:25.000
Well, hydrogen only has one proton in its nucleus; so that one electron that has to jump up has a certain pull on it from this positive.
00:46:25.000 --> 00:46:38.000
Well, helium has 2 protons in its nucleus; therefore, it is harder for that one electron: now it has to struggle twice as hard to jump up to the sixth level.
00:46:38.000 --> 00:46:49.000
Well, because it has to struggle twice as hard (in other words, we have to put more energy to push it up to the sixth level), when it drops back down, it releases more energy.
00:46:49.000 --> 00:47:06.000
That is the explanation: that is it--it is that simple: you can draw out a picture, or you can simply say, "Because helium has 2 protons and a greater positive charge in its nucleus, the atom, in order to make the initial transition from n2 to n6...more energy has to be put in."
00:47:06.000 --> 00:47:11.000
"Therefore, when it drops back down from level 6 to level 2, it is going to emit more energy."
00:47:11.000 --> 00:47:15.000
That is the answer to section #3.
00:47:15.000 --> 00:47:22.000
OK, thank you for joining us here at Educator.com; we will see you next time for a continuation of the free response questions; goodbye.