Calculate the concentration of all species in a 0.155 M solution of H₂CO₃.

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Accepted Answer

Hey there, welcome back. Alright, so in this practice problem we're given a di protic acid, H two A. This is just a generic way to say die protic acid. Remember diplomatic assets have to aesthetic hydrogen is which we have right there. Right? And as because it's a di protic acid, it's going to have to um equilibrium constants, right? K one and K A two. So we have the values for that. What we need to calculate here is the ph of 20. molar solution of H two A. We also need to calculate the concentration of H two A and a two minus for the solution at equilibrium. So, first we're going to be able to find the ph of this dye protic acid as we can see the K. A. The first one is times 10 to the negative six. Right? So we know it's a weak acid here. Um So the concentration here we're going to be able to find but we do need to do a nice table. And then after we find ph we're gonna be able to calculate the concentration of um H two A. And then after that concentration of its blast conjugate base right after it has lost both of the hydrogen. So because this is a die protic acid we are going to need to do to um ice tables because we do need to do this in two steps. We can't um you know, write an equation and lose two hydrogen all at the same time. It has to be a stepwise Process here. So let's go ahead and write out the equation. So we have H two a and of course this is going to be Aquarius because it's Equus right? It's going to be in water. And remember water is usually liquid. So we are going to have an equilibrium because we're dealing with a weak acid here and first we're going to lose just one hydrogen, right? So after the hydrogen is gone, is going to be a church A But because H plus has associated H is going to be a chait minus, still equals and then water is going to be acting as a weak base and it's going to accept that H plus from the dye protic acid from the weak die protic acid. And we're going to have that H +30 plus iron, the hydro Nehemiah on and it's going to be Aquarius. Alright, so let's go ahead and constructor ice table. So we have I. C. E. Alright, so as always with equilibrium we're going to ignore liquids and solids so we don't need to fill out anything for water. Now we do have an initial concentration of this dye protic acid and it is 0.700 molar, So .07 00 moller. Now that similarity and then initially we don't have any of the products made yet. So it's just going to be zero for both of them for for the H A minus for the conjugate base and the hydro nia my on. now with change for C. We're going to lose reactant, we're gaining products so we're going to Lose H two A. It's going to be negative X. And then for the product it's going to be plus X. For H a minus and then plus X for H +30. Plus. Okay, and then at equilibrium, we're going to basically have um for the reactant, what we started with minus what we lost. So it was going to be 300.7 00. Just realize that it's still in polarity. I'm not going to include that in their minus X. And then for the products for both of the products, we're going to have X and X at equilibrium. Okay, So first thing that we need to do is actually solve for X. And here we can see that the hydro Nehemiah N is X. And also the H a minus X. So it doesn't matter which one they're the same excess. We need to solve for X. Because what is ph ph is just negative log of the hydro nia my on, right? So once we have X we can figure figure out the ph and then from that we can figure out the concentration of our reactant, right? Because if we have X, we just take 0.700 minus the X. Okay, for the first part. So let's go ahead and do that. We're going to use the first K. A. Oops. So for the first reaction we're using K. A. one. Let me try that again. So okay, A one which is 4.2 times 10 to the negative sex. Right? So let's go ahead and take that K. And write the equilibrium expression. Okay, A one Is given to us and that is 4. times 10 to the negative six. And it's going to equal two products. Overreacting. So our products here X. And X. So we can just write X squared. Right? And then over reacting just one reactant and it is zero. Hold on a 2nd 0.700 minus X. So now we're solving for X. But here we have X squared and then negative X. On the bottom. Remember we can use um this rule of 500 where we take the initial concentration of our reactant whether it's an acid or a base. So here we have an acid. So you're going to take the concentration of that which is 0.700. And then you're going to divide it by its K A K A one. Okay. Not K. Two. But the K one because K one is the one that's associated with H. Two. A. So once you do that, go ahead and divide those numbers 0.7 by K. A. One which is 4.2 times 10 to the negative six. And when you do that you actually see that you get a number that is greater than 500. Okay, we don't really need that value. Just make sure that it is greater than 500. If it is, then we can actually go ahead and just um ignore that negative X. On the bottom. So the K one expression just becomes X squared over 10.700. Now it's much easier to find X. Right, Alright, so now let's go ahead and do the math here. So we're going to go ahead and multiply both sides by .0700. And on the left side we're going to get 2. times 10 to the negative seven. And then it's gonna equal to X squared because we just want X by itself. Not X squared. We're going to go ahead and square root both sides. Right, let's go ahead and find that number. All right, so, square root of both sides, we're going to get X And X is gonna equal to 5.4 Times 10 to the -4. Okay, so X. Here we said, corresponds to the concentration of H 30 plus of hydrogen ion at equilibrium. Alright, so, since we have that to find p H R. First answer here, we're going to take negative log of the hydro nia mind concentration. Right, That's the the equation here. So let's go ahead and plug that into a ph equals to negative log of 5. Times 10 to the -4. Okay, so once you put that into your calculator, we should get a ph of 3.27. Okay, so the ph of H two a. With the concentration of .0 700 moller is 3.27. Okay, so 3.27. So looking at the answer choices, we have 3.7 as answer A. Or D. So we need to go ahead and obviously keep answering the question. So next we want the concentration of H two A. Now H two A is right here and our first reaction and again we can go ahead and find this concentration at equilibrium because we already know what X is. So we just need to do one more step. Alright, so concentration of H two A equals two. Now this is at equilibrium equals 2.700 minus X. And X. Here. Let's go ahead and plug in the X. So x is 5.4 Times to the -4. Right? So when you do that subtraction, you get 0. .0695. And of course it's in my polarity polarity of each to a. Alright, so that's equilibrium. So what we started was what we started with was . Moller initially. And then after equilibrium after You know it lost one hydrogen the v equilibrium basically what's left over of that type product acid is .695. So as you can see almost all of it is still intact because it's weak. Okay, so that is our second answer. So let's go ahead and take a look at our answer choices. So it looks like option A here has to be our correct answer because we have our ph of 3.27 we have our concentration of the dye product asset at equilibrium is 0.695 moller. The other one does not match. But how do we actually find the concentration of A 2 -? That is not in our first equation. Right? So what we need to do here is of course you can go ahead and just pick a but if you want to know how to so for that let's go ahead and keep doing that. Right, let's keep going. So now we're going to start with H a minus. That is going to be our initial because we're taking um what is already basically um ionized, right H a minus is going to be the conjugate base of the dye protic acid, but it is going to also act as a as a weak acid and still lose the second hydrogen. Of course it's going to be much, much harder for it to lose its it's even weaker now, but we're going to go ahead and take that and put it in water. So we have H A minus Aquarius put it in water. So this is our Oops, 20. Is there a second equation Still a equilibrium? And I was going to finally lose the second hydrogen and become a minus but -2. Right? So there is that iron right there. That's the concentration that we want to go ahead and solve for. And of course water here on the left side is going to pick up that H class and become a cheerio plus and now we do the ice shirt just like we did previously for the first reaction. Okay, so I. C. E. Alright so initially of H A minus, how much do we have if we take a look here in the previous reaction, H A minus um at equilibrium was equal to X. And X was equal to 5.4 times 10 to the negative four. Right? So that is going to be what we're starting with. That's the initial concentration of H. A. So we have 5.4 Times 10 to the -4. And of course smaller as well. Water, we ignore. And then um initially we don't have any A two minus yet. But what about Hydro New Mayan? Well, we actually do have some already starting um in that reaction right, we're still dealing with the same solution. So we already have X concentration of H 30 plus, which is the same thing as the concentration of H a minus. So it's also 5.4 times 10 to the 94 as the initial. So let's go ahead and write that. So 048 to minus. And then 5.4 times 10 to the negative four Mueller initial for H 30 plus. Now again change, we're going to lose reactant two minus X. Gain products plus X plus X for both of the products. And then at equilibrium we're going to have for the reactant, what we started, what, what we started with minus what we lost. So 5.4 times 10 to the negative four minus X for the reactant. And then for the products for a two minus, we're just going to have X at equilibrium. And then for hydro Nehemiah and we're going to have 5.4 time stands in the negative four plus X. Okay, so now these are our concentrations at equilibrium. And now since we're dealing with this second reaction where we are organizing the second hydrogen, we have to look at K. A. Two. Okay. And K two, if we scroll up here, it's given to us, it is going to be 6.8 times 10 to the -9. Okay, so let's go ahead and write the K two expression now. Okay, a two equals to 6. Times 10 to the -9. And then the expression becomes again, products are reacting to the same thing but we just have different values. So here we have x Times 5.4 Times 10 to the -4 plus X. So make sure you put 5.4 times 10 to the negative four plus X. And parentheses and then multiplied by X. And then divided by and again for the reactant we have 5.4 times standard ng and negative four minus X. And that also goes on the bottom and parentheses. Alright, so now we're solving for X. We want to go ahead and solve for X. Why? Because X is going to be the concentration of a two minus and that is what we want to find equilibrium. The concentration of that final conjugate base of this dye protic acid. Right? We want its concentration. So we need to find X. But right here we have a lot of excess and this looks very intimidating. Very complicated. So let's go ahead and see if we can actually get rid of some of these exes. Okay, so for the bottom remember we're going to use the same role. We're going to go ahead and take the initial concentration of our asset and in this case for the second reaction we're dealing with a che minus. Right? So you take 5.4 times 10 to the negative four And you're going to divide it by its K which is KA two. Okay too. And we actually get a very large number is much greater than the 500 as you'll see. Which means that we can actually go ahead and ignore that negative X. On the bottom. What about the top one? Well, the same thing. Okay, because that H is it does not want to come off of H a minus because it is so weak. So little amount of it comes off that it doesn't even really matters. So that plus X. It's so so small. That minus X. And plus X. Number is very tiny. So we can actually ignore both of these, the positive X. On the numerator and the negative X. In the denominator. So this becomes much easier to deal with. Okay, so what we're going to be left off with is that we have 5.4 times 10 to the negative four being divided by 5.4 times 10 to the negative four. So those two numbers will actually cancel out and we're just going to have that one X. That was being multiplied on the top on the numerator. Ok, so this just becomes 6.8 times 10 to the negative to the negative nine, which is K two equals two X. So that is actually X. And that is what's going to happen with dye product assets with weak democratic acids once they go all the way and lose all of their hydrogen the concentration of the hydro knee. Um iron is so very small that that is being lost. That the concentration of the leftover of the conjugate base is just gonna equal to the K. A. Okay, because X. Is going to be super small. And that second reaction, so X. Here equals to of course to the concentration of a two minus. And folks, that is it. That is going to be our final answer. So let's go ahead and see if we did this right? So 6.8 times 10 to the negative nine and here we have six times 80 times 10 to the negative nine molar. So yes, that is correct. So our our correct answer here is option A because the ph matches what we got, the concentrations of both the dye protic acid at equilibrium and the concentration of um the conjugate base on the very last step of that dye protic acid also matches with what we got. Alright folks, I know this is kind of long because anytime you're dealing with diplomatic acids and there are they're going to be losing their hydrogen. Not altogether, but step by step. So you do unfortunately have to do an ice chart two times for di protic acid. Okay. Alright folks, that is it. Thank you so much for watching and we'll see you in the next video

Calculate the concentration of all species in a 0.155 M solution of H₂CO₃.

Verified Solution

Key Concepts

Acid-Base Chemistry

Dissociation Equilibrium

Concentration Calculations

Calculate the concentration of all species in a 0.155 M solution of H2CO3.

Verified Solution

Key Concepts

Acid-Base Chemistry

Dissociation Equilibrium

Concentration Calculations

Calculate the concentration of all species in a 0.155 M solution of H₂CO₃.