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Calculate the concentration of the acid [a] and the conjugatebase [b] in a 0.5M solution of sodium acetate buffer at pH 4.7.Express your answer in molar (M) to two ...

Question

Calculate the concentration of the acid [a] and the conjugatebase [b] in a 0.5M solution of sodium acetate buffer at pH 4.7.Express your answer in molar (M) to two decimal places. (pKa ofacetic acid = 4.7)

Calculate the concentration of the acid [a] and the conjugate base [b] in a 0.5M solution of sodium acetate buffer at pH 4.7. Express your answer in molar (M) to two decimal places. (pKa of acetic acid = 4.7)



Answers

A buffer is composed of formic acid and its conjugate base, the formate ion. (a) What is the pH of a solution that has a formic acid concentration of $0.050 \mathrm{M}$ and a sodium formate concentration of $0.035 \mathrm{M} ?$ (b) What must the ratio of acid to conjugate base be to increase the pH by 0.5 unit?

This question is pretty straightforward. If we know the two concentrations, the weak acid and the weak base, then pH will be equal to PK of formic acid. We can look that up in the back of the book today is 1.8 times tend the negative four. So we'll take the negative log of it to get the P K A plus the log of the base concentration, which is thesis odium form eight, divided by the acid concentration, which is formic acid, and we get 3.59 So if we're gonna have a ratio of acid to base that gives us a pH 0.5 units higher than this on the Ph is going to be 4.9 That will be equal to the PK a plus the log of the ratio of base toe acid. Do a little bit of algebra, and we get a base toe acid ratio of 2.21 or take the reciprocal. The question is asking for it the acid to base ratio and we get 0.45

So for this question, we are going to calculate the pH of a solution of acetic acid and sodium acid. So first of all, this is gonna be what's called a buffer solution. And what a buffer solution is is a solution of an acid and its consequent base. In this case, the acid is acetic acid, and the contra get base is going to be the estate. And so what a buffer solution does is when you add a bit of strong acid or strong base, it resists pH. It resists change in pH. So if you if you're adding acid or base, the pH is not going to change very much, so only gonna change a little bit. So the first thing that we can do is we can make a nice table, and what I stands for is the initial concentrations of all of these the change in the concentration of the things in the solution and the equilibrium concentration of all of these compounds. So, for water, water is the solvent in this reaction. So the concentration of water isn't gonna change because there is no concentration. It's just water. It's just stands alone. So the initial concentration of acetic acid is going to be what is given in the question. It's gonna be 0.50 multi. So I'm gonna write that right That right here to your 0.5 more and the initial concentration of our asked and I on is going to be the concentration of sodium acid ate the solution. So zero point 07 five smaller And this is the initial concentration is before the reaction I've written here has happened. So we've put things, we put everything in the solution. We have our acetic acid, we have our estate and ions and we haven't made any. We haven't made any hydro, hydro knee and ions yet, so that's going to be zero for the concentration for that. And so what's gonna happen is when the when they reaction here that I've written here occurs, we're gonna have a change in the concentration, so will represent that using X. So on the left side, we're gonna have minus X because we're this reaction is going to the right. It's an equilibrium so it can go left or right. But we are going to denote it, going to the right and then over here, the change is going to be positive acts. The change in the concentration of estate and the change in the concentration of hydro neon ions is also going to be X. And so the equilibrium concentration of each of these is just going to be the some of the initial and the change. So our acetic acid is going tohave. Equilibrium concentration of 0.5 Moller minus X minus are changed in the concentration that occurs when this reaction shifts deep delivery. And then for acetic for the estate and I on it's going to be 0.75 plus X and for the hydro knee, Um, I on It's going to me X because it's just zero plus facts. And so what we can do is weaken right and equal on equation for the equilibrium constant of acetic acid using these equilibrium concentrations that we just found. So the equation for the equilibrium constant is just going to be the concentrations of things on the left side of the equation over the things on the right side of the equation. So on the left side of the equation, we have our hydro knee, um, ions, the concentration of which is just X. And we have our Castillo and the concentration of that it's going to be zero point 075 plus x 075 plus X. I need to extend my line a little bit. And on the left side of the equation, we just have our acetic acid. So the concentration of that is going to be 0.50 minus X. And so the first thing that we can do to solve this equation for X, which is going to give us the concentration of the hydro knee, um ions is we can just cross these out. And the reason for that is that this is since it's a buffer solution, the concentrations of acetic acid and acid. Tate and I aunts are not really gonna change that much, and you can kind of see this is a very small number and these are very large numbers. So X is gonna be sort of a small number. So if we cross out these exes, it's still gonna be a really good estimate for the the ratios of the concentrations of asked and ions and and acetic acid. So to solve for acts we just Now what we can dio is we can just say that this is equal to x times the ratio of the concentrations. So that's going to be just 1.5 and then I'm going to erase a little bit to make some space up here. But what we're gonna do next is divided 1.8 times 10,000 negative, fifth by 1.5. And that is going to give us the concentration of hydro knee, Um, and of hydro knee, um, ions in this solution, and that works out to be that works out to be 1.8 times 10 to the negative. Fifth, divided by five is 1.2 is may. That's gonna be one point, Teoh times 10 to the negative fifth. And so that is so This number here is the concentration of hi Droney, um, Catalans. And so that is going to give us if we take the negative log of this number, that is gonna be our pH. Cause that's gonna be the concentration of hydrogen ions in the solution. And so that's going to tell us how acidic basic this solution is. So if we take the negative log of 1.2 times 10 to the negative fists and that is going Teoh equal, that's gonna equal four point nine to which is our final pH for this solution.

All right, So for this question, were given that we have 0.1 mostly leader of acetic acid and 0.25 most cheerleader off sodium acid Tate. The city cast is gonna be our acid. And so the masked he's gonna be your base, the concentrations of them. So we can put this into the Henderson Hospital equation to find the pH of our solution P h is equal to and the PK of ah, acetic acid buffer is 4.76 plus the lock 0.25 over play. One log of 0.25 over 0.1 is equal. 2.398 0.398 post plus 4.76 gives us a pH 5.16

So normally when we have a weak acid Kentucky based system, we apply the Henderson Hasselbach equation, but in this case we have the opposite of that and we have a base and weak base in the conjugate acid. So you have to apply the essentially the similar equations for conjugate basis and uh contact assets and week basis. So we're given information that the ph of this uh week based conjugate acid solution is 8.8 and the ah the P O. H is 14 minus the ph, Which means that the Ph. H. for this is 5.112. And were given information that our initial concentration of our conjugate acid Is equivalent to 0.25 molars. Our initial concentration of our Uh weak base is equivalent to 0.40 molars. So now we can essentially use this to find the KB for the week base. This becomes a 0.25 over 0.40. So we can plug in and solve for the PKB in this case. Mhm. Mhm. And we find that essentially the uh K. B. In this case is equivalent to 4.8 Times 10 to the -6. And this piece of information will be useful later. Now we're giving information that we're going to stress the system by adding hcl. So hcl in this case only can react with the base since the Conte acid is already predominated and a double permission wouldn't be favorable. So here we have our base reacting with H plus leads to we can assume this is a complete reaction since there's a significant difference in the relative pks of the two compounds. So this leads to the formation of B. H. Plus. So through this process the concentration of base will decrease while the concentration of our conjugate acid will increase. So we're given information that the the volume of our solution solution is 0.25 leaders. And since we know that concentration is expressed as the number of moles of our volume, we can find that the number of moles of base that we have initially is equivalent to .4 times .25, Which is equivalent to 0.10 moles. and by adding 0.0020 moles of a strong acid. This would be subtracted and become So minus this quantity here. This would become 0.0080 moles. And we can express this as a concentration by dividing over the volume. So this would be the new concentration of the base. So we can call this concentration based prime. However, you're essentially decreasing the concentration of your weak base but you're increasing the concentration of your you're going to get acid in this case. So we have to turn our contract acid into moles and add on the amount added. So Maybe 0.25 Squared Which is 0.0625. So this becomes plus 0.002. So I'd be $.25 times .25 leaders. So that be .0625. Then we're adding on this. So plus 0.002. And I made a numerical mistake here, that should be 0.0098 months. And by adding on this, this would become 0.0645 malls over the volume for the new concentration of R H B plus prime. So we can see that once we plug this into the Henderson House, uh the modified Henderson Hasselbach equation, the volumes are essentially going to cancel out. So we don't even need to use the volumes technically speaking and now we have to basically use our information for the KB before so we remember that we take the negative log in the KB to find the PKB Scroll back down. Then we have two plus the log, loss of concentration, loss of concentration or moles in this case is fine as well and we can find that the p O H in this case is equivalent to 5.14 Ph is 14 minus thoughts and we can find that the ph is equivalent to 8.86 which makes sense considering we added strong asset to a buffer system. So accordingly the ph will decrease and this is our final answer.


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