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Analytical Chemistry

Learn the toughest concepts covered in your Analytical Chemistry class with step-by-step video tutorials and practice problems.

8. Monoprotic Acid-Base Equilibria

Weak Acid-Base Equilibria

WEAK ACIDS and WEAK BASES are weak electrolytes that do not completely ionize in solution, but instead form an equilibrium. 

Weak Acid-Base Equilibria
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example

Weak Acid-Base Equilibria Calculations 1

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So here it asks us, what is the original polarity of a solution of weak acid with a K. Of 4.7 times 10 to the negative three and a ph of 4.12 at 25 degrees Celsius. Now, the key here is realizing that we're dealing with a weak acid and remember we know it's a weak acid because it's K. Value is less than one with a weak acid. We're going to say that our generic weak acid we can say represents H A weak acids, weak bases react with water in order to establish equilibrium. Remember the acid here, according to the Bronston Laurie definition, will donate an H plus. So it's going to become here a minus plus H +30 plus. So we have initial change, equilibrium water is a liquid. Remember liquids and solids are not included within an ice chart here. The we're we're told to find the original polarity. So that's our formal concentration. We don't know what it is. So we'll just place F. Here. These initially are zero. Remember we're losing our reaction over time in order to produce products. So this will be minus X plus X plus X. Bring down everything F minus X plus X plus x. Knowing this, we set up the equilibrium expression as K equals products. Overreacting so it's K equals a minus times H 30 plus, divided by H A. Again, we're ignoring water because it's a liquid R. K. is told to be 4.7 times 10 to the -3 at equilibrium, both of my products are equal to x. So if they're multiplying each other, that's gonna be X squared on top divided by f minus x. What we have to do here is we have to figure out that formal concentration, they tell us the ph of the solution, Remember if we know ph then we know H plus or H +30 plus concentration? That's because H +30 plus equals 10 to the negative ph And once we find H30 plus, this will represent X. Because at equilibrium H 30 plus is equal to X. With that X variable, you can plug it into here and plug it into here. So we'll have all the variables we need and the only missing one will be our formal concentration which will have to solve for. Alright, so we're gonna plug in 10 to the negative 4.12 when we do that, that gives me an answer of 7.59 times 10 to the negative five. So again this is equal to our X. Variable. So we're gonna have here is 4.7 times 10 to the minus three equals X squared which is 7.59 times 10 to the negative five, which will be squared divided by F. Which we still don't know minus X. Which is 7.59 times 10 to the negative five. Alright, so all we need to do again is figure out our formal concentration. So cross multiply these two. So this will distribute into here and distribute into here. So when we do that, that's going to give me at this point 4.7 times 10 to the -3 times f minus now. This times this gives me 3.565 times 6531 times 10 to the minus seven equals I'm gonna square this X on top. So that gives me 5.7544 times 10 to the negative nine. Remember we need to isolate our formal concentration. So we're just trying to isolate F. Here. So you're going to add 3.5653, 1 times 10 to the -7 on both sides. So this cancels out. Bring down 4.7 times 10 to the -3 f. equals when we added those two numbers together. It gives me 3.6-2 8. 6 times 10 to the -7. Isolate your formal concentration. So divide both sides by 4.7 times 10 to the -3. So here at the end our formal concentration equals 7.71 Times 10 to the -5 Molar. So that that represents our original or formal concentration of our weak acid. So remember things to look out for if we have a weak acid we'll have a k. Less than one. If we have a weak base will have a K. B less than one here. If they're telling us ph that can give us a 30 plus. If they gave us p O. H, then that will give us O. H minus. These are key things that you need in order to figure out on whether they're asking for the concentration of the original weak acid or one of its products.
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example

Weak Acid-Base Equilibria Calculations 1

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So here it says you are seeking to identify an unknown mono protic acid by determining its K. A. Value. Here we're told a 6.5 times 10 to the negative two molar solution of this unknown mono protic acid has a ph of 2.12 to determine the K. Of this unknown acid. Alright, so we know that we're dealing with a weak acid in this case because strong acids have K values that are equal to infinity because there's so much greater than one. So our generic form of a weak acid is H. A weak acids, weak bases react with water and solution because it's the acid. We know that it would donate an H. Plus to water and therefore create a minus plus H +30. Plus. We know that we'd be dealing with some type of ice chart to show this equal ceramic Equilibrium process. Water is a liquid liquid and cells are not included within an ice chart. So the initial concentration is 6.05 times 10 to the negative to moller. Initially we don't have anything for the product. So they're both zero. We lose react ints in order to create products, bring down everything 6.5 times 10 to the minus two minus X plus X plus x. Alright, so now we're looking for K. A. K A. Is equal to products overreacting. So the setup is is quite similar to the example. We saw up above. So a minus times H 30 plus divided by H. A. Remember we're looking for. Kay now. So at equilibrium both our products are X. So that's X squared over 6.5 times 10 to the minus two minus X. Like up above. If we know ph that means we know the concentration of H. 30 plus because H 30 plus is equal to 10 to the negative ph so when we do that That's gonna give me 7.5509 times 10 to the -3. And again this equals acts like we set up above. So plug that number in for X. So 7.5509 times 10 to the -3 squared Divided by 6.05 times 10 to the -2 -7.5509 times 10 to the -3. When we plug that in. What you'll get as your answer is 1.08 Times 10 to the -3 for my K. value. Remember equilibrium expressions K. K. B. They don't have any units. So it's just that number. Looking at our options option D would have to be the correct choice. So remember both questions on this page. Example one and two. Although asking different things, the setup is quite similar when dealing with the week after week base we have an equation that's occurring at equilibrium with this we set up an ice truck just to help us organize the equilibrium expression for K. A. So that we can know that K equals products over reactant. and then just solve for the missing variable. In this case, the missing variable was r. K. Value.
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example

Weak Acid-Base Equilibria Calculations

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here. It tells us a weak acid has a. K. A. Of 5.35. What is the hydro knee? Um Ion concentration in a 0.10 molar solution of this weak acid. Now we know that we're dealing with a weak acid. And the general generic formula of a weak acid is H. A. Remember weak acids and weak bases react with water within solution. So this weak acid will react with water. It is the acid. So that means water is the base. Remember, according to Bronson Laurie theory, the acid donates an H. Plus. So what we're gonna create is a minus plus H. +30. Plus. We're dealing With an ice chart because we're dealing with a weak species anytime we're dealing with a weak acid, a weak base, we set up an ice chart. Remember in a nice chart we ignore solids and liquids. So this water which is a liquid is ignored. Our initial concentration is said to be .10 moller. We're not given initial concentration for products or they're both zero. Remember, we lose react ints in order to make products bring down everything. Since we're dealing with a weak acid, we want to use our acid dissociation constant or K. A. K. Is equal to products. Overreact ints. So our products multiply together, divided by our reaction on the bottom. Again, we ignore water because it's a liquid here. The issue is we don't have K. We have P. K. So we're gonna have to convert P. K. Into K. A. Here K. A. Equals 10 to the negative P. K. A. We plugged that PK Thou given to us of 5.35 n. That gives me 4.467 times 10 to the negative six. So that's my K. A. Which I'm gonna plug over here at equilibrium. Both my products are X. So X. Times X. Is X squared divided by what's on the bottom which is 0.10 minus X. Here we have this minus X. Although this is on a local chemistry, I want to be as precise as possible. If that X variable is insignificant enough, we can ignore it to determine if we can ignore it. We do the 5% approximation method. Now the 5% approximation method says that if I take my initial concentration of my weak species and I divided by its association constant which is K. In this in this case if I divide the initial by the K. A. And the ratio gives me a value greater than 500 then I can ignore that minus X. So my initial concentration is 0.10 moller we're gonna divide it by the K. We just found. So 4.467 times 10 to the negative six. When I do that it gives me 22,387 0.2 As a value. So that number is definitely greater than 500. So according to the 5% approximation method I can ignore that -1. There All we do now is we solve for X. So we're gonna multiply both sides by .10. So we're gonna have 4.467 times 10 to the -7 equals X squared We just want X. So we're gonna take the square root of both sides here. So when I take the square root of both sides here I get x equals 4.6 6. 8 times 10 to the -4 Molar. When I sold for X. Here, this X. Gives me this sex which is equal to H 30. Plus. Remember when we're setting up a nice chart for a weak acid or a weak base anytime we find X. That represents either my H 30. Plus concentration or my O. H minus concentration. Since I'm dealing with an acid here, it's gonna give me the H 30 plus concentration which is my hydro ni um ion concentration based on our answer option F would be the correct choice. So just remember some of the fundamentals that we went over in terms of this question. If we're dealing with a weak acid or weak base in order to find hydro knee, um ion concentration or hydroxide ion concentration, we'll have to set up some type of ice chart in order to solve for X. Once we do we'll have that number. If we wanted to go a step further, we could have taken the negative log of this concentration and that would have given me ph here Were not asked to find pH were just asked to figure out what the concentration is of hydro name ion. So we just stopped there for X being 6.68 times 10 to the negative for Mueller.
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Problem

The pH of an aqueous 0.10 M nitrite ion is 8.17. What is the base dissociation constant of the base?

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