Here it asks us what is the original molarity of a solution of weak acid with a Ka=4.7×10-3 and a pH of 4.12 at 25 degrees Celsius. Now, the key here is realizing that we're dealing with a weak acid. Remember, we know it's a weak acid because its Ka value is less than 1. With a weak acid, we're going to say that our generic weak acid, we can say represents weak acids, weak bases react with water in order to establish the equilibrium. Remember, the acid here, according to the Brønsted-Lowry definition will donate an H+. It's going to become here, A- plus H3O+. We have initial change equilibrium. Water is a liquid. Remember, liquids and solids are not included within an ICE chart. Here, we're told to find the original molarity. That's our formal concentration. We don't know what it is, so we'll just place f here. These initially are 0. Remember, we're losing our reactant over time in order to produce products. This will be -x plus x plus x. Bring down everything, f-x plus x plus x. Knowing this, we set up the equilibrium expression as Ka=A-×H3O+f-x. So it's Ka=A-×H3O+ divided by Again, we're ignoring water because it's a liquid. Our Ka is told to be 4.7×10-3. At equilibrium, both of my products are equal to to x. If they're multiplying each other, that's going to be x2 on top divided by f-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 H3O+ concentration. That's because H3O+=10-pH. Once we find H3O+, this will represent x because at equilibrium, H3O+ is equal to x. With that x variable, you can plug it into here and plug it into here. We'll have all the variables we need and the only missing one will be our formal concentration which we'll have to solve for. All right. We're going to plug in 10-4.12. When we do that, that gives me an answer of 7.59×10-5. So again, this is equal to our x variable. So what we're gonna have here is 4.7×10-3=x2which is7.59×10-5,which will be squared divided byfwhich we still don't know minusxwhich is7.59×10-5. Alright. So all we need to do again is figure out our formal concentration. So, cross-multiply these 2. So this will distribute into here and distribute into here. So when we do that, that's going to give me at this point 4.7×10-3×f-now this times this gives me3.56531×10to the minus 7 equals I'm gonna square this x on top so that gives me5.7544×10-9. Remember, we need to isolate our formal concentrations. We're just trying to isolate f here. You're going to add 3.56531×10to the minus7 on both sides. So this cancels out. Bring down 4.7×10-3f=when we added those two numbers together, it gives me3.62286×10to the minus7. Isolate your formal concentration, so divide both sides by 4.7×10-3. So here, at the end, our formal concentration equals 7.71×10-5 molar. That there represents our original or formal concentration of our weak acid. Remember, things to look out for. If we have a weak acid, we'll have a Ka less than 1. If we have a weak base, we'll have a Kb less than 1. Here, if they're telling us pH, that can give us H3O+. If they gave us pOH, then that will give us OH-. These are key things that you need in order to figure out whether they're asking for the concentration of the original weak acid or one of its products.

- 1. Chemical Measurements1h 50m
- 2. Tools of the Trade1h 17m
- 3. Experimental Error1h 52m
- 4 & 5. Statistics, Quality Assurance and Calibration Methods1h 57m
- 6. Chemical Equilibrium3h 41m
- 7. Activity and the Systematic Treatment of Equilibrium1h 0m
- 8. Monoprotic Acid-Base Equilibria1h 53m
- 9. Polyprotic Acid-Base Equilibria2h 17m
- 10. Acid-Base Titrations2h 37m
- 11. EDTA Titrations1h 34m
- 12. Advanced Topics in Equilibrium1h 16m
- 13. Fundamentals of Electrochemistry2h 19m
- 14. Electrodes and Potentiometry41m
- 15. Redox Titrations1h 14m
- 16. Electroanalytical Techniques57m
- 17. Fundamentals of Spectrophotometry50m

# Weak Acid-Base Equilibria - Online Tutor, Practice Problems & Exam Prep

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

## Weak Acid-Base Equilibria

### Weak Acid-Base Equilibria Calculations 1

#### Video transcript

### Weak Acid-Base Equilibria Calculations 1

#### Video transcript

Here it says you are seeking to identify an unknown monoprotic acid by determining its Ka value. Here, we're told a 6.05×10-2 molar solution of this unknown monoprotic acid has a pH of 2.122. Determine the Ka of this unknown acid. All right. We know that we're dealing with a weak acid in this case because strong acids have Ka values that are equal to infinity because they're so much greater than 1. Our generic form of a weak acid is this:

Weak acids and weak bases react with water in solution. Because it's the acid, we know that it would donate an H^{+} to water and, therefore, create OH-H3O^{+}. We know that we'd be dealing with some type of ICE chart to show this equilibrium process. Water is a liquid. Liquids and solids are not included within an ICE chart. The initial concentration is 6.05×10-2 molar. Initially, we don't have anything for the products so they're both 0. We lose reactants in order to create products. Bring down everything, 6.05×10-2-x, x, x. Now we're looking for Ka.

Ka is equal to products over reactants. So the setup is quite similar to the example we saw up above. A-H3O^{+} divided by A-. Remember, we're looking for Ka now. So at equilibrium, both our products are x so that's x2 over 6.05×10-2-x. Like up above, if we know pH, that means we know the concentration of H_{3}O^{+} because H3O^{+} is equal to 10-pH. So when we do that, that's gonna give me 7.5509×10-3. Again, this equals x like we set up above. Plug that number in for x.

So 7.5509×10-3 squared divided by 6.05×10-2-7.5509×10-3. When we plug that in, what you'll get as your answer is 1.08×10-3 for my Ka value. Remember, equilibrium expressions, Ka, Kb, 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. Remember, both questions on this page, examples 1 and 2, although asking different things, the setup is quite similar. When dealing with a weak acid or a weak base, we have an equation that's occurring at equilibrium. With this, we set up an ICE chart just to help us organize the equilibrium expression for Ka so that we can know that K equals products over reactants and then just solve for the missing variable. In this case, the missing variable was our Ka value.

### Weak Acid-Base Equilibria Calculations

#### Video transcript

Here it tells us a weak acid has a Ka5.35. What is the hydronium 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. Now 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. According to theory, the acid donates an H plus. What we're going to create is A^{-} plus H_{3}O^{+}. We're dealing with an ICE chart because we're dealing with a weak species. Anytime we're dealing with a weak acid or a weak base, we set up an ICE chart. Remember, in an ICE chart, we ignore solids and liquids. This water, which is a liquid, is ignored. Our initial concentration is said to be 0.10 molar. We're not given initial concentrations for our products, so they're both 0. Remember, we lose reactants 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 Ka. Ka=productsreactants so our products multiply together divided by our reactant on the bottom. Again, we ignore water because it's a liquid. Here the issue is we don't have Ka, we have pKa. We're going to have to convert pKa into Ka. Here, Ka=10-pKa. We plug that pKa value given to us of 5.35 in. That gives me 4.467×10-6. That's my Ka which I'm going to 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 analytical chemistry, we 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. The 5% approximation method says that if I take my initial concentration of my weak species and I divide it by its dissociation constant, which is Ka in this case, if I divide the initial by the Ka and the ratio gives me a value greater than 500, then I can ignore that minus x. My initial concentration is 0.10 molar. We're going to divide it by the Ka we just found, so 4.467×10-6. When I do that, it gives me 22,387.2 as a value. That number is definitely greater than 500. According to the 5% approximation method, I can ignore that minus x there. All we do now is we solve for x. We're going to multiply both sides by 0.10. We're going to have 4.467×10-7=x2. We just want x, so we're going to take the square root of both sides here. So when I take the square root of both sides here, I get x equals 6.68×10-4molar. When I solve for x here, this x gives me this x which is equal to H_{3}O^{+}. Remember when we're setting up an ICE chart for a weak acid or a weak base, anytime we find x that represents either my H_{3}O^{+} concentration or my OH^{-} concentration. Since I'm dealing with an acid here, it's going to give me the H_{3}O^{+} concentration which is my hydronium ion concentration. Based on our answer, option F would be the correct choice. Just remember some of the fundamentals that we went over in terms of this question. If we're dealing with a weak acid or a weak base, in order to find the hydronium 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 take the negative log of this concentration and that would have given me pH. Here, we're not asked to find pH. We're just asked to figure out what the concentration is of hydronium ion. So we just stop there for x being 6.68×10-4molar.

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

^{-16}

^{-11}

^{-6}

^{‑5}

^{-3}