Titrations: Weak Base-Strong Acid - Video Tutorials & Practice Problems
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1
concept
Before the Equivalence Point
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1m
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We now take a look at a weak base strong acid titration. Now this type of titration has the weak base as the titrate, and the strong acid as the titrant. Remember, when it comes to the titrant, it has to be a strong species. So by default the weak base would have to be the tight trait. Now recall when a weak species react with a strong species, we use an ICF chart. Here, ICF stands for initial change final. Now in the first journey on our weak base strong acid titration, we'll take a look at calculations before the equivalence point. Now in this part of the titration the weak base is greater than the moles of the strong acid. We're going to say, as the strong acid neutralizes the weak base, some weak acid or conjugate acid is formed. And as it's forming, we'll have some weak acid conjugate acid with still some present weak base, helping us to create a buffer. So before the equivalence point, because we have the formation of a buffer, we'll be able to utilize the Henderson Hasselbalch equation in order to calculate the pH of the solution. So now that we laid down the groundwork and ideas, let's click on the next video and let's see how we let's see how we calculate the pH of a solution before the equivalence point.
2
example
Titrations: Weak Base-Strong Acid Example
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5m
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Here it says to calculate the pH of the solution resulting from the titration between 25 ml's of a 0.100 molar chloric acid solution, and 50 mals of a 0.100 molar ammonia solution. Here we're told the kB value of ammonia is 1.75 times 10 to the negative 5. Alright. So we're going to use the steps 1 to 3 to help set up the ICF chart. Now, Now if you don't remember the steps of 1 to 3, make sure you go back and take a look at my video dealing with the titration between a weak acid and a strong base. We're going to say here that we have a weak species and a strong species, so we have to set up an ICF chart. We're gonna say that the strong species has to be set as a reactant. So here, chloric acid has to be a reactant. It reacts with its chemical opposite. What's the opposite of an acid? A base. The ammonia is the base. Now, using the Bronston Loring definition of acids and bases, the acid will donate an h plus to the base. So it donates 1 to ammonia to give us ammonium ion, and then what we have left here is the chlorate ion. Now in an ICF chart, we only care about 3 things. We only care about the weak acid, or in this case, conjugate acid, same thing. We only care about the conjugate base or weak base, and whatever is strong. The fourth thing we ignore it. Now in an ICF chart, the units have to be in moles, which is liters times molarity. So divide the ml's by a 1,000, multiply by the molarity. So that'll give us 0.00 25 moles of our strong acid, 0.0050 moles of our weak base, 0 initially of this conjugate acid slash weak acid. Look at the reactant sign. The smaller moles will subtract from the larger moles. So point 0025 or minus for both. We know we're dealing with an ICF, which is initial change final. So at the end, we'll have 0.0025, 0 left of this. Remember, based on the law of conservation of mass, matter is either created nor destroyed, it just changes form. Which means whatever we lose on the reactant side, we're gaining on the product side. So we're gaining 0.0025 here. So we're gonna say here, at the end what do we have? We have a weak acid and conjugate base remaining. So that leads us to step 4. The Henderson Hasselbalch equation is used for a buffer to find the pH of a solution, and that's what we have, we have a buffer. Now using the final row, use the moles of the weak acid and its conjugate base to find the pH. Now the Henderson alphabet can be observed 2 different ways, where it could be pH equals pKa plus log of conjugate base over weak acid. Or to simplify things for us, we can use this other equation when given k b, where pH equals p k b plus log of conjugate acid over weak base. So since we're giving Kb in the question, let's use the second version. So pH equals pKa, pKb plus log of conjugate acid over a weak base. So we're gonna say here equals negative log of 1.75 times 10 to the negative 5, plus log of 0.0025.0025. When you plug that in, you're gonna get 4 point 76 as the pH for a solution that's before the equivalence point.
3
Problem
Problem
Calculate the pH of the solution resulting from the mixing of 75.0 mL of 0.100 M NaC2H3O2 and 75.0 mL of 0.30 M HC2H3O2 with 0.0040 moles of HBr.
A
3.26
B
4.27
C
3.87
D
6.23
4
Problem
Problem
In order to create a buffer 7.321 g of potassium lactate is mix with 550.0 mL of 0.328 M lactic acid, HC3H5O3. What is the pH of the buffer solution after the addition of 300.0 mL of 0.100 M hydrobromic acid, HBr? The Ka of HC3H5O3 is 1.4 × 10−4.
A
3.16
B
2.96
C
4.74
D
4.35
5
concept
At the Equivalence Point
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1m
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Now, in our discussion of the titrations dealing with weak base to strong acids, we're now going to take a look at calculations around the equivalence point. We're gonna say in this part of the titration, the moles of weak acid is equal to the moles of strong base. We're gonna say because they're equal in amount, we're gonna say the weak base and strong acid have been neutralized, and if they've been completely neutralized, what we're gonna have left is some weak acid or conjugate acid remaining. Right? And we're gonna say since we're at the equivalence point, we could find the equivalence volume of our titrant. To do that, we use the formula molarity of the acid, m acid, times v acid, equals m base times v base. This would help us to determine the equilibrium volume of our titrant. Alright. So we use it if we need to find the volumes of both the titrate and titrant. So keep this handy when necessary. Alright, so now let's take a look at calculations where we have to calculate the pH at the equivalence point.
6
example
Titrations: Weak Base-Strong Acid Example
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7m
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Calculate the pH of the solution resulting from the titration of 25 ml's of a 0.100 molar chloric acid solution, and 50 ml's of a 0.050 molar ammonia solution. Here we're told that kB of ammonia is 1.75 times 10 to the negative 5. Alright. So here we're gonna use the steps 1 to 3 to set up our ICF chart. Here with our ICF chart, the strong species has to be a reactant. So we're gonna have chloric acid as a reactant, It reacts with its chemical opposite, so this strong acid will react with the weak base. Following the Bronsted Lowry definition of an acid in the base, the acid donates an h plus to the base, so we create the ammonium ion, and then we have the chlorate ion as a byproduct. Here we only care about the weak acid, the conjugate base, and whatever is strong. The 4th species, you ignore. Now here we're dealing with an ICF, so initial, change, final. With an ICF chart, we need the units to be in moles. Moles equals liters times molarity, so divide the ml's by a1000 and and multiply by the molarity to give us the moles of each, so this comes out to be 0.0025 moles. 0.0025 moles, this initially is 0. Now look on the reactant side, the smaller moles will subtract from the larger moles. Here, since the moles are the same, they completely neutralize one another, giving us 0 at the end for both. Now whatever we lose on the reactant side, we gain on the product side, so we're gonna gain 0.0025 moles here. At the end, what do we have left? We only have weak acid remaining. And how do we find the pH of a weak acid? By utilizing an ICE chart. So now we're going to have to set up an ICE chart. Now using the final rule, determine the concentration of the weak acid or conjugate acid, however you want to look at it. We do this by dividing its final moles by the total volume using the chemical reaction. So we have 0.0025 moles of ammonium ion, and we're gonna divide it by the total volume. The total volume is 25 ml and plus 50 ml, so 75 ml. Divide that by a 1,000, and that'll give us our liters. So when we do that, we're gonna get as our liters 0.033 molar of the ammonium ion. Okay. So there goes the molarity. Now if all that is left is a weak species, then set up a nice chart, having it reacting with water. So here, this ammonium ion will react with water. Again, following the Bronsted and Lore definition, the acid donates an h plus to the base. So ammonium ion donates an h plus, it becomes an h three as a result. Water gains an h plus to become h three o plus. Here we're dealing with initial change equilibrium. So we're gonna plug in the initial concentration. In an ICE chart, we have no solids and liquids. Our products initially are 0. We lose reactants to make product. Now, using the equilibrium row, so this row here, set up the equilibrium constant expression with, since we're dealing with an acid now, we're using k a, and we'll use k a to solve for x. Now check if a shockwave can be utilized to avoid the quadratic formula. So what we do here, the shortcut we're gonna utilize is the 500 approximation method. So here we're going to say if the initial concentration to your k a is greater than 500, we can ignore the minus x. Within this question, we're given k b. Remember that k w equals k a times k b, so k a equals k w divided by k b. So 1.0 times 10 to the minus 14, divided by 1.75 times 10 to the negative 5. That gives me a k b value of 5.71 times 10 to the negative 10. Plug in the initial concentration here. When we do that, we get an answer of 5.775 times 10 to the 7. So we just found the ratio is much greater than 500, so we can ignore the minus x within our equilibrium expression. Our equilibrium expression itself is k a equals products over reactants. So 5.71 times 10 to the negative 10 equals x squared divided by 0.033 minus x. Again, because our ratio was greater than 500, I can ignore this minus sex here and avoid the quadratic formula. Alright. So I'm gonna work the math up here. So we have 5.71 times 10 to the negative 10 equals x squared divided by 0.033. Cross multiply these 2 together, x squared equals 1.8843 times 10 to the negative 11. Square root, square root, x equals 4.34 times 10 to the negative 6. When you find your x, it either gives you h three zero plus or o h minus. So look at your equation in the ice chart to see which one you found. X here gives us h 3 o plus. So we just found out what h 3 o plus is, and because of that we can determine our pH. So pH remember is equal to the negative log of h 3 o plus. So take that number you just found and plug it in. When we do that, we get a pH of 5.36 for the pH of this particular solution at the equivalence point. So that would be our final answer.
7
Problem
Problem
Consider the titration of 100.0 mL of 0.100 M CH3NH2 with 0.250 M HNO3 at the equivalence point. What would be the pH of the solution at the equivalence point? The Kb of CH3NH2 is 4.4 × 10−4.
A
8.10
B
11.79
C
2.21
D
5.90
8
concept
After the Equivalence Point
Video duration:
53s
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In our discussion of the titration between a weak base and a strong acid, we're now looking at calculations after the equivalence point. Now when we get to the equivalence point and beyond, we no longer have a buffer. It's been completely destroyed. So in this case though, we're gonna say since we're after the equivalence point, we're gonna say in this part of the titration the moles of weak base is less than the moles of strong acid. Here, we're going to say there will be an excess strong acid remaining after it has neutralized the weak base. Now this is a good thing, because if you have excess of a strong species it becomes that much more easy to determine the overall pH of the solution. Because remember, strong species ionize completely, so there would be no need for an ice chart.
9
example
Titrations: Weak Base-Strong Acid Example
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3m
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Here it says to calculate the pH of the solution resulting from the titration between 125 ml's of a 0.100 molar chloric acid solution and 50 ml's of a 0.050 molar ammonia solution. Here we're told the kv value of ammonia is 1.75 times 10 to the negative 5. Alright. So here what we're gonna do is remember, the strong species has to be a reactant. It reacts with its chemical opposite, so it reacts with the ammonia molecule. Following the Bronsted Lowry definition of acids and bases, the acid will donate an H plus to the base, creating the ammonium ion, and then we'd have chlorine ion as a biproduct. In an ICF, we only care about 3 things: the weak acid, its conjugate base, and whatever is strong. We're dealing with an ICF which is initial, change, final. In an ICF we need moles, Moles equals liters times molarity, so divide the ml's by a 1,000 to get liters, and multiply them by the molarity, you'll have the moles of your strong acid and lead base. So when we do this, we get 0.0025 moles or 2 225 25 moles. And then here we're gonna have what? We're gonna have 0.0125 moles. We have nothing of the weak acid. Here this is ignored. Look at the reactant side, the smaller moles will subtract from the larger moles. So we're gonna subtract 0.0025, subtract 0.0025. So at the end, this will be 0. Here, we'll actually have some strong acid remaining. Based on the law of conservation of mass, whatever we lose on the reactant side we gain on the product side, so plus 0.0025. At the end, what do we have? We have strong acid and we have weak acid remaining. The strong acid will have a much larger impact on the overall pH, so we focus on that. So we're gonna say, using the final row, determine the concentration of the strong acid, divide its final moles by the total bond used in the chemical reaction. So we have 0.010 moles of chloric acid divided by its total volume. What's the total volume? Well, we have a 125 ml here and 50 ml here. So together that's a 175 ml. You would divide that by a1000 to get liters. Okay. So now we have the liters on the bottom. So this gives us a molarity of 0.0571 molar. Now recall the concentration of the strong acid will be equal to the H+H3O plus concentration. So because we know the concentration, we can find pH, because remember, pH is the negative log of h plus concentration. So plug that in, Okay, so that's gonna give me 1.24 as my pH for this solution after the equivalence point.
10
Problem
Problem
A solution contains 100.0 mL of 0.550 M sodium nitrite, NaNO2. Find the pH after the addition of 180.0 mL of 0.400 M HClO4. The Ka of HNO2 is 4.6 × 10−4.
A
11.17
B
1.22
C
3.85
D
12.78
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