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 tit Trent. Remember when it comes to the tight trend, it has to be a strong species. So by default, the weak base would have to be the tight trend. Now recall when a weak species react with a strong species, we use an IC F chart here IC F stands for initial change fine. Now, in the first journey on our weak based 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, then the moles of the strong acid. And 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 Hasselback equation in order to calculate the ph of the solution. So now that we laid down the groundwork ideas, let's click on the next video and let's say 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 MLS of a 0.100 molar chloric acid solution and 50 mls of a 0.100 molar ammonia solution. Here, we're told the KB value of ammonia is 1.75 times 10 to the negative five. All right. So we're gonna use the steps 1 to 3 to help set up the IC F chart. 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 said as reacted. So here, chloric acid has to be reacted. It reacts with its chemical opposite. What's the opposite of an acid? A base? The ammonia is the base. Now using the Bronson la definition of acid and bases, the acid will donate an H plus to the base. So it donates one to ammonia to give us ammonium ion. And then what we have left here is the chlorate iron. Now, in an IC F chart, we only care about three things. We only care about the weak acid or in this case, kate acid, same thing, we only care about the conjugate base or weak base and whatever is straw. The fourth thing we ignored. Now, in an IC F chart, the units have to be in moles which is liters times molarity. So divide the MLS by 1000 multiply by the molarity. So that'll give us 0.0025 moles of our strong acid, 0.0050 moles of our weak base zero. Initially, of this conjugate acid slash weak acid, look at the reacting sign, the smaller moles will subtract from the larger moles. So 0.0025 or minus for both. We know we're dealing with an IC F which is initial change final. So at the end, we'll have 0.00250 left of this. Remember based on the law of conservation of mass matter is neither 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 four, the Henderson hasselback equation is used for a buffer to find the page of a solution. And that's what we have. We have a buffer. Now using the final roll, use the moles of the weak acid and its conjugate base to find the PH. Now the Henderson alphabet can be observed two different ways where it could be PH equals PK A plus log of conjugate base over weak acid. Or to simplify things for us, we can use this other equation when given KB or Ph equals PKB 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 PK A and PKB plus log of conjugate acid over weak base. So we're gonna say here equals negative log of 1.75 times 10 to the negative five plus log of 0.0025 0.0025. When you plug that in, you're gonna get 4.76 as the ph four 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, strong acids, we're now gonna 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 equivalent volume of our titrate to do that. We use the formula molarity of the acid and acid times the acid equals M base times the base. This would help us to determine the equilibrium volume of our titrate, right. So we use it if we need to find the volumes of both the titrate and t trend. So keep this handy when necessary. All right. 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 MLS of a 0.100 molar chloric acid solution and 50 mls of a 0.050 molar ammonia solution. They were told the KB of ammonia is 1.75 times 10 to the negative five. All right. So here we're gonna use the steps 1 to 3 to set up our IC F chart here with our IC F 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 Bronson Laurie 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'd have the corrate ion as a bro. Here, we only care about the weak acid, the conjugate base and whatever is strong. The fourth species you ignore. Now, here we're dealing with an IC F. So initial change, final with an IC F chart, we need the units to be in moles, moles equals liters times molarity. So divide the MLS by 1000 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 zero. Now look on the reactant side, the smaller moles or subtract from the larger moles here, since the moles are the same, they completely neutralize one another. He even got zero 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 gonna have to set up an ice chart. Now using the final role, determine the concentration of the weak acid or conjugate acid. However, you wanna 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 MLS and plus 50 MLS. So 75 amounts divide that by 1000 and that'll give us our leaders. So when we do that, we're gonna get as our leaders 0.033 molar of the ammonium ion. OK. So there goes the molarity Now, if all that is left is a weak species and set up a nice chart having it reacting with water. So here this ammonium ion will react with water again, following the Bronston Laurie definition, the acid donates in H plus to the base. So ammonium ion donates in H plus, it becomes an H three. As a result, water gains an H plus to become H +30 plus. Here we're dealing with initial change equilibrium. So we're gonna plug in the initial concentration in a nice chart with no solids or liquids. Our products initially are zero, 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 to solve for X. Now check if a show it 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 KB. Remember that KW equals K A times KB. So K A equals KW divided by KB. So 1.0 times 10 to the minus 14, divided by 1.75 times 10 to the negative five. That gives me a KB 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 seven. 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 could ignore this minus X here and avoid the quadratic formula. All right. 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 two 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 six. When you find your X, it either gives you H 30 plus or oh minus. So look at your equation in the ice chart to see which one you found X here gives us H 30 plus. So we just found out what H 30 plus is. And because of that, we can determine our Ph so ph remember is equal to the negative log of hr 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 p of this particular solution at the equivalence plant. 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 mold of strong acid. Here, we're gonna say there will be an excess strong ass 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 page 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
Video duration:
3m
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Here, it says to calculate the ph of the solution resulting from the titration between 100 and 25 MLS of a 0.100 molar chloric acid solution and 50 MLS of a 0.050 molar ammonia solution. Here, we're told the KB value of ammonia is 1.75 times 10 to the negative five. All right. So here, what we're gonna do is remember, the strong species has to be reacted, it reacts with its chemical opposite. So it reacts with the ammonia molecule following the brassy lai definition of acids and bases. The acid will donate an H plus to the base, creating the ammonium ion. And then would have chy ion as a bike in an IC F. We only care about three things. The weak acid, its conjugate base and whatever a straw we're dealing with an IC F which is initial change fine. In an ICF we need moles, moles equals liters times molarity. So divide the MLS by 1000 to get liters and multiply them by their molarity, you'll have the moles of your strong acid and weak base. So when we do this, we get 0.0025 moles or two 22525 moles. And then here we're gonna have what we're gonna have uh 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 zero here will 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 rope determine the concentration of the strong acid divide its final moles by the total vol used in the chemical reaction. So we have 0.010 moles of chloric acid divided by its total volume. What's the total volume? But we have 100 and 25 MLS here and 50 MLS here. So together that's 100 and 75 MLS, you would divide that by 1000 to get leaders. OK. So now we have the leaders on the bottom. So this gives us a molarity of 0.0571 molar. Now recall the concentration of the strong ass will be equal to the H plus or H +30 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. Yeah. So that's gonna give me 1.24 as my Ph for the solution after the equivalent plan.
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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