Titrations: Weak Acid-Strong Base - Video Tutorials & Practice Problems
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1
concept
Before the Equivalence Point
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Now this type of titration has the weak acid as the titrate and the strong base as the tight Trent. Remember your tight trend is always going to be a strong species with what you typically will see. So the weak acid has to be the titrate. Now, when a weak species in the form of a weak acid reacts with a strong species. In this case, in the form of a strong base, we use what's called an IC F chart. IC F stands for initial change. Fine. Here we're going to say the IC F chart is used to calculate the final amounts of compounds. The units of an IC F chart will be in molds. And here you need to remember that moles equals leaders times molarity. Now, at this present point, we're gonna look at titrations under this category that happened before the equivalence point. Now, before the equivalence point, we're gonna say in this part of the titration, the moles of weak acid is greater than the moles of strong base. And we're gonna say here as a strong base neutralize the weak acid, some conjugate base is going to be formed. So keep in mind when we take a look at the questions under this type or this portion of our titration.
2
example
Titrations: Weak Acid-Strong Base Example
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Has considered the titration of 75 mls of 0.0300 molar of pyruvic acid, which has a K of 4.1 times 10 to the minus three with uh with 12 mls of 0.0450 molar of potassium hydroxide. Here, we need to calculate the ph. All right. So what we have here is pyruvic acid which has a cave of less than one would say weak acid. And then potassium hydroxide is a strong base. We have a weak species mixing with a strong species. So that's why in step one, we set up an IC F chart. So remember we set up an IC F chart. If it's weak and strong, mixing together here with an IC F chart, the strong species is set up as a reactant. The, the, the strong species then reacts with its chemical opposite. So our strong species is the potassium hydroxide. Koh it reacts with its chemical opposite. What's the opposite of a base? An acid? So it would react with the pyruvic acid following brasini definitions of acids and bases. The acid would donate an H plus that H plus would go to the oh minus of potassium hydroxide to create water, which will be a liquid. And then here what's left would be the potassium and pyruvate, which is C three H 303 aqueous. Here we have our equation. Now this is an IC F truck which remember stands for initial change, final. We know that with an IC F chart, our units need to be in moles. Remember, moles equal leaders times molarity. So you would divide these MLS by 1000 to get liters and then multiply them by their molarity. And you'd have the moles of your weak acid and your strong base. So when we do that, we get here 0.00225 moles of pyruvic acid. And for koh we get 0.00054 moles. So using the initial role, we place the given amounts in moles in an IC F chart, we only care about three things. We only care about the weak acid and it's conjugate base and then whatever the strong thing is, whether it be a strong mass or a strong base. So here we don't have any information on the conjugate base. So this is zero. Initially, we only care about three things and this is our fourth thing. So we don't care about it besides it's water, which is a liquid in an IC F chart just like an ice chart. We ignore solids and liquids. Now, this is where things get a little bit different. So this is a deviation from what we typically do with an ice chart with an IC F chart for step three, using the change rope, look at the reactants subtract from their initial amounts by the smaller mole amount. So if we take a look here, we have these two moles as the initial amounts for our reactants. The moles of the strong base are smaller in amount. So we subtract both of them by the smaller amount. Now, using the law of conservation of mass, whatever you lose as a reactant, you gain that amount to products. So we're losing 0.00054 moles on the reactant side. So we're gaining that on the product side, remember matter is neither created nor destroyed. It just changes forms. So it's not that we're actually losing. You don't say that you're destroying the acid by that amount. You're basically saying that you're transforming it into your conjugate base. All right. So now after we do the subtracting, what do we have left? We have 0.00171 moles of this zero of our strong base. And we have 0.00054 of our conjugate base. This takes us to step four, the Henderson Hava equation for a buffer. He used for a buffer to find the page of a solution using the final roll, use the moles of the weak acid and conjugate base to find the Ph since we're dealing with K A within the question, we've used this form of the Henderson Hasselback equation which is PH equals PK A plus log of conjugate base over weak acid. So we come back over here. So we're gonna say PH equals PK A. So that's negative log of RK A which is 4.1 times 10 to the minus three plus log of our conjugate base. So 0.00054 of this conjugate base divided by the weak acid, which is our pyruvic acid. 0.00171. When we do that, we get 1.89 as the Ph for this particular buffer solution.
3
Problem
Problem
In order to create a buffer 7.510 g of sodium cyanide is mixed with 100.0 mL of 0.250 M hydrocyanic acid, HCN. What is the pH of the buffer solution after the addition of 12.0 mL of 0.300 M NaH? Ka = 4.9 × 10−10.
A
6.82
B
10.01
C
8.52
D
10.17
4
concept
At the Equivalence Point
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1m
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In our discussion of the titration between a weak ass and a strong base, we now take a look at how to calculate Ph at the equivalence point. Here, we're going to say in this part of the titration, the moles of weak acid is equal to the moles of strong base. Now, since they're equal to one another, we're going to say here, the weak acid and strong base have been neutralized and only the conjugate base remains. So this is what's true at the equivalence point. We can also say at the equivalence point, we can deal with equivalent volume of our titrate. Remember, we can utilize the equation of M acid times V acid equals M base times V base in order to determine the volume of our T, right. So we can utilize this equation if they don't give us the volume of our T. Now that we've set up this picture of what exactly is happening at the equivalence point. Click on to the next video and let's see a problem where we have to determine the Ph
5
example
Titrations: Weak Acid-Strong Base Example
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8m
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He has considered the titration of 75 MLS, a 0.0300 molar of pyruvic acid with a K A of 4.1 times 10 to the minus three with 50 mls of 0.0450 molar of potassium hydroxide. Here, we need to calculate the ph. All right. So we have a weak acid reacting with a strong base. We have a weak species and a strong species, which means we need to set up an IC F chart. Now, remember with an IC F chart, whatever is strong has to be set as a reactant. So the potassium hydroxide is a reactant. It will react with its chemical opposite, which is the pyruvic acid. Following the Bronson Nori definition, the acid donates an H plus to the base. So the H plus from the acid will combine with the oh minus of the base to give us water. And then the other species combined to give us our conjugate base. Remember this is an IC F chart which is initial change fine with an IC F chart. We need our units to be in moles, moles equals liters times molarity. So divide the MLS by 1000 to get liters and then multiply by the molarity. So that'll give us the moles of our weak acid and strong base. So we have 0.00225 0.00225. In an IC F chart, we only care about three things. The weak acid, its conjugate base and whatever the strong species is. The fourth species. In this case, what we ignore. Now here, our conjugate base initially is zero. Look on the reactant side, the smaller moles will subtract from the larger moles. In this case, they both have the same amount of moles. So they neutralize each other entirely. So at the end, we have zero of them, but based on the law of conservation of mass matter is neither created nor destroyed. So we're gonna gain this much for the product. At the end of this, what do we have? We have only conjugate based remaining conjugate base here would represent a weak base. So remember how do we find the ph of a weak base? We'll have to eventually utilize an ice chart. So now we're gonna have to do an ice chart afterwards. All right. So step four, using the final role to determine the concentration of the conjugate base. So we have 0.00225 moles of our conjugate base, we divide its final moles by the total volume used in the chemical reaction. So what is our total volume used in a chemical reaction. Well, we have 75 MLS of our acid, 50 mls of our base, which is 100 and 25 MLS total, dividing it by 1000 gives us 0.125 L. So our new concentration is 0.018 molar. Now set up an ice chart for the conjugate base, which is our weak base that has it reacting with water. It is an ionic base to ignore the neutral metal C. Here, I ignore the potassium ions and neutral. It reacts with water. Again, following the bronci definition. It is a base. Now, so water is gonna be the acid, the acid donates in H plus to regenerate the pyruvic acid and to give us our hydroxide ion. Now we're dealing with an nice chart. So that's initial change equilibrium. So here our initial amount is 0.018 molar in an ice chart. We ignore solids and liquids. So water is ignored. Our products initially are zero, we lose reactants to make product. So bring down everything. Now using the equilibrium row, set up the equilibrium constant expression with, since it's a base that we're dealing with, we use KB and we're gonna use KB to solve for X. Now, here we're gonna check to see if a shortcut can be utilized in order to avoid the quadratic formula. So here we're going to look to see if I could do the 500 approximation rule. Here, this is when the concentration of initial concentration, which is the 0.018 molar and we divided by, in this case, KB, if that ratio happens to be something greater than 500 I'll be able to ignore the minus X within my equilibrium expression to avoid the quadratic formula. So here, let's set that up. We're gonna say here we need KB. They gave us K A in the very beginning, beginning of the question. Remember that KW equals K A times KB. So KB will equal KW divided by K A. So that will be KB equals 1.0 times 10 to the minus 14 divided by the K A within the question which is 4.1 times 10 to the minus three KB. I'm gonna write it over here in this box. So when I do KW divided by K A, my KB comes out to be 2.14 times 10 to the negative 12. So now that I have my KB, I do initial concentration divided by KB. So my initial concentration is 0.018 molar, 2.4 times 10 to negative 12. This gives me a value of 7.38 times 10 to the nine. So that's a number way bigger than 500. So I can ignore the minus X within my equilibrium expression. So equilibrium expression is KB equals products over reactants which will be X squared over 0.018 minus X. Again, since our ratio is greater than 500 I can ignore this minus X and avoid this quadratic formula. So here's gonna be equal to X squared over 0.018 cross multiply these two X squared equals 4.32 times 10 to the minus 14, take the square root of both sides. So when I take the screw to both sides, X equals 2.078 times 10 to the minus seven. Now when you find X that can either give you H 30 plus or oh minus. So look at the equation for the ice chart and see which one you have for the equation for the ice chart. X here will give me oh minus. So we just found the concentration of oh minus, which means I just found out what POH is. So plug it in for oh minus. So that equals 6.68. If I know what POH is, I know what Ph is because they're connected together by this formula. Ph plus POH equals 14. So we're gonna say here, Ph equals 14 minus po so 14 minus 6.68. So this equals 7.32. So the Ph of the solution with all this work we found is 7.32.
6
Problem
Problem
Consider the titration of 75.0 mL of 0.60 M HNO2 with 0.100 M NaOH at the equivalence point. What would be the pH of the solution at the equivalence point? The Ka of HNO2 is 4.6 × 10−4.
A
11.73
B
8.13
C
5.87
D
1.35
7
concept
After the Equivalence Point
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47s
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In this part of our weak acid, strong base titration, we've gone beyond the equivalence point. So we've completely neutralized our weak acid and we're just continuously adding more strong base. So here we're gonna say in this part of the titration, the moles of weak acid is less than the moles of strong base. Remember once we reach the equivalent point and beyond, we no longer have a buffer here, we're going to say there will be excess strong base remaining after it has neutralized the weak acid, right? So we're just gonna have this excess of strong base, which is a good thing because if you have an excess of a strong species, it becomes that much more easy to find the ph at the end of the reaction.
8
example
Titrations: Weak Acid-Strong Base Example
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4m
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Here, it says, consider the titration of 75 mls of 0.0300 molar of pyruvic acid. Here, the K A is 4.1 times 10 to the minus three. And we're reacting it with 75 mls of 0.0450 molar of potassium hydroxide. Calculate the ph, all right. So we have a weak species reacting with a strong species. So we know we need to set up an IC F chart. The strong species has to be set as a reactant. So potassium hydroxide has to be reacted. It reacts with its chemical opposite. So it reacts with the pyruvic acid. Now, following the bras and lai definition, the acid would proton or donate an H plus to the base that will give us water as a by-product as the H plus combines with the oh minus and then the potassium will combine with the pyruvate to give us potassium pyruvate. Now we're gonna say that this is an IC F. So that's initial change fine with an IC F chart. We need our units to be in moles, divide your MLS by 1000 to get liters because remember, moles equals liters times molarity. So when we do that, we're gonna get 0.00 225 moles of our weak acid. And then we're gonna have 0.003375 of our strong base, we have zero of our conjugate initially. And the fourth species we don't care about. Now, here, look at the reactants, the smaller moles subtract from the larger moles. So minus 0.00225 minus 0.00225. So at the end, we have no weak acid remaining, but we're gonna have some strong baseline, whatever we lose on the reactant side, we gain on the product side based on the law of conservation of mass. Now, at the end, what do we have, we have strong base conjugate base. The strong species will have a larger impact on our overall ph So focus on the strong species. So we have here using the final row, determine the concentration of the strong base divide its final moles by the total volume using a chemical reaction. So what do we have here? We have 0.001125 moles of potassium hydroxide divided by the total volume. The total volume is 75 MLS, 75 ML. So that's 100 and 50 MLS when you divide that by 1000 that gives you 0.150 L. So that'll give me my new concentration which is 0.0075 molar for my strong bits. Now this is important because we call the concentration of the strong base will be equal to the concentration of oh minus. So we just found out the concentration of oh minus is this number. If we know oh minus concentration, that's good because that help me find POH uoh equals negative log of oh minus. So plug that number in. So that's gonna give me 2.12. If I know P A um POH, then I know Ph because of this formula. So Ph if we rewrite it equals 14 minus POH so it's 14 minus 2.12 which comes out to 11.88. So this will represent the Ph of my solution after the equivalence plant.
9
Problem
Problem
Calculate the pH of the solution resulting from the mixing of 55.0 mL of 0.100 M NaCN and 75.0 mL of 0.100 M HCN with 0.0090 moles of NaOH.
A
12.06
B
11.74
C
1.89
D
2.26
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