Weak Acid-Strong Base Titrations - Online Tutor, Practice Problems & Exam Prep

In weak acid-strong base titrations, the pH changes significantly at various points. Initially, an ICE chart is used to find the pH of the weak acid. As strong base is added, an ICF chart helps calculate pH until the equivalence point, where moles of acid equal moles of base, resulting in a basic solution. Beyond the equivalence point, excess strong base dictates pH. The Henderson-Hasselbalch equation is crucial for buffer calculations, while the relationship between pH, pKa, and concentrations of weak acid and conjugate base is essential for understanding titration dynamics.

Weak Acid-Strong Base Titrations

So now we're going to take a look at weak acid-strong base titrations. Now we're going to say here that our weak species, which is our acid, will behave as the analyte, and our strong base, which we'll slowly add to it, represents our titrant. Now we're going to say whenever you titrate a weak species, in this case, the acid, with a strong species, in this case, the strong base, the fact that we have an acid and a base mixing together means that we're going to have to use the ICF chart. Now when we say ICF, ICF stands for Initial, Change, Final. Now we're going to use the following roadmap, to determine the pH at different points in our titration between a weak acid and a strong base. Now, before we even begin all of this, we first look at the equivalence volume. Now remember, we calculate the equivalence volume, \( v_e \), in order to determine the volume of titrant required to reach the equivalence point. In this case, the titrant again is our strong base. Remember, the equivalence point is where your moles of acid equals moles of base. And remember that moles itself equals liters times molarity. So if we take a look here, it says the titration of 300 ml of 0.100 molar nitrous acid with 0.30 molar potassium hydroxide. So nitrous acid represents our weak acid, potassium hydroxide represents our strong base. We're going to say here that \( m_{\text{acid}} \times v_{\text{acid}} = m_{\text{base}} \times v_{\text{base}} \). So the molarity of my acid is 0.100 molar, though its volume is 300 milliliters, the molarity of my strong base is 0.30 molar, and we're looking for the volume of our base, our equivalence volume in this case. Divide both sides by 0.30 molar, here you'll see that molarities cancel out and we'll be left with milliliters as our volume. So when we work that out, we'll get 100 ml's of KOH is required in order to reach the equivalence point. Now let's look at the next step, this is before any strong base has been added:

Before any strong base has been added, we just have a weak acid initially. Here we have the titration of 300 ml of 0.100 molar nitrous acid with 0 ml of 0.300 molar KOH. Here, there is no strong base being added, so again, we just have a weak acid by itself. Remember, when we have a weak acid or a weak base by itself, we need to utilize an ICE chart in order to determine our pH. Now, an ICF chart represents Initial, Change, Final, but an ICE chart represents Initial, Change, Equilibrium. Remember, in these types of ICE charts, our units will be molarity. Because no strong base is being added, we don't need this volume initially. All we require is the initial molarity of my weak species. We plug that in. Remember, weak acids and weak bases react with water within an ICE chart. So this nitric acid would react with the water. Remember, an acid is a proton donor. So \( \text{HNO}_2 \) donates an \( \text{H}^+ \) to give us \( \text{NO}_2^- \) at the end, which is our nitrite ion, and then \( \text{H}_2\text{O} \) accepts an \( \text{H}^+ \) to become \( \text{H}_3\text{O}^+ \). Our change here, we lose reactants in order to create products. So our reactant here will be minus x, our products here that are being formed are plus x. Initially no information is given on them, so initially there's 0. We bring down everything for our equilibrium line, so 0.100 minus x, and then plus x and plus x comes down. Now at this point, we would say that since we're dealing with a weak acid, we have to use our acid dissociation constant. Remember, weak acids use \( K_a \), weak bases, which we'll see later, deal with \( K_b \). Now \( K_a \) is the equilibrium constant for the weak acid. It equals products over reactants, so we know that'll be x times x, which is \( x^2 \), divided by our initial concentration, minus x. If we can isolate our x variable, that x variable will give us \( \text{H}_3\text{O}^+ \), which is the same thing as \( \text{H}^+ \). If we know the concentrations of \( \text{H}^+ \), we can take the negative log of that to find pH. Now there are going to be situations where we can ignore this minus x here. In those situations where we can ignore that minus x, we can avoid the quadratic formula. In cases where we cannot ignore that minus x, we keep it in, and we set up all the math that we need to do, and from there, we use a quadratic formula. Now in order to determine if we can keep that minus x or not, we do what I call a 5% approximation. So basically, we have our initial concentration of our weak acid divided by its \( K_a \) value. If this ratio gives us a value greater than 500, then we can ignore the minus x. Now in this case, the initial concentration of weak acid is 0.100, and the \( K_a \) of our nitrous acid from your book is approximately \( 7.1 \times 10^{-4} \). When you plug that in, you'll see that it gives you 140.8. Here it's not greater than 500, so we'd have to keep this minus x and use the quadratic formula in order to isolate x. So if we work this out we'd have, \( K_a \) so I'm going to take this expression here. So \( K_a \) is \( 7.1 \times 10^{-4} \) equals \( x^2 \) divided by \( 0.100 - x \). So you would have if you work this out, you'd have \( 7.1 \times 10^{-4} \times (0.100 - x) = x^2 \). Then, we're going to say here, you're going to distribute. When you work that out, you're going to get \( 7.1 \times 10^{-5} - 7.1 \times 10^{-4}x = x^2 \). This x has the highest power, so it's the lead term. When you rearrange this, it gives you \( x^2 + 7.1 \times 10^{-4}x - 7.1 \times 10^{-5} \). Then we're going to say here, the quadratic formula remember is negative b, plus or minus, square root of b squared minus 4ac over 2a. Okay. You would plug into that formula. So bring it down, it'd be negative \( 7.1 \times 10^{-4} \), plus or minus \( 7.1 \times 10^{-4} \) squared minus 4 times 1 times, don't forget the negative sign for c, negative \( 7.1 \times 10^{-5} \) divided by 2 times 1. Now because of this plus-minus point here, you'd get 2 x values, one that's positive and one that's negative. But here's the thing, at equilibrium, you're only allowed to have a positive value, so we would disregard the negative x and go with the positive x answer. That would give us x equals 0.00808 molar. Again, in this situation, x would be equal to \( \text{H}_3\text{O}^+ \), which is equal to \( \text{H}^+ \). Taking the negative log of that number would give you my pH, which is 2.09. So that would be the long version of this reaction. So again, at this point, we don't have any strong base being added. We just have a weak acid by itself, so we must utilize an ICE chart in order to isolate our pH. Now, click on the next video, and let's start slowly adding our strong base and see what happens in our titration.

Weak Acid-Strong Base Titrations

So remember, it's going to require, on our part, 100 ml of potassium hydroxide in order to reach the equivalence point. In this question here, we have only 50 ml of KOH. So that means we're dealing with calculations before the equivalence point. Now, remember, once our acid and base begin to mix, we transition away from the ice chart and we move to the ICF chart to determine the pH. And remember, in an ICF chart, the units have to be in moles. Moles itself equals liters times molarity. We divide these ml's here by 1,000 to change them into liters and then multiply them by their molarities. That would give me the moles of each of these species. So here are moles of nitrous acid, would give me 0.030 moles, and then here, this would give me the moles of KOH, 0.015 moles. We don't have any of our nitrate ion not nitrite ion, so, we start initially at 0. Like in ice chart, ICF charts ignore water because it's a liquid. It's not really affecting our pH value, so we could ignore it. Now we look at the reactant side. We know on the reactant side, the smaller moles, which is our limiting amount, will deduct from the larger moles. So our smaller moles are these 0.015 moles. They subtract from themselves because they're being used to neutralize the acid. We subtract them from the acid as well. At the end, we'll have left 0 of my strong base, and I'll have some of this weak acid remaining. Based on the law of conservation of mass or matter, we know that matter can't be created nor destroyed. It just changes forms. So whatever I'm losing on the reactant side, I'm actually gaining as this conjugate base on the product side. So at the end of this, what we have is we have weak acid and we have conjugate base remaining. Remember, weak acid conjugate base means we have a buffer. And if we have a buffer, then we utilize the Henderson Hasselbalch equation in order to calculate our pH. So if we look here, remember our Henderson Hasselbalch equation is pH equals pKa plus log of the conjugate base amount divided by the weak acid amount.

We said earlier that the so negative log on pKa equals negative log of Ka, and we said earlier that the Ka of nitrous acid is 7.1−4 according to your book, and then it's log of the conjugate base amount divided by the weak acid amount. When you plug that in, that gives you 3.15 as your pH. So you're seeing as we're slowly adding strong base, we see that our pH is increasing. And, remember, before we've reached the equivalence volume of the strong base titrant, that means we will still have a buffer and we can utilize the Henderson Hasselbalch equation to calculate our pH. This particular example, we see that at the end, we have equal amounts of my weak acid and conjugate base. This happens when we're at the half equivalence point, and it's important to remember that at the half equivalence point, pH equals pKa. That's because log of 0.015 divided by 0.015 is really log of 1. Log of 1 is equal to 0. So at the half equivalence point, that's the half of the volume we needed to get to the equivalence point pH equals pKa. Now that we've seen these types of calculations, let's move on to what happens when we're at the equivalence point between a weak acid and a strong base.

Weak Acid-Strong Base Titrations

So when we reach the equivalence point between a weak species and a strong species, we're actually going to have to utilize 2 charts. We'd start off with an ICF chart because, again, we have an acid and a base titrating. But at the end, we'll have a product which is weak in itself and would need an ICE chart in order to determine pH and pOH or pOH. So if we take a look here at the equivalence point, we're going to say, at the equivalence point of a weak acid strong base titration, whatever is stronger will dictate what kind of pH I have. Since the base is what's strong, it's going to create a solution that is basic. And if the solution is basic, that means it'll be greater than 7 when we have our standard temperature of 25 degrees Celsius. So here we divide our mL's by 1000 to get liters of each, and when you multiply them together, you'd see that we have actually the same moles for both, which makes sense because at the equivalence point, we have equal moles of our acid and our base. They would subtract from one another, so at the end, we'll have both of them being 0. Through the law of conservation of mass, we'd be gaining this amount on the product side. So at the end what we have is conjugate base. This conjugate base, it comes from a weak acid, so it's a little bit stronger, but in actuality, it's still a weak base. Okay. So it's still a weak base. And remember, with weak acids and weak bases, we need to use an ICE chart in order to find our pH.

So what we do here is we'd have to now set up an ICE chart, with this NO₂^-. Remember, in an ICE chart, the units have to be in molarity. So we take the moles of what's left of our conjugate base, and divide it by the total liters used within the titration, which would be the 0.300 liters of the weak acid plus the 0.100 liters of a strong base. When we do that, we get a new concentration of 0.075 molar of our nitrite ion. So we plug that here. Now remember, in an ICE chart, our weak acid or weak base reacts with water. Since it's a base, it's going to be a proton acceptor, so water will donate an H⁺ to it, giving us nitrous acid and OH^- as products. Now we ignore water because liquids and solids are ignored in ICE charts. Initially, both of these products are 0. We lose reactants to make products. We bring down everything for the equilibrium line. So that's our filled in ICE chart. Now because we're dealing with a base, bases utilize K_b. We know what the K_a of nitrous acid is, and we're going to use that to find K_b. K_b equals K_w divided by K_a. So remember K_w is 1.0×10-14, divided now by 7.1×10-4, that gives me a value of 1.41×10-11. That's our K_b, which we'll plug into here. At this point, in order to isolate x, we're going to say K_b equals x² divided by the concentration of our conjugate base, which is our weak base. So we're going to do 1.41×10-11 equals x² divided by 0.075. Multiply these 2 together, that gives me x² equals 1.0575×10-12. To get just x, we take the square root of both sides, so x equals 1.0283×10-6.

Now remember, when we find x, x here will give us either H₃O⁺ or OH^- based on the equation in the ICE chart. Here, x is giving us OH^-. Which means that if I take the negative log of that, I'll get pOH. So that comes out to 5.99. And if I know pOH, I know pH, because pH is 14 minus pOH, which is 8.01. So again, we're seeing that we have the gradual increasing of our pH as we start adding more and more of our strong base So at the equivalence point, we found out that our pH is 8.01.

Now that we've seen this, click on the next video. Let's see what happens when we go a little bit beyond the equivalence volume. So we're going to go a little bit above 100 mL of our potassium hydroxide.

Weak Acid-Strong Base Titrations

We've now reached the point where we've gone beyond the equivalence volume of potassium hydroxide. So here, after the equivalence point of a weak acid-strong base titration, we will have excess strong base remaining. As usual, we divide these milliliters by 1,000 to get liters, multiply them by their molarity which gives us the moles of each. We see that we have the moles for nitrous acid and potassium hydroxide. Remember, on the reactant side, you use the smaller moles, which is our determining or limiting amount, and subtract them from both, that gives us 0 left of the weak acid, but we're still going to have some strong base remaining. Here we'd have plus, but here we'd say that the strong base has a much greater impact on the overall pH, so we're just going to go with that amount. Since it's a strong base, if I can find its concentration, then I can use negative log of that concentration to find pOH. So we have 0.009 moles of potassium hydroxide. We divide it by the total volume of 0.300 liters plus 0.130 liters, that equals 0.021 molar KOH. And when we take the negative log of that, that's going to give us pOH, which comes out to 1.68. And if we know pOH, then we know pH because pH is 14 minus pOH. So that's 12.32. These represent the different points of our weak acid-strong base titration. So just remember, when we have only weak acid by itself, we utilize an ICE chart. But once we start adding a strong base titrant, we switch over to an ICF chart. And depending on where in the titration you are, we can take different approaches to find either pH or pOH.

Weak Acid-Strong Base Titrations Calculations 1

So here, let's take a look at this question. It says, consider the nitration of 75 ml of 0.0300 molar of H₃C₃O₃. It's given as having a K_a of 4.1 times 10 to the minus 3, and it's being titrated with 30 ml of 0.0450 molar KOH. Here, we're told to calculate the pH. Since its K_a is less than 1, we know that this represents a weak acid, and we know that potassium hydroxide is a strong base. We're going to write them here as our reactants. Remember the acid is going to be a proton donor. It's going to donate a proton to OH^- to give us water, and what we'll have left over here as our conjugate base is KH₂C₃O₃. Since we have an acid and a base titrating one another, we utilize an ICF chart. In this ICF chart, we're going to ignore water, which is a liquid. Remember, in an ICF chart, we use moles as our units. Moles equals liters times molarity. So divide these ml's by 1000, multiply by their molarities gives us the moles of our weak acid which is 0.00225 moles. Here KOH is 0.00135 moles, and we're not given any initial information on our conjugate base, so initially it's 0. Now remember, looking at the reactant side, the smaller moles, which is our limiting amount, we'll subtract from the larger moles. So subtract 0.00135 subtract 0.00135. So when we do that, we're going to have 0 left of this. For our weak acid, we're going to have 0.00090 moles. And remember, conservation of mass, this side would increase 0.00135. Now look and see. At the end of this titration, we have left weak acid. We have left its conjugate base. Remember, weak acid conjugate base means that we have a buffer. And with a buffer, that means we have to utilize the Henderson-Hasselbalch equation. So we're going to say here that pH equals pKa, which is the negative log of K_a, plus log of the conjugate base amount, So there goes my conjugate base amount, oops, divided by our weak acid amount. Okay. So that equals 2.56. Now remember, because we still have a buffer in terms of this titration, that means we're dealing with calculations before the equivalence point. And because we still have a buffer, we can utilize the Henderson-Hasselbalch equation to find our pH. Now that we've seen this one, click on to the next video and take a look at the practice question below.

Weak Acid-Strong Base Titrations Calculations 1

So here we're dealing with a calculation at the equivalence point. Here it says consider the titration of 50 ml of 0.150 molar hydrofluoric acid with 0.100 molar sodium hydroxide at the equivalence point. What would be the pH of the solution at the equivalence point? We're told here that our \( K_a \) value for hydrofluoric acid is \( 3.5 \times 10^{-4} \). Now we have our weak acid reacting with a strong base. Remember the weak acid is a proton donor, so \( H^+ \) will combine with \( OH^- \) to give us water. The \( Na^+ \) will combine with the \( F^- \) to give us NaF. We have here, because we're mixing acid and base, we know we're utilizing an ICF chart. And now I'm writing this pretty small because, remember, we said that at the equivalence point between a weak acid and a strong base, we first have to do an ICF chart followed by an ICE chart. Alright. So here, water is ignored. Now we divide these ml's here by 1,000 to get the liters, which we multiply by molarity to get the moles. So it's going to give me 0.0075 moles of HF. Since we're at the equivalence point, we have equal moles of our weak acid and strong base. So even though I don't have the volume of NaOH, which I'll need eventually though, I don't need it because I know at the equivalence point both of these have the same moles. We don't have any initial information on NaF, so initially it's 0. Now remember on the reactant side the smaller moles subtract from the larger moles. Since they're both the same, we're going to get 0 at the end for both. They completely neutralize each other. But remember on this side we have the creation of this many moles of the conjugate base, NaF. Now remember, this is our conjugate base. A conjugate base here is a weak base. And because it's a weak base, in order to find its pH, we'd have to use an ICE chart. Now, in order to figure out, in order to use an ICE chart, we need to use the units of molarity. So we need the molarity of this conjugate base. Now we don't really need the Na there, it's just a spectator ion, so it's just \( F^- \) that's important. We need its concentration. So we need the moles of it that's left, which is the same number, but we need the total volume, the volume of both the HF and the NaOH. Now to figure out the volume of the base, remember, at the equivalence point, their moles are equal. Because they're at the equivalence point where their moles are equal, you can say \( M_{acid} \times V_{acid} = M_{base} \times V_{base} \). Plug in the concentration of the acid times its ml equals molarity of the base, and we're looking for its volume. Again, we need the volume of the base because we need the total volume in order to figure out the final the new concentration of \( F^- \). So molarities cancel out here, so I'll have my volume of my base, which comes out to be 75 milliliters. So now I'm gonna come over here and say that my total volume is 0.050 liters of the weak acid, 0.075 liters of the strong base. Together, that gives me a concentration of 0.06 molar for my fluoride ion. Now we can bring it down and utilize an ICE chart. Remember, weak acids and weak bases react with water within an ICE chart. Remember also that bases are proton acceptors, so water is going to act as the acid and donate an \( H^+ \) to \( F^- \). That gives us back HF and the creation of \( OH^- \). So we have initial, change, equilibrium. In an ICE chart, we don't care about solids and liquids, so water is ignored. Bring in the initial concentration of the \( F^- \). Our products initially are 0. We lose reactants to make products. Now because we have a base here, a weak base, remember, weak bases utilize \( K_b \). So there's a lot of conversions that are going on in this problem. We have the \( K_a \), we need \( K_b \). So remember, \( K_b = \frac{K_w}{K_a} \). So if you plug that in correctly in your calculator, make sure you put parentheses around it, you get \( 2.9 \times 10^{-11} \). So that is our \( K_b \) value for \( F^- \). Bring that down, so \( K_b = \frac{\text{products}}{\text{reactants}} \), so \( 2.9 \times 10^{-11} = \frac{x^2}{0.06} \). Remember, at the equivalence point we could ignore the minus x here because it's just \( K_b = \frac{x^2}{\text{concentration of our conjugate base which is our weak base}} \). Alright. Now multiply these 2 together, so that gives me \( x^2 = 1.74 \times 10^{-12} \), take the square root of both sides, so \( x = 1.32 \times 10^{-6} \). Now remember, at this point, when we find x, x will either give me \( H_3O^+ \) or \( OH^- \). Go back and look at your ICE chart. In the ICE chart, we see here that x gives us \( OH^- \). So, since I know the concentration of \( OH^- \), that means if I take its negative log, that gives me pOH. And if I know pOH, I know pH because \( \text{pH} = 14 - \text{pOH} \). So that comes out to be 8.12. So here our answer is reasonable because we know that when it comes to a weak acid and a strong base titration at the equivalence point, our solution should be basic. So we should expect to get a pH above 7. Alright. So just remember, within the different points of our titration, between a weak acid and a strong base, you have to remember when to utilize an ICE chart, when to utilize an ICF chart, or in this case, utilizing first an ICF chart followed by an ICE chart.

Here’s what students ask on this topic:

What is the equivalence point in a weak acid-strong base titration?

The equivalence point in a weak acid-strong base titration is the stage where the moles of the weak acid equal the moles of the strong base added. At this point, the solution is typically basic because the strong base dominates the reaction. The pH at the equivalence point is greater than 7, as the conjugate base of the weak acid hydrolyzes to form OH^- ions. To calculate the equivalence volume (V_e), use the formula:

$m_{\mathrm{acid}}\times V_{\mathrm{acid}}=m_{\mathrm{base}}\times V_{\mathrm{base}}$

How do you calculate pH before adding any strong base in a weak acid-strong base titration?

Before adding any strong base, the pH of the weak acid solution is determined using an ICE chart. The weak acid dissociates in water, and the equilibrium concentrations are used to find the pH. The steps are:

1. Write the dissociation equation: HA + H₂O ⇌ H₃O⁺ + A^-

2. Set up the ICE chart with initial concentrations, changes, and equilibrium concentrations.

3. Use the acid dissociation constant (K_a) to solve for [H₃O⁺]:

$K$_a=x2c-x

4. Calculate pH: pH = -log[H₃O⁺]

What is the Henderson-Hasselbalch equation and how is it used in titrations?

The Henderson-Hasselbalch equation is used to calculate the pH of a buffer solution, which contains a weak acid and its conjugate base. The equation is:

$\mathrm{pH}=\mathrm{pK}$_a+log([A^-][HA])

In titrations, it is used when a buffer is formed before reaching the equivalence point. The pH can be calculated by knowing the concentrations of the weak acid (HA) and its conjugate base (A^-).

How do you determine pH at the equivalence point in a weak acid-strong base titration?

At the equivalence point in a weak acid-strong base titration, the pH is determined by the hydrolysis of the conjugate base formed. Follow these steps:

1. Calculate the concentration of the conjugate base (A^-) formed.

2. Set up an ICE chart for the hydrolysis reaction: A^- + H₂O ⇌ HA + OH^-

3. Use the base dissociation constant (K_b) to find [OH^-]:

$K$_b=x2c-x

4. Calculate pOH: pOH = -log[OH^-]

5. Convert to pH: pH = 14 - pOH

What happens to pH beyond the equivalence point in a weak acid-strong base titration?

Beyond the equivalence point in a weak acid-strong base titration, the pH is determined by the excess strong base. The steps are:

1. Calculate the moles of excess strong base (OH^-).

2. Determine the total volume of the solution.

3. Find the concentration of OH^-: [OH^-] = moles of OH^- / total volume

4. Calculate pOH: pOH = -log[OH^-]

5. Convert to pH: pH = 14 - pOH

This results in a high pH, indicating a basic solution.