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

In a weak base-strong acid titration, the weak base (analyte) reacts with a strong acid (titrant), causing a decrease in pH. The equivalence point occurs when moles of acid equal moles of base, leading to the formation of a conjugate acid, resulting in a pH less than 7. Calculating pH involves using ICF charts for moles and ICE charts for weak species. The Henderson-Hasselbalch equation can be applied for buffer solutions formed during the titration. Understanding these concepts is crucial for mastering acid-base reactions and titration techniques.

Weak Base-Strong Acid Titrations

In these next set of videos, we're going to take a look at weak base strong acid titrations. Now, in a weak base strong acid titration, we're going to say that our weak base will represent our analyte, so our beginning material. And our strong acid will just be our titrant, which we're adding to it. So, we should expect, as we're adding strong acid to our weak base, we should expect our pH to drop because we're adding strong acid to it. Now we're going to say whenever you titrate a weak species, which in this case is a weak base, and with a strong species, which in this case is a strong acid, the fact that we're doing an acid and a base mixture means that this titration will require an ICF chart.

Now when we say ICF, ICF stands for initial change final. And whereas an ICE chart requires molarity as the units, an ICF chart requires moles as the units. Now here we're going to follow a road map for the different points within a weak base strong acid titration. Now before we start things off with this titration, sometimes it's important to know what the equivalence volume is. We're going to say we calculate the equivalence volume, which we'll abbreviate as BE, in order to determine the volume of titrant required to reach the equivalence point.

Remember, at the equivalence point, your moles of acid will equal your moles of base. And remember, moles itself equals liters times molarity. So, if we take a look here, it says the titration of a 100 ml of 0.100 molar ammonia with 0.20 molar hydrochloric acid. So at the equivalence point, our moles are equal to each other. Moles are liters times molarity or volume times molarity.

So we can say here m_acid⋅V_acid=m_base⋅V_base. Here I'm not specifying if I want my volume in liters or milliliters, so we can keep these milliliters here. Plug in the molarity of our acid, which is 0.20 molar of hydrochloric acid, we don't know its volume to get to the equivalence point, we know here that we have 0.100 molar of our base, ammonia, and we have 100 ml of it. Divide both sides here by 0.20 molar, molarities cancel out and we have the volume of our acid, which equals 50 ml. Now, this is important to know because as we're slowly adding the amount of strong acid to our weak base, based on the volume added, we'll know if we're dealing with titrations at the equivalence point, before the equivalence point, or after the equivalence point.

Now before any of this strong base on a strong acid is added, we essentially just have a weak base by itself. We're going to say here that a weak base or weak acid requires an ICE chart in order to determine its pH. So here they're telling us we have the titration of 100 ml of 0.100 molar ammonia with 0 ml of 0.20 molar hydrochloric acid. We have 20 ml here I mean, 0 ml here of HCl, which means it's not a factor in our calculations. We only have the weak base.

Because the acid hasn't been added yet, we don't need this volume just yet. So all we bring down into the ICE chart is molarity. Water is a liquid and remember, in an ICE chart, we ignore liquids and solids. Remember that, if this is the base, water will behave as the acid. Acids act as proton donors, so water would donate an H⁺ to NH₃ to give us NH₄⁺ and water would become OH^-.

Now initially, we don't have any of those products, so initially there's 0 molarity. Now looking at the change line, remember, we lose reactants in order to create products. So we'd have minus x here, plus x, plus x. Bringing down everything, we'd have 0.100 minus x, plus x and plus x. Now with a weak base, we use k_b, our base dissociation constant.

We'd say here k_b=x2C_initial-x. Because we're dealing with a base here, when we find x, that gives us OH^-. If I know the concentration of OH^-, I can determine pOH, because it equals negative log of OH^-. And then if I know pOH, then I know pH, because pH equals 14 minus pOH. We're going to say here that this minus x that is part of my equilibrium expression, it can be ignored if it is not significant.

And the way I determine if it's significant or not, I use what's called the 5% approximation method. All that is is I take the initial concentration of our weak base in this case and divide it by its k_b value. If this ratio here is greater than 500, then I could ignore the minus x within my equilibrium expression, thereby going away from the quadratic formula and making the solution faster to get. So if we took the initial concentration here and we did 0.100 molar of ammonia and divided by its k_b in your book the k_b of ammonia is approximately 1.76 times 10 to the negative 5. When we do that we get 5,681, a ratio that's definitely greater than 500.

That means I can ignore the minus x here and avoid the quadratic formula. So I would take this k_b, plug it in, equals x squared over the initial concentration of my weak base. Multiply these together, so 1.76 times 10 to the negative 6 equals x squared. Take the square root of both sides, so x here will equal 1.33 times 10 to the minus 3 molar, and because x here gives me OH^-, that means I know what concentration of OH^- is. By taking the negative log of that, that gives me pOH.

So that comes out to 2.877, and if I know pOH, I know pH, which is 14 minus pOH. So that equals 11.123. So we haven't started to add any of our strong acid just yet. So right now, we're dealing with a simple ICE chart. Click on the next video and let's see what starts happening to our pH as I slowly begin to add my strong acid titrant.

Weak Base-Strong Acid Titrations

So here, we're finally going to start adding our strong acid titrant. Remember, we're going to require 50 ml of hydrochloric acid in order to reach our equivalence point. We're going to say once our acid and base begin to mix, we use an ICF chart to determine the pH. Here, we only have 20 ml of hydrochloric acid being added, not quite enough to reach the equivalence point. That's why we're at this point where it's before the equivalence point.

Now here's the titration of 100 ml of 0.100 molar ammonia with 20 ml of 0.20 molar hydrochloric acid. Now remember, in an ICF chart, the units have to be in moles, and moles equal liters times molarity. We're going to divide these mls by 1000 and multiply by the molarity. That's going to give me the moles of my weak base, which can be seen as conjugate base, and the moles of my strong acid. We don't have any of our weak acid or conjugate acid, so initially it's 0.

Remember, look on the reactant side. The smaller moles will subtract from the larger moles. So what we'll have left is 0.006 moles of this conjugate base, which can be seen as a weak base. We'll have 0 left of our strong acid. Based on the law of conservation of mass, matter can't be created nor destroyed, it just changes form, so this increases by that same amount.

Okay. So here at the end we have weak acid, which can be seen as conjugate acid, whichever way you look at it, and realize at the end of this process what do we have? Well, we have a weak acid, we have a conjugate base, and remember with those 2 that helps to give us a buffer. If we have a buffer, then we utilize the Henderson Hasselbalch equation. Here it'll be pH=−log⁡Ka,becausethat’swhatpKais.

At the moment, all we have is the K_b of NH₃ as 1.76×10−3. Remember, K_a=K_w÷K_b, so that'll be 1.0×10−14 divided by 1.76×10−5. That comes out and gives us 5.68×10−10. We plug that in, plus log of our conjugate base, or weak base, which is NH₃, divided by our weak acid or conjugate acid depending on how you look at it, of ammonium ion. So that gives me at the end a pH of about 10.746.

So we can see that we started out with a pH of just over 11, but as we slowly start adding our strong acid titrant, our pH begins to decrease. And we can expect as we add more and more strong acid, that's going to continue. Now that we've seen the calculations before the equivalence point, let's move on to at the equivalence point.

Weak Base-Strong Acid Titrations

So now we're dealing with a titration at the equivalence point. At the equivalence point of a weak base strong acid titration, the solution is acidic. This means its pH will be less than 7 when we're dealing with the temperature of 25 degrees Celsius. Here we have 100 ml of 0.100 molar ammonia being titrated with 50 ml of 0.200 molar HCl. Remember, when I divide these ml's by 1,000 and multiply them by their molarities, that gives me the moles of both compounds.

Doing that, we see that we have 0.100 of both. They will completely neutralize one another, so at the end we have 0 of both, but we're going to have plus 0.100 of ammonium ion. Ammonium ion represents our conjugate acid or weak acid. Remember, when you have a weak acid, you have to utilize an ICE chart in order to determine pH, and the units have to be in molarity in an ICE chart.

To calculate the new molarity of our weak acid, we take 0.100 moles of ammonium ion, and divide it by the total volume, which is the volume of NH₃ (0.100 liters) plus the volume of HCl (0.050 liters). When we do that, we get a molarity of 0.667 molar of ammonium ion. We take that molarity and plug it into our ICE chart, showing that at the equivalence point, there is a lot more work involved because we first have to do an ICE chart due to our titration. What's left as a product will be a weak species and therefore requires an ICE chart in order to solve for pH.

In an ICE chart, water, which is a liquid, is ignored. The products here initially are 0. In an ICE chart, we say that we lose reactants to make product. Bring down everything. Now, we have to isolate x, because x here will give me H3O+.

We utilize this: Ka is our acid dissociation constant of our weak acid. We calculated that earlier as 5.68 × 10^-10 which equals x² divided by the concentration of my weak acid. After plugging in the values, we find that x² = 3.789 × 10^-10. Taking the square root of both sides, x equals 1.95 × 10^-5. This is equal to my H3O+, and because of that, we can just take the negative log of that number and that will give us our pH, which comes out to 4.71.

We can see here that our pH continues to drop as we have more and more strong acid titrant. At the equivalence point, both our weak base and strong acid have been completely neutralized, and as a result of their neutralization, we have the creation of a conjugate acid or weak acid. Because it's an acid, it's going to give us a pH less than 7. Now that we've seen how to tackle this type of titration at the equivalence point, move on to the next video where we're talking about going beyond the equivalence point.

Weak Base-Strong Acid Titrations

We're now dealing with titrations after the equivalence point. So we're going to say after the equivalence point of a weak base strong acid titration, we will have excess strong acid remaining. Remember, it required 50 ml of hydrochloric acid to reach the equivalence point and now we're at 60 ml. As before, since we're having an acid and a base titrate one another, we have to use an ICF. Divide the ml by a 1,000 to get liters and multiply them by their molarities gives us these moles.

Again, look at the reactant side. The smaller mole total will subtract from the larger one. So what we have left at the end is 0 of my conjugate base or weak base, but I'll still have some strong acid remaining. I'll also have some weak acid remaining. Now the strong acid has a much bigger presence in changing the pH, so we're going to go with the strong acid amount.

We're going to say here, we need to find its concentration, so the concentration of HCl equals the moles of it left divided by the total liters, so that would be 0.100 liters + 0.060 liters. So that comes out to 0.125 molar of HCl. Now that I have the molarity of a strong acid, remember, then it's simply the negative log of that concentration to find our pH. So it'll be -log(0.125) and that gives me 0.903 as my final pH. So, notice from the very beginning of our titration of this weak base, we first initially started with a pH just above 11 when it was the weak base by itself.

As we gradually added more strong acid, we've seen that our pH decreases to this point right here after the equivalence point where it's just below 1. Remember the steps and remember the spots within this titration and whether we're using an ICE chart or an ICF chart or a combination of the 2, and you'll be able to determine the pH of this type of titration.

Weak Base-Strong Acid Titrations Calculations

So here are example states. Consider the titration of 50 ml of 0.150 molar of methylamine. It has a Kb=4.4×10-4 with 75 ml of 0.200 molar hydrochloric acid. Here, they tell us to calculate the pH. So we have a weak base here since its Kb value is less than 1 and it's being titrated by this HCl.

So, we're going to write this down, so CH3NH2 + HCl, the acid will protonate or give over an H+ to make it CH3NH3+. What's left behind is Cl-. Since we have an acid and a base titrating one another, it's initial, change, final. We don't care about this Cl-, which is a neutral ionic salt. We're going to remember, if we're dealing with an ICF, we need the units in moles. Divide the ml's by 1,000 to get liters, multiply them by their molarities will give us the moles of both the methylamine and the hydrochloric acid.

So those numbers would come out to be 0.0075 moles and 0.015 moles. We don't have anything initially of this compound, so it's 0. Now, looking at the reactant line, we're going to say the smaller moles will subtract from the larger moles. So at the end, we'll have 0 left of our weak base, and we'll have some strong acid remaining. Now, we're going to say over here it'd be plus that amount of moles.

At the end, what we have remaining is strong acid and we have weak acid. Now, remember the strong acid is going to have a bigger effect on the pH, so we're going to focus on that instead. We need its concentration, so we take the moles of it left divided by the total volume; 0.050 liters of methylamine plus 0.075 liters of our hydrochloric acid. That gives me 0.060 molar of HCl. And because I have the molarity of a strong acid, I can just take the negative log of that concentration, and that'll give me pH.

So pH=-log(0.060), which comes out to 1.22. We can see here since we're left with some strong acid at the end, this must be a question dealing with after the equivalence point. Because remember, after the equivalence point, we'll have some excess strong acid remaining. Now that we've seen this example, continue to the next question, and let's see if you can apply the concepts that we've gone over in terms of a weak base-strong acid titration.

Weak Base-Strong Acid Titrations Calculations

So here in this practice question, it says, calculate the pH of the solution resulting from a mixing of 75 ml of 0.100 molar Sodium Acetate and 75 ml of 0.150 molar acetic acid with 0.0040 moles of perchloric acid. Here, we're told the k value of acetic acid is \(1.8 \times 10^{-5}\). Alright. So we're coming into a situation where there are actually 3 species all mixing together. But remember, the point of these types of titrations is we're adding a strong titrant, to a weak species.

Our strong titrant here would be the strong acid. That strong acid we know is going to be a reactant. An acid will react with a base, so this strong acid would have to react with this conjugate base. So this conjugate base would have to also be a reactant. The acid will proton will protonate or give an H⁺ to the acetate ion creating acetic acid over here as a product, plus, we'd have what's left as NaClO₄.

That would represent a neutral salt, so we don't really care about it. Because we have an acid and a base involved in this titration, we do an ICF chart. Remember, in an ICF chart, we use moles as our units. So we divide these ml's by 1000, multiply it, by their molarities gives us moles of each. So for this sodium acetate, I'd have 0.0075 moles.

They tell us the moles already of this perchloric acid, so that's 0.0040 moles. And then these liters multiplied by this molarity gives me 0.01125 moles. Now remember, we look at the reactant side. I know there's stuff on the product side, but we've always looked at the reactant side because that's what's involved in the titration. On the reactant side, the smaller moles will subtract from the larger moles.

So at the end we'll have 0.0035 moles of this conjugate base, we'll have 0 left of our strong acid, Here on this side it would increase by this many moles, so then we have 0.01525 moles. At the end what do we have? We have conjugate base remaining, we have weak acid remaining. So we have a buffer, and because we have a buffer we use the Henderson-Hasselbalch equation. So pH equals pKa plus log of conjugate base over weak acid.

Remember, pKa is just the negative log of Ka, so bring down this Ka value plus log of our conjugate base over our weak acid. So as a result of this, when we plug it in, that's gonna give me a value of 4.11 as the pH. So in this question, because we still have a buffer at the end of this titration, we must be dealing, with a titration before the equivalence point. So remember, before the equivalence point, we still have the presence of a buffer, so utilize the Henderson-Hasselbalch to isolate your pH at the end.

Here’s what students ask on this topic:

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

The equivalence point in a weak base-strong acid titration occurs when the moles of the strong acid (titrant) added equal the moles of the weak base (analyte) initially present. At this point, the weak base has been completely neutralized by the strong acid, resulting in the formation of a conjugate acid. The pH at the equivalence point is less than 7 because the solution is acidic. This is due to the presence of the conjugate acid, which partially dissociates in water to produce H₃O⁺ ions, lowering the pH.

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

Before the equivalence point, the pH is calculated using an ICF (Initial, Change, Final) chart to determine the moles of the weak base and the strong acid. The remaining weak base and its conjugate acid form a buffer solution. The pH can be calculated using the Henderson-Hasselbalch equation:

$\mathrm{pH}=pKa+\frac{\log }{\frac{\mathrm{[A^-]}}{\mathrm{[HA]}}}$

where [A^-] is the concentration of the conjugate base and [HA] is the concentration of the weak acid. This equation helps determine the pH of the buffer solution formed during the titration.

Why is the pH at the equivalence point of a weak base-strong acid titration less than 7?

The pH at the equivalence point of a weak base-strong acid titration is less than 7 because the solution contains the conjugate acid of the weak base. When the weak base is completely neutralized by the strong acid, the resulting solution contains the conjugate acid, which partially dissociates in water to produce H₃O⁺ ions. This increase in H₃O⁺ concentration lowers the pH, making the solution acidic. The exact pH can be calculated using an ICE chart and the acid dissociation constant (K_a) of the conjugate acid.

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

After the equivalence point, the pH is determined by the excess strong acid remaining in the solution. First, calculate the moles of excess HCl by subtracting the moles of the weak base from the total moles of HCl added. Then, find the concentration of the excess HCl by dividing the moles of excess HCl by the total volume of the solution. Finally, calculate the pH using the formula:

$\mathrm{pH}=-\log \left(\mathrm{[H⁺]}\right)$

where [H⁺] is the concentration of the excess HCl. This gives the pH of the solution after the equivalence point.

What is the role of the Henderson-Hasselbalch equation in weak base-strong acid titrations?

The Henderson-Hasselbalch equation is used to calculate the pH of buffer solutions formed during a weak base-strong acid titration. When a weak base and its conjugate acid are present in significant amounts, they form a buffer solution. The equation is:

$\mathrm{pH}=pKa+\frac{\log }{\frac{\mathrm{[A^-]}}{\mathrm{[HA]}}}$

where [A^-] is the concentration of the conjugate base and [HA] is the concentration of the weak acid. This equation helps determine the pH of the buffer solution, which is crucial for understanding the pH changes during the titration process.