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

In titrations involving a weak base and a strong acid, the pH calculation varies depending on the stage of the titration. Before the equivalence point, the weak base is in excess, creating a buffer solution, allowing the use of the Henderson-Hasselbalch equation to determine the pH. At the equivalence point, the moles of weak base and strong acid are equal, and the pH can be found using the formula:

$M_1{V}_1=M_2{V}_2$

where M and V represent molarity and volume, respectively. After the equivalence point, the strong acid is in excess, and its complete ionization simplifies the pH calculation, eliminating the need for an ICE (Initial, Change, Final) chart. Understanding these stages is crucial for accurate pH determination during acid-base titrations.

Before the Equivalence Point

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 titrate. 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 titrate. 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 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 than the moles of the strong acid. We're going to say as the strong acid neutralizes the weak base, some weak acid more 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 Hasselbeck equation in order to calculate the pH of the solution. So now that we laid down the ground working 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.

Titrations: Weak Base-Strong Acid Example

Here it says to calculate the pH of the solution resulting from the titration between 25 MLS of a .100 molar chloric acid solution and 50 mills of a .100 molar ammonia solution. Here we're told that KB value of ammonia is 1.75×10-5. All right, so we're going to use the steps one to three to help set up the ICF chart. Now, if you don't remember the steps of one to three, 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 going to say that the strong species has to be set as a reactant. 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 Bronsted-lowry definition of acid 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 three things. We only care about the weak acid, or in this case conjugate ask the 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 molds, which is liters times molarity. So divide the MLS by 1000, multiply by the molarity. So that will give us 0025 moles of our strong acid, 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 0.0025 or minus from both. We know we're dealing with an ICF which is initial change final, so at the end we'll have .00250 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 going to say here at the end what do we have? We have a weak acid and conjugate based remaining, so that leads us to Step 4. The Henderson Hasselbeck equation is used for a buffer to find the pitch 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 a pH. Now the Henderson houseboat can be observed two different ways where it could be pH equals PKA plus log of conjugate based 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 PKA on PKB plus log of conjugate acid over week base. So we're going to say here equals negative log of 1.75×10-5 plus log of point 0025.0025. When you plug that in, you're going to get 4.76 as the PH-4 solution. That's before the equivalence point.

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

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 K_a of HC₃H₅O₃ is 1.4 × 10⁻⁴.

• A. 3.16
• B. 2.96
• C. 4.74
• D. 4.35

At the Equivalence Point

Now in our discussion of the titrations dealing with weak bases and strong acids, we're now going to take a look at calculations around the equivalence point. We're going to say in this part of the titration, the moles of weak acid is equal to the moles of strong base. We're going to say because they're equal in amount.

We're going to say the weak base and strong acid have been neutralized. And if they've been completely neutralized, what we're going to have left is some weak acid or conjugate acid remaining, right? And we're going to say since we're at the equivalence point, we could find the equivalence volume of our titrate.

To do that, we use the formula molarity of the acid. M acid × V acid= M base × V 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 titrate. 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.

Titrations: Weak Base-Strong Acid Example

Calculate the pH of the solution resulting from the titration of 25 MLS of a .100 molar chloric acid solution and 50 MLS of a .050 molar ammonia solution. It were told the K_B of ammonia is 1.75 * 10^-5. All right, so here we're going to use the steps one to three to set up our ICF chart here with our ICF chart. The strong species has to be a reactant, so we're going to 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⁺ 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 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 0. Now look on the reactant side the smaller moles of subtracting the larger moles. Here, since the moles are the same, they completely neutralize one another, giving us zero at the end for both. Now whenever we lose on the reactant side, we gain on the product side. So we're going to 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 going to divide it by the total volume. The total volume is 25 MLS and plus 50 MLS, so 75 MLS. Divide that by 1000 and that will give us our liters. So when we do that, we're going to get as our liters 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 an ICE chart having it reacting with water. So here this ammonium ion will react with water. Again following the Bronsted-Lowry definition, the acid donates an H⁺ to the base. So ammonium ion donates an H⁺. It becomes an H₃. As a result, water gains in H⁺ to become H₃O⁺. Here we're dealing with initial change equilibrium, so we're going to plug in the initial concentration in a nice chart. We ignore solids and liquids. Our products initially are 0. We lose reactants to make product nap 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 shortcut can be utilized to avoid the quadratic formula. So what we do here, the shortcut we're going to 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 * 10^-14 / 1.75 * 10^-5. That gives me a K_B value of 5.71 * 10^-10.

Plug in the initial concentration here. When we do that, we get an answer of 5.775 * 10⁷. 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 * 10^-10 = X² / 0.033 - X. Again, because our ratio is greater than 500, I could ignore this minus X here and avoid the quadratic formula. All right, so I'm going to work the math up here. So we have 5.71 * 10^-10 = X² / 0.033. Cross multiply these two together X² = 1.8843 * 10^-11 square root.

Square root X = 4.34 * 10^-6. When you find your X, it either gives you H or OH minus. So look at your equation in the ICE chart to see which one you found. X here gives us H₃O⁺, so we just found out what H₃O⁺ is and because of that we can determine our pH. So pH, remember is equal to the negative log of H₃O⁺. 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.

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 K_b of CH₃NH₂ is 4.4 × 10⁻⁴.

• A. 8.10
• B. 11.79
• C. 2.21
• D. 5.90

After the Equivalence Point

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 going to say since we're after the equivalence point, we're going to 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.

Titrations: Weak Base-Strong Acid Example

Here it says to calculate the pH of the solution resulting from the titration between 1:25 MLS of a 100 molar chloric acid solution and 50 mills of a .050 molar ammonia solution. Here we're told that KB value of ammonia is 1.75×10-5. All right, so here what we're going to do is remember the strong species has to be 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 to the base, creating the ammonium ion and then we'd have chloride ion as a bipartite. In an ICF, we only care about three 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 mills 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, 222525 moles, and then here we're going to have what we're going to 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 going to subtract 0.0025, subtract 0025 S. At the end, this will be 0. Here we'll actually have some strong assets 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 going to say, using the final row, determine the concentration of the strong acid. Divide its final molds by the total volume 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 125 mills here, 50 mills here, so together that's 175 MLS. You would divide that by 1000 to get liters.

OK, 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 plus or 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. So that's going to give me 1.24 as my pH for this solution after the equivalence point.

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 K_a of HNO₂ is 4.6 × 10⁻⁴.

• A. 11.17
• B. 1.22
• C. 3.85
• D. 12.78

Here’s what students ask on this topic:

What does the midpoint of the titration of a weak acid with a strong base correspond to?

The midpoint of the titration of a weak acid with a strong base corresponds to the point where exactly half of the weak acid has been neutralized by the strong base. At this stage, the amount of the weak acid (HA) is equal to the amount of its conjugate base (A^-) that has been formed. This is significant because it is the point where the pH of the solution is equal to the pK_a of the weak acid. The pK_a is a measure of the strength of the acid; it's the pH value at which the acid is half dissociated. Therefore, at the midpoint of the titration, the buffer capacity of the solution is at its maximum, meaning the solution can resist changes in pH upon the addition of small amounts of acid or base. This is due to the Henderson-Hasselbalch equation, which describes the relationship between pH, pK_a, and the ratio of the concentrations of the conjugate base and the weak acid.

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

In a strong acid-weak base titration, the pH at the equivalence point is typically less than 7. This is because the weak base being titrated has a conjugate acid that is a weak acid. When the strong acid (like HCl) reacts with the weak base (like NH₃), the products are a salt and water. However, the salt formed in this reaction partially dissociates in water to produce the weak conjugate acid of the base.

Since the conjugate acid is weak, it does not ionize completely in water, and it releases some hydrogen ions (H⁺) into the solution. This presence of extra H⁺ ions causes the solution to be acidic. Therefore, unlike the equivalence point in a strong acid-strong base titration, which is neutral (pH 7), the equivalence point in a strong acid-weak base titration is on the acidic side of the pH scale, typically below 7. The exact pH value will depend on the specific weak base used and its dissociation constant (K_b).

Why does the equivalence point for a weak base-strong acid titration occur at a pH < 7.00?

In a titration involving a weak base and a strong acid, the equivalence point is where the amount of acid added equals the amount of base present. A weak base does not completely dissociate in water, meaning it does not produce many hydroxide ions (OH^-). When a strong acid is added, it completely dissociates into hydrogen ions (H⁺), which are highly reactive. These hydrogen ions readily react with the hydroxide ions from the weak base to form water.

At the equivalence point, all of the weak base has been neutralized by the strong acid, but because the base is weak, the conjugate acid formed in the reaction is slightly stronger and can donate hydrogen ions to the water, creating a slightly acidic solution. This means that the pH at the equivalence point will be less than 7, reflecting the presence of excess hydrogen ions in the solution. The exact pH will depend on the strength of the weak base and the temperature of the solution.