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

Understanding the pH during various stages of a weak acid-strong base titration is crucial for predicting the outcome of the reaction. Initially, the moles of weak acid exceed those of the strong base, leading to the formation of a conjugate base as neutralization occurs. The ICE (Initial, Change, Final) chart is a valuable tool for calculating the final concentrations in moles, using the relationship:

moles=liters×molarity

At the equivalence point, the moles of weak acid are equal to the moles of strong base, indicating complete neutralization with only the conjugate base remaining. The volume of titrant can be determined using the equation:

M_acid×V_acid=M_base×V_base

if not provided. Beyond the equivalence point, the reaction features an excess of strong base, simplifying the process of finding the pH as the system no longer behaves like a buffer. This understanding is essential for accurate titration analysis and pH calculation in chemical reactions.

Before the Equivalence Point

Now this type of titration has the weak acid as the titrate and the strong base as the titrate. Remember, your titrate is always going to be a strong species is what you typically will see. So the weak asset has to be the titrate. Now when a weak species in the form of the weak acid reacts with a strong species, in this case in the form of a strong base, we use what's called an ICF chart.

ICF stands for Initial Change Find. Here we're going to say the ICF truck is used to calculate the final amounts of compounds. The units of an ICF chart will be in moles. And here you need to remember that moles=liters×molarity. Now at this pressing point, we're going to look at titrations under this category that happened before the equivalence point.

Now before the equivalence point. We're going to say in this part of the titration the moles of weak acid is greater than the moles of strong bass. And we're going to 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.

Titrations: Weak Acid-Strong Base Example

Here it says consider the titration of 75 MLS of 0.300 molar off pyruvic acid, which has a K_a of 4.1×10−3 with 12 mills 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 K_a of less than 1, so it's a 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 ICF chart.

So remember we set up an ICF chart if it's weak and strong, mixing together here with an ICF chart, the strong species is set up as a reactant. 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 that pyruvic acid. Following Bronsted-Lowry definitions of acids and bases, the acid would donate an H⁺ that H⁺ would go to the OH⁻ 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 C₃H₃O₃ aqueous.

Here we have our equation. Now this is an ICF chart, which remember stands for Initial Change Final, and we know that with an ICF chart our units need to be in moles. Remember moles equal liters times molarity, So you would divide these MLS by 1000 to get liters and then multiply them by their molarities 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 row, we place the given amounts in moles and in an ICF chart, we only care about three things. We only care about the weak acid and its conjugate base and then whatever the strong thing is, whether it be a strong acid or a strong base.

So here we don't have any information on the conjugate base. So this is 0. 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 ICF 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 ICF chart. For Step 3. Using the change row, 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 4. The Henderson-Hasselbalch equation for a buffer is used for a buffer to find the pH of a solution. Using the final row, 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-Hasselbalch equation, which is pH equals pK_a plus log of conjugate base over weak acid. So we come back over here. So I'm going to say pH equals pK_a. So that's negative log of our K_a, which is 4.1×10−3 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.

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? K_a = 4.9 × 10⁻¹⁰.

• A. 6.82
• B. 10.01
• C. 8.52
• D. 10.17

At the Equivalence Point

In our discussion of the titration between a weak acid and a strong base, we now take a look at how to calculate pH at the equivalence point. Here we're going to say 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 asset 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 equivalence volume of our titrate. Remember, we can utilize the equation of M⁢acid⁢times⁢V⁢acid=M⁢base⁢times⁢V⁢base in order to determine the volume of our titrate, right. So we can utilize this equation if they don't give us the volume of our titrate.

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 in the pH.

Titrations: Weak Acid-Strong Base Example

Here it is considered the titration of 75 MLS of 0.0300 molar of pyruvic acid with AKA of 4.1×10−3 with 50 mills of 0.450 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 ICF chart. Now remember, with an ICF 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 Bronsted-lowry definition, the acid donates an H plus to the base. So the H 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 ICF chart which is initial change final and with an ICF chart we need our units to be in moles. Moles equals liters times molarity. So divide the MLS by 1000 negative liters and then multiply by their molarities so that will give us the moles of our weak acid and strong base. So we have point 00225.00225 in an ICF chart. We only care about three things, The weak acid, it's conjugate base, and whatever the strong species is.

The 4th species in this case. What do we ignore now? Here are conjugant base Initially is 0. 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 matters either created or destroyed. So we're going to 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 going to have to do an ice chart afterwards.

All right, so Step 4, using the final row, determine the concentration of the conjugate base. So we have .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 the chemical reaction? Well, we have 75 mills of our acid, 50 mills of our base, which is 125 mills total. Dividing it by 1000 gives us 125 liters, so our new concentration is .018 molar. Now set up an ice chart for the conjugate base, which is our weak base that has a reactor with water. It is an ionic base so ignore the neutral metal cation here. I ignore the potassium ions in neutral. It reacts with water.

Again following the Bronsted-lowry definition, it is a base now, so water is going to be the acid. The acid donates an H plus to regenerate the pyruvic acid and to give us our hydroxide ion. Now we're dealing with an ice chart, so that's initial change equilibrium. O here our initial amount is 01/8 molar in a nice chart. We ignore solids and liquids, so water is ignored. Our products initially are 0. 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 going to use KB to solve for X. Now here we're going to 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 a .018 molar, and we divide it by, in this case KB, and 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 going to say here we need KB. They gave us KA in the very beginning. Beginning of the question. Remember that kW equals KA times KB, so KB will equal kW divided by KA. So that'll be KB equals 1.0×10−14 divided by the KA within the question, which is 4.1×10−3 KB. I'm going to write it over here in this box.

So when I do kW divided by KA, my KB comes out to B 2.14×10−12. So now that I have my KB, I do initial concentration divided by KB. So my initial concentration is 01/8 molar 2.4×10−12. This gives me a value of 7.38×109, 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 would be X2/0.018-X. Again, since R ratio is greater than 500, I can ignore this minus X and avoid this quadratic formula. So here's going to be equal to X2/0.018. Cross multiply these two X2=4.32×10−14. Take the square root of both sides.

So I want to take the square root of both sides X=2.078×10−7. Now when you find X, that can either give you H3O 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. Axial will give me OH minus. So we just found the concentration of OH, 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 going to say here pH equals 14 minus POH, so 14 -, 6.68, so this equals 7.32. So the pH of the solution with all this work we found is 7.32.

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 K_a of HNO₂ is 4.6 × 10⁻⁴.

• A. 11.73
• B. 8.13
• C. 5.87
• D. 1.35

After the Equivalence Point

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 going to 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 equivalence 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 going to 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.

Titrations: Weak Acid-Strong Base Example

Here it says consider the titration of Sony 5 MLS of 0.0300 molar of pyruvic acid. Here the KA is 4.1×10-3 and we're reacting it with 75 M LS 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 ICF chart. The strong species has to be set as a reactant, so potassium hydroxide has to be a reactant. It reacts with its chemical opposite, so it reacts with the pyruvic acid.

Now, following the Bronsted-Lowry definition, the acid would protonate or donate an H to the base. That will give us water as a byproduct as the H combines with the OH^- and then the potassium will combine with the pyruvate to give us potassium pyruvate. Now we're going to say that this is an ICF, so that's initial change, final with an ICF 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 going to get 0.00225 moles of our weak acid and then we're going to have 0.003375 of our strong base. We have zero of our conjugate initially and the 4th species we don't care about. Now here, look at the reactants. The smaller moles subtract from the larger moles, so minus OH 0.00225, minus 0.00225. So at the end, we have no weak acid remaining, but we're going to 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 used in 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 MLS, so that's 150 MLS. When you divide that by 1000, that gives you 150 liters.

So. So that'll give me my new concentration, which is 0.0075 molar for my strong bings. Now this is important because recall the concentration of the strong base will be equal to the concentration of OH^-. So we just found out the concentration of OH^- is this number. If we know OH concentration, that's good because that will help me find pOH. So pOH equals negative log of OH^-. So plug that number in. So that's going to give me 2.12. If I know pOH then I know pH because of this formula. So pH if we rewrite it equals 14 minus pOH, so it's 14 - 2.12 which comes out to 11.88. So this will represent the pH of my solution after the equivalence point.

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

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 weak acid is equal to the amount of its conjugate base formed in the solution. This is significant because it is the point at which the pH of the solution equals the p_aK_a of the weak acid. The p_aK_a is a measure of the acid's strength; it is the pH at which the acid is half dissociated. Therefore, at the midpoint of the titration, the solution acts as a buffer, and the pH changes very little with the addition of small amounts of strong base or acid. This is also the point where the buffer capacity is at its highest. Understanding this concept is crucial for accurately determining the p_aK_a of the weak acid and for preparing effective buffer solutions.

What is the relative pH at the equivalence point of the titration of a weak acid with a strong base?

The pH at the equivalence point of a titration involving a weak acid and a strong base will be greater than 7. This is because the weak acid (HA) reacts with the strong base (like NaOH) to form a salt (such as NaA) and water. The salt produced in this reaction typically dissociates into its ions (Na⁺ and A^-). The A^- ion, which is the conjugate base of the weak acid, undergoes hydrolysis with water to form OH^- (hydroxide ions) and HA (the weak acid).

Since the hydrolysis of the conjugate base (A^-) produces OH^- ions, the solution becomes basic. Therefore, the resulting solution at the equivalence point has a pH that is above 7, indicating a basic solution. The exact pH value depends on the strength of the weak acid and the concentrations of the acid and base used in the titration.

In the titration of a weak acid with a strong base with a 1:1 ratio, when does pH equal the pKa?

In a titration of a weak acid with a strong base, the pH equals the pK_a at the point known as the half-equivalence point. This is the stage where exactly half of the amount of the weak acid has been neutralized by the strong base. At this point, the concentration of the weak acid (HA) is equal to the concentration of its conjugate base (A^-). This creates a buffer system, and according to the Henderson-Hasselbalch equation, when [HA] = [A^-], the pH of the solution is equal to the pK_a of the weak acid. This is a key concept in understanding buffer systems and acid-base titrations, as the pK_a gives you insight into the acid strength and the pH at which the buffer system can effectively resist changes in pH.