Titrations and Titration Curves - Online Tutor, Practice Problems & Exam Prep

Redox titration curves illustrate the concentration changes of the analyte and titrant as titrant volume is added. The equivalence volume is calculated using the equation: $CV=CV$. At the equivalence point, equal moles of analyte and titrant form a precipitate, and the concentration is determined using the solubility product constant (Ksp). Post-equivalence, excess titrant concentration is calculated to find the analyte concentration, demonstrating the principles of redox reactions and equilibrium.

Redox Titration Curves

We've talked about titrations as well as titration curves in the past. Now we're dealing with redox titration curves. We're going to say here a redox titration curve follows the change in either the analyte, which is coined your titrant, or the titrant itself's concentration as a function of the titrant's volume. So as we're adding our titrant volume by volume, we're going to see a change in the concentration of either species. Now with all titrations, whether they be acid-based titrations or redox titrations, we should always first determine what our equivalents volume of our titrant will be.

So here we have the titration of 50 ml of 0.100 molar sodium chloride with 0.100 molar silver nitrate. Here would produce our precipitation reaction. What happens here is that Na⁺ would combine with the NO₃^- here. But based on solubility rules, that would give us an aqueous compound. Here, we're concerned with the silver ion combining with the chloride ion to give us silver chloride. Remember, when dealing with a precipitation reaction, we're really talking about K_sp of the ionic compound. So remember, K_sp deals with the solid ionic compound breaking up into its ions.

Here in this equation, what I've done is we've reversed the reaction. So now, my products are reactants and my reactant here is now a product to show the formation of the solid. Here, this K represents our formation constant. It is actually the inverse of my K_sp. So K here is actually 1 over K_sp. Remember here, when we reverse the reaction, we get the inverse of our original equilibrium constant. So for this original breakdown of my ionic solid, we had K_sp. Because I reversed the reaction, it now becomes 1 over K_sp, which is represented by this formation constant here.

The fact that it is a number much greater than 1 tells me that the formation of this solid is highly favorable, which makes sense because based on solubility rules, silver when it combines with the chloride ion definitely forms a solid precipitate. Remember, in terms of calculating the equivalence volume, here we'd say the molarity of my analyte, which we'll say is a times volume of my analyte equals the molarity of my titrant times equivalence volume of my titrant. And when we're calculating the equivalence volume, we're looking for the volume of the titrant. So you'd plug in 0.100 molar of my analyte times its volume equals We know that the first compound is the analyte because we're seeing the titration of this with this. So we're adding this to it. We have 0.100 molar times the equivalence volume of my titrant. Divide both sides by 0.100 molar. So my equivalence volume of my titrant equals 50 ml. So that's the first step in terms of our redox titration. Gathering all this information, overall, will help us determine what our titration curve would look like once we've calculated all different points in terms of the titration.

Now that we've covered equivalence volume, move over to the next video where we take a look at calculations before we reach the equivalence point.

Redox Titration Curves

So before we reach the equivalence point, we have an excess of analyte. At this point, we have an excess of analyte. We need 50 mL of our titrant to get to the equivalence point. Here, we're only using 20 mL. At this point, we're dealing with an excess of analyte, so when we're calculating the concentration of our analyte, we're actually calculating the amount that is left unreacted. That's what we're calculating at this point. Now, here, _Cl^- happens to be our analyte. _Ag⁺ is our titrant. It is the ion that's going to connect with the _Cl^- to form the precipitate. Here we'd say that this represents our concentration of analyte. Here, we'd say its concentration equals its initial moles minus the moles of titrant added divided by the total volume. Now, I wrote moles here and remember moles equal liters times molarity. But you could just leave it in milliliters if you want. Remember the word 'of' in between two numbers means multiply. So I could have kept it in millimoles which would just be milliliters times molarity. If we're using liters times molarity, then at the end, we'd have moles on top and we'd have to use liters on the bottom. And that would give me molarity. If I choose to keep the mL's, it'd be milliliters times molarity which would give me millimoles. So you'd have millimoles on top, milliliters on the bottom, which ratio-wise doesn't change anything proportionally. So this would still give you molarity at the end. Here we'll just keep the mL's. So, we're going to do 50 mL times 0.100 M of NaCl. So that's going to give me 5 millimoles of my analyte minus 20 mL times 0.100 M, 2 millimoles of my titrant divided by the total volume. So that's 50 mL plus 20 mL. Okay, so that's going to give me my new concentration of unreacted analyte. So that comes out to 0.042857 M.

Here, we're talking about it in terms of a titration curve. So we're going to do pCl^-. Remember, 'p' here just means negative log. So we're taking the negative log of the chloride ion concentration. Plugging in what we just calculated, here that gives me 1.37. So before we reach the equivalence point, at this exact moment, based on the amount of titrant used, this would be the negative log of my analyte concentration. As we continue to add more and more titrant, we should expect this value to increase.

Now that we've talked about before the equivalence point, move over to the next question and take a look at what happens when we're at the equivalence point. Click over to the next video and see what we do in terms of that situation.

Redox Titration Curves

So at the equivalence point, we have equal moles of our analyte and our titrant. Here, we've formed our precipitate. It exists in equilibrium with the ions itself. To figure out the concentration of our particular analyte, we actually have to use the solubility product constant of the precipitate. We're going to have to use the K sp of my silver chloride solid. Here, remember how do we do that? We have our silver chloride solid. It breaks up into ions. It breaks up into silver ion, aqueous plus chloride ion, aqueous. Here, solids are ignored.  We're going to say at equilibrium, we only have one silver ion here which is x and one chloride ion here which is x. If there were 2 here, we would have to put a 2 here and that would have an effect on my equilibrium expression. Remember, K sp is just like any other equilibrium constant. It equals products over reactants, but here my reactant is a solid, so we're gonna ignore it. So K sp would just equal products. So it would equal Ag⁺ times Cl⁻. So we'd say K sp , which is this number, equals x times x. So it equals x². Let's say that there were 2 here, then K sp would equal actually, bring the 2 down here, it would equal x times 2x². Because remember, whatever the coefficient is, that becomes the power. So just in those cases, that's what we would do for those situations. Now, what we need to do is we need just to isolate x because once we know what x is, that gives me actually the concentration of Chloride Ion. So take the square root of both sides here. That gives me x equals 1.34 times 10⁻⁵ molar. That equals the concentration of my chloride ion. Now we just say again, p of Cl equals negative log of my chloride ion concentration. So that equals 4.87. So as we expected, we see an increase in the negative log of my chloride ion as I add more and more titrant. Now finally, we're going to go to the point where we are after the equivalence point and we need to determine what our final p of chlorine would be in that situation. So click onto the last video and see what we do in this situation.

Redox Titration Curves

After the equivalence point, what we now have is an excess of our titrant, and in order to figure out the concentration of our analyte, we first need to determine what the concentration of our titrant would be at this point. So we're gonna say here silver ion represents our titrant. So it's moles of the titrant added or millimoles to make it easier. So multiply 70 ml's times 0.100 molar. That gives us 7 millimoles minus millimoles of our analyte. So multiply these 2 gives me 5 millimoles divided by the total concentration which would be 50 ml's plus 70 ml's. Ounce. When we plug that in, that gives me 0.167 molar.

Now that we have the concentration of our excess analyte, we're going to use that value with K_sp in order to figure out the concentration of my analyte. So we're going to take this equilibrium expression here that we've seen earlier. So we're going to say K_sp equals the concentration of my titrant times the concentration of my analyte. Plug in the value that we have for K_sp, which was given as: 1.8 × 10 - 10 , Here, this would be 0.0167 molar and we're looking for the analyte concentration. Divide both sides by 0.0167 molar. Cancel, cancel. We'd get our concentration here for chloride ion as being: 1.07784 × 10 - 8 molar.

Take the negative log of that number would give me p of cl. So here, that would give me 7.97 when I take the negative log of this concentration. So we can see that taking the negative log of my concentration has increased as I go through the steps for the redox titration. Now, here if we were to construct basically a chart, we’d say here, we’d have the volume of our titrant which would be in this case, the silver nitrate. Here, we’d have the negative log of our analyte, which in this case is the chloride ion, and we would see from our calculations that we saw that pCl was increasing and what we’d expect to see is if we did a bunch of points, we’d see it gradually increase and then you’d see it hit a volume where there's a sharp increase meaning that we've reached the equivalence volume where there's an equal moles of our titrant and our analyte reacting with one another, and then eventually it’d level off. If we’re to use an indicator, that indicator would change colors at a precise moment as it reacts with the titrant. And that’d be somewhere in the close to the middle or so of this redox titration curve. Sometimes it may not be the exact middle, But again, you’d use an indicator to find more or less where the approximate range of the end point would be.

Another thing that we could use is we could use, potent metric titrations in which we can basically use readings in terms of voltage to give us information on the different types of solution used within this redox titration. From this information, we could possibly identify our unknown analyte if this were done in a lab. And just realize very similar to things that we’ve done in the past in terms of acid based titrations, but now we’re using it in terms of redox titrations. So just remember, we apply the same principles that we’ve done before in the past where we're trying to determine what our equivalence volume is to start. And then from there, determining are we at the equivalence point, before the equivalence point or after the equivalence point to know what set of equations to utilize to find our final answer.

Redox Titration Curves Calculations

So here it says to calculate the fluoride ion concentration from the titration of 130 ml of 0.120 molar potassium fluoride with 150 ml of 0.100 molar barium chloride. Here, we're told that the solubility product constant of barium fluoride is 1.5 times 10 to the negative 6. Alright. So we're titrating potassium fluoride with barium chloride. So what we need to do first is determine what our equivalence volume will be. This will help me determine, am I at the equivalence point, before the equivalence point, or after the equivalence point? So, m_analyte ⋅ v_analyte = m_titrant ⋅ v_titrant . We have here 0.120 molar times 130 ml equals 0.100 molar times the equivalence volume of my titrant. Divide both sides by 0.100 molar. So my equivalence volume is 156 milliliters. So I need 156 milliliters of the titrant to reach the equivalence point. Here, I only have 150 ml. So that means we're dealing with calculations before the equivalence point. And if we're dealing with the calculations before the equivalence point, that means that we have an excess of analyte. So we're going to see the concentration of our analyte. How do we know who's the analyte? The analyte in this case would be fluoride ion because we're titrating with barium chloride. Barium is part of the ionic solid with the Ksp. So, barium is the titrant, which means my fluoride ion is the analyte. It's what's being titrated by the barium chloride. So here, concentration of my analyte here would equal moles or millimoles of analyte minus millimoles of titrant divided by total volume. So multiply milliliters times molarity to get the millimoles of each. So that's 15.6 millimoles of fluoride ion minus 15 millimoles of barium ion divided by 280 milliliters. So here, that would give me a concentration of 2.14 × 10 - 3 molar of fluoride ions. So that will represent the concentration of my fluoride ion that's remaining before I've reached the equivalence point. If we wanted to go a step further, you could have just done after that negative log of this concentration to find Here, we're not being asked to do that. We just have to figure out the concentration which is this original value here. Now that we've seen this, attend to the practice question that's left on the bottom the page and come back and see if your answer matches up with mine.

Calculate the pCN from the titration of 40.0 mL of 0.060 M NaCN with 20.0 mL of 0.050 M AgC₂H_­3O₂. The solubility product constant of AgCN is 2.2 x 10^-16.

• A. 1.53
• B. 0.023
• C. 1.45
• D. 1.63
• E. 0.029

Here’s what students ask on this topic:

What is a redox titration curve and how is it different from an acid-base titration curve?

A redox titration curve tracks the concentration changes of the analyte and titrant as the titrant volume is added. Unlike acid-base titration curves, which focus on pH changes, redox titration curves focus on the redox potential or concentration of specific ions. The equivalence point in redox titrations is determined by the stoichiometric reaction between the analyte and titrant, often involving a change in oxidation states. The curve typically shows a sharp change at the equivalence point, similar to acid-base titrations, but the calculations often involve redox potentials and solubility product constants (K_sp).

How do you calculate the equivalence volume in a redox titration?

The equivalence volume in a redox titration is calculated using the equation:

$C_1{V}_1=C_2{V}_2$

where C₁ and V₁ are the concentration and volume of the analyte, and C₂ and V₂ are the concentration and volume of the titrant. By rearranging the equation to solve for V₂, you can determine the volume of titrant needed to reach the equivalence point. For example, if you have 50 mL of 0.100 M NaCl and titrate with 0.100 M AgNO₃, the equivalence volume is 50 mL.

What happens at the equivalence point in a redox titration?

At the equivalence point in a redox titration, the moles of analyte and titrant are equal, resulting in the formation of a precipitate or a change in oxidation state. For example, in the titration of NaCl with AgNO₃, Ag⁺ ions react with Cl⁻ ions to form AgCl precipitate. The concentration of the analyte is determined using the solubility product constant (K_sp) of the precipitate. The K_sp expression for AgCl is:

$K_{\mathrm{sp}}=x^2$

where x is the concentration of Ag⁺ and Cl⁻ ions at equilibrium.

How do you determine the concentration of the analyte after the equivalence point in a redox titration?

After the equivalence point, there is an excess of titrant. To determine the analyte concentration, first calculate the excess titrant concentration. For example, if 70 mL of 0.100 M AgNO₃ is added to 50 mL of 0.100 M NaCl, the excess Ag⁺ concentration is:

$\frac{7-5}{50+70}=0.0167$

Then, use the K_sp of AgCl to find the analyte concentration:

$K_{\mathrm{sp}}=C_{\mathrm{titrant}}{C}_{\mathrm{analyte}}$

Solving for the analyte concentration gives:

$C_{\mathrm{analyte}}=\frac{K_{\mathrm{sp}}}{C_{\mathrm{titrant}}}$

What is the role of the solubility product constant (Ksp) in redox titrations?

The solubility product constant (K_sp) plays a crucial role in redox titrations involving precipitates. K_sp is used to determine the concentration of ions in a saturated solution at equilibrium. For example, in the titration of NaCl with AgNO₃, Ag⁺ and Cl⁻ ions form AgCl precipitate. The K_sp expression for AgCl is:

$K_{\mathrm{sp}}=x^2$

where x is the concentration of Ag⁺ and Cl⁻ ions at equilibrium. By knowing K_sp, you can calculate the ion concentrations at the equivalence point and beyond, helping to construct the titration curve and understand the reaction dynamics.