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

In strong acid-strong base titrations, the equivalence point occurs when equal moles of acid and base react, resulting in a neutral solution with a pH of 7 at 25°C. Before reaching this point, the pH can be calculated using an ICF chart, where moles are determined from molarity and volume. After the equivalence point, excess strong acid or base dictates the pH, which can be calculated using the formula pH = 14 - pOH. Understanding these concepts is crucial for mastering acid-base reactions and titration techniques.

Strong Base-Strong Acid Titration

So in this video, we're going to take a look at strong base strong acid titrations. Now we're going to say here that in a strong base strong acid titration, we're going to see that the initial compound represents our analyte and we're going to make this second structure our titrant. Now we're going to say whenever we have a strong acid or strong base, we never have used an ICE chart. Remember, ICE charts are used predominantly for weak acids and weak bases because they don't dissociate 100%. We are using a nice chart to determine what their concentrations are at equilibrium.

And from there, we can figure out pH or pOH. Now whenever we titrate 2 species, like in this case, a strong acid and a strong base, then we use what's called an ICF chart and it uses the units of moles. Now remember, an ICE chart stands for initial change equilibrium. And when we have 2 strong species titrating one another, we're going to use an ICF, and this stands for initial change final. Now the following roadmap is what we're going to use whenever we have a strong acid to strong base titration.

So first, we start out with the equivalence volume, the volume that's necessary to get to the equivalence point between these two substances. Now we're going to say we calculate the equivalence volume V_e in order to determine the amount of titrant or volume of titrant required to reach the equivalence point. Here we have the titration of 150 ml of 0.100 molar NaOH, which is a strong base, with 0.050 molar nitric acid. Here, nitric acid is acting as our titrant. We're adding it to the NaOH.

Now remember, at the equivalence point, we have equal moles of our acid and our base. At the equivalence point, we could say that m_acid, so the molarity of the acid, times the volume of the acid equals molarity of the base times volume of the base. Remember we said at the equivalence point our moles are equal? Well, remember moles equals molarity times volume. So by multiplying these 2, that's giving me the moles of acid, and by multiplying these 2, that's giving me the moles of base.

This equation is saying that our moles of acid equals our moles of base, which is true at the equivalence point. We're going to plug in these values. So the molarity of our base, NaOH, is 0.100. The molarity of our acid is 0.050 molar, and we don't know what its volume is. Divide both sides by 0.050 molar, see here that our molarities will cancel out and we'll be left with milliliters which are the units for our volume. So when we plug that in, that gives us an answer of 300 milliliters. That means it'll take 300 milliliters of our nitric acid in order to reach the equivalence point in terms of this titration. Now we determine the equivalence volume.

Now let's talk about before we've added any strong acid to our mixture. We're going to say here before any of the strong acid titrant is added, we only have a strong base initially. Remember, strong acid, strong base dissociate completely, so their concentrations can help us determine the amount of H⁺ or OH⁻ present. So here we have the titration of 150 ml of 0.100 molar sodium hydroxide with 0 mls of 0.050 molar nitric acid. Notice here we have no strong acid volume being added.

So all we have within our jar, within our container, is strong base. Now remember, because it's a strong base, it dissociates completely. So it's going to break up into Na⁺ and OH⁻ completely. Because it's a strong electrolyte, the concentration of NaOH is equal to the concentration of each ion. So now we have the concentration of OH⁻ our hydroxide ion.

If you know the concentration of your hydroxide ion, you can calculate pOH because remember, pOH equals negative log of OH⁻. We plug in this concentration of 0.100. That gives me a pOH of 1. If we know what pOH is, we know what pH is because pH equals 14 minus pOH. So we plug in this one that was for pOH and what we'll see is that our pH equals 13.

So here we're just going through the different steps of our titration between this strong base and this strong acid. At this point, all we have is strong base, so we can easily find the pOH of that solution. Now, as we move on to the next video, we'll talk about what starts to happen as I slowly add my strong acid to my strong base.

Strong Base-Strong Acid Titration

So remember, it's going to take us 300 ml of nitric acid to reach the equivalence point. So until that point, we're going to be dealing with titrations before the equivalence point. Now once our acid and base begin to mix, we use an ICF to determine the pH. Now this is true no matter which one is the analyte and which one is the titrant. As soon as an acid and base mix together, a titration is occurring and you have to utilize the ICF chart.

Now, if we take a look at this example, it says a titration of 150 ml, a 0.100 molar sodium hydroxide with 120 ml of 0.050 molar nitric acid. We can see that we're at 120 ml, so we haven't reached the 300 ml necessary to reach the equivalence point. Remember also that when we're dealing with an ICF chart, the units will be in moles. Moles itself equals liters times molarity. Remember, this is just a rearrangement of the molarity equation.

If we're trying to isolate moles, we multiply both sides by liters, and that's why moles equals liters. Now dividing these by 1,000 will give us our liters of sodium hydroxide and nitric acid, so that's 0.150 liters, "of" means multiply, so times 0.100 molar, this is 0.120 liters times 0.050 molar. Multiplying those gives me the moles of NaOH, which I can plug into my ICF chart, and moles of nitric acid. Now in our ICF chart, the two compounds that are involved in the titration are reactants. In this case, it's a strong acid and a strong base.

Remember, an acid acts as a proton donor, so the HNO3 will donate a H⁺ to the OH^- combining together gives us water. Then the Na⁺ and the NO3^- combine together gave us NaNO₃. This represents our conjugate base. Now that we have our initial moles, we're going to go to change. Now in an ICF chart, we look at the moles of the reactants and the smaller amount of moles represents our limiting amount.

It will subtract from the larger moles. At the end, we'll have 0 left of the strong acid and when we subtract these two numbers, we'll have 0.009 moles of our strong base. But remember the law of conservation, matter is neither created nor destroyed. We're not really losing that amount. What's happening, it's being reformed into our conjugate base here as a product.

So whereas we lose 0.006 moles as reactant, it's just being reformed. So we're gaining 0.006 moles over here as product. Bring this down. At the end of this, what we have remaining is strong base and conjugate base. Now, the conjugate base is not going to affect the pH as far as the strong base will.

So again, if we have a strong species remaining, that's what we care about. So we're going to ignore this conjugate base amount. So before we've reached the equivalence point, we don't have quite enough strong acid to completely neutralize our strong base. Because we have strong base remaining, we're going to have to determine its new concentration and from that we can determine our pOH. So we take a look here.

We're going to say the new concentration of my strong base abbreviated as SB equals moles left after the titration divided by total liters used. So we have here 0.006 moles of NaOH remaining divided by our total volume. Our total volume would be this 0.150 liters plus this 0.120 liters. Together, that'll give me a concentration of 0.033 molar of NaOH. Now because I have the molarity of a strong base, I can just take the negative log of that concentration to give me pOH.

So pOH equals negative log of 0.033, so that gives me 1.48 as my pOH. If I know pOH, I know pH because pH equals 14 minus pOH. So that comes out to 12.52. So at this point, before we've reached the equivalence point, we still have excess strong base remaining, so we should expect a pH that's above 7. Here, we've worked it out, and we see that it's 12.52.

Now what we're going to see next is what happens when we get to the equivalence point in this titration of a strong base with a strong acid. So click on the next video and let's see what that entails.

Strong Base-Strong Acid Titration

We've now reached our equivalence point. So we've used the 300 ml of nitric acid necessary to reach this step. Now we're going to say here at the equivalence point of a strong base-strong acid titration, the solution is neutral and that's because we have equal moles of our strong acid and strong base which completely neutralize one another. We're going to say since our solution is neutral, our pH would automatically be equal to 7. Remember, this is true at 25 degrees Celsius.

Remember that if our temperature changes, that affects our \( K_w \) constant and, therefore, the definition of neutral would change to a number other than 7. So, again, a neutral solution, its pH is equal to 7 at exactly 25 degrees Celsius. Now, here we have the titration of 150 ml of 0.100 molar NaOH with our equivalent volume of 300 ml of 0.050 molar nitric acid. So if we divide these by 1000 to get liters and we multiply them by their molarities, we see that we have equal moles of our strong base and our strong acid, which makes sense because at the equivalence point we have equal moles of both. Now remember, we look on the reactant side, the smaller moles will subtract from the larger moles.

Since they're both the same, they completely eliminate one another. So at the end, we have 0 moles of both. Through the use of the law of conservation of mass, we know that we'll have 0.015 moles of this conjugate base. But here's the thing, when a strong acid and strong base completely neutralize each other, the conjugate base that's formed represents a neutral salt. This is further justification to explain why our pH is 7 and why our solution overall is neutral.

So, again, remember, when we have a strong acid and a strong base with one another and they're at the equivalence point, our pH overall will equal 7. Now that we've seen what happens at the equivalence point, let's look and see what happens when we go after the equivalence point.

Strong Base-Strong Acid Titration

We've now gone beyond the equivalence point. Remember, we needed 300 milliliters of nitric acid, and at this point, we've used 310 ml. Now we're going to say that after the equivalence point of a strong base-strong acid titration, we will have excess strong acid remaining. As before, we divide these ml's by 1,000 to get liters, multiplying them by their molarities gives us the moles of NaOH and HNO₃. So here, as always, we look at the reactant side and remember the smaller moles, which are our limiting moles, will subtract from the larger ones.

When I do that, I'll have no base left, but I'll have just a little bit of my strong acid, and that's all that's really necessary. Here again, following the law of conservation of mass, we'd actually be increasing on this side, but that doesn't matter because we talked about how that's a neutral salt, so it doesn't influence the pH. But here, we're going to say that we're going to take this amount of moles of our strong acid and find its concentration. So we're going to say the new concentration of our strong acid will be the moles of it left divided by the total liters. So we have here 0.005 moles of nitric acid divided by the total volume, which is 0.150 liters, plus 0.310 liters.

When we plug that in, that'll give you a molarity of about 0.001087 molar HNO₃. Since it's a strong acid, when I take the negative log of that, that'll give me pH. So pH equals negative log of H⁺, which is the negative log of that concentration I just wrote. It gives me approximately a pH of 2.96 as my final answer. So as we've been slowly adding strong acid to our strong base, we can see that our pH has decreased.

We started at a pH much higher than 7, and now at this part, since we have excess strong acid remaining, we're going to have a pH much lower than 7. Now correction in terms of this neutral salt, if we don't take into consideration things such as activity coefficients, then we can say here that this neutral salt won't influence our pH at all. But here we're not talking about that type of topic. So we're going to say at this point, we care more about the fact that we have excess strong acid than we do that neutral salt, that conjugate base. But these are all the steps that we need to take when we're doing a strong base, strong acid titration. Now that we've seen every step along the way, let's start doing some additional example and practice questions when it comes to these types of titrations.

Strong Base-Strong Acid Calculations

So here it states, calculate the pH of the solution resulting from the titration of 150 ml of 0.20 molar sodium hydroxide with 80 ml of 0.15 molar hydrobromic acid. Alright. So we have here our strong base, and it's being titrated by this strong acid. Remember, when we have an acid and a base undergoing titration, we should set up an ICF (Initial Change Final) chart. So we start out with our NaOH plus HBr, titrating one another.

Now we're going to say here that, remember your acid is a proton donor, so it would donate an H⁺ and OH^-, and that would create water. The Na⁺ and the Br^- left behind will combine together, and that would produce our conjugate base. Remember, we have ICF, which is initial, change, final. Water is a liquid. Remember, liquids and solids are not involved in ICE charts or ICF charts.

Remember an ICF chart uses moles as the units, so we'll divide these ml's by 1,000 to change them into liters, and then multiply by their molarities. When we do that, we'll get 0.030 moles of NaOH, 0.012 moles of HBr. We're not given any information on NaBr, so initially it's 0. Now, looking at the reactant side, the smaller mole total, which is our limiting amount, we'll subtract from the larger moles. So we're going to subtract 0.012 from both.

So, at the end, we'll have no strong acid remaining, and we'll have a little bit of strong base remaining. Through the law of conservation of mass, this side would increase, But here, we're only concerned with the strong species because it'll have a greater impact on the overall pH, so we have excess strong base remaining within this titration. That means that we are before the equivalence point. We haven't added enough strong acid to completely neutralize and overtake the strong base. So now we need the concentration of NaOH.

So we're going to take the moles left and divide it by the total volume used, so that would be 0.150 liters plus 0.080 liters. So when we work that out, we should get 0.078 molar in your calculator. Now that we have a concentration of our strong base, we take the negative log of that, and that'll give us pOH. So it'd be negative log of 0.078, which would give me 1.11 as my pOH. If I know my pOH, then I know my pH, because pH equals 14 minus pOH.

So that comes out as 12.89. So that's the setup for this first example. Remember, we're just implementing the changes that we know in terms of our strong base, strong acid titration. Follow the steps to determine where you are in terms of your titration in reference to the equivalence point. You'll know what you'll have at the end, and then take the necessary steps in order to find your pOH or pH at the end.

Now that we've seen this problem, move on to the next one.

Strong Base-Strong Acid Calculations Practice

So we now have this practice question. It states, calculate the pH of the solution resulting from the titration of 100 mL of 0.30 molar lithium hydride with 150 mL of 0.40 molar hydroiodic acid. We have our strong base initially, and we are titrating it with this strong acid. These are our initial reactants. We're going to say here that we know Li+I− will combine together to give us our conjugate base.

Here, we're not going to see our typical water as a product. What's going to happen here is this H- will combine with this H+, and we actually get H2 gas, which would mean if this reaction took place we'd have the evolution of gas. Now we're going to say that this is ICF because it's an acid and base titrating one another. Just remember that's initial, change, final. We know that the units in an ICF chart have to be in moles, so divide your mL's by 1,000 to get liters and multiply by the molarities to obtain our moles.

This will give me 0.030 moles, and then this will give me 0.060 moles. H2 is a gas that's going to escape the solution, so we don't have to worry about it here. We're going to say now we have lithium iodide initially is 0. Looking on the reactant side, the limiting moles will subtract from the larger moles, so it'll be 0.030 subtracted from 0.060. At the end, we have none of the strong base remaining, and we still have some of our strong acid.

Through the law of conservation of mass, we have the increase here, so we'll have this many moles of our conjugate base, but again we're focusing on the strong acid remaining. Since we have a strong acid remaining, we have to find its concentration now. The concentration of our strong acid equals the moles left behind after the titration is complete divided by the total volume. The total volume would be 0.10 L plus 0.150 L. When we plug that into our calculators, you should get 0.12 molar HI.

Since it's a strong acid, when I take the negative log of that concentration, that'll give me pH. So that comes out to 0.92. Very acidic. So again, in this case, because we have excess strong acid left at the end, that means we are after the equivalence point.

Once we find the concentration of our strong acid, just take the negative log, and you'll find your pH.

Strong Acid-Strong Base Titration

We now will take a look at strong acid, strong base titrations. In this case, when I use the term "strong acid, strong base titration," I'm making our strong acid our analyte, and it's going to be titrated by our strong base. So our strong base will be our titrant. Here, we provided the roadmap to help guide us on the methods to calculate pH at different points of the titration. First of all, it's important to know the equivalence volume when dealing with this type of titration.

So, we're going to say here, calculate the equivalence volume, \( V_e \), in order to determine the volume of titrant required to reach the equivalence point. Here we have 120 mL of 0.250 molar hydrochloric acid, which is our strong acid, with 0.200 molar KOH, which is potassium hydroxide, our strong base. So, we say it's going to be \( m_{\text{acid}} \times V_{\text{acid}} = m_{\text{base}} \times V_{\text{base}} \). We plug in the molarity of our strong acid, so that's 0.250 molar, the volume of our strong acid, which is 120 mL, equals the molarity of my strong base, and then we don't know the volume of the base here. So, what we do here is we're going to divide both sides by 0.200.

So that's going to cancel out the molarities here, and we'll be left with volume of the base in milliliters. When we work this out, the volume of the base, which will represent my equivalence volume, equals 150 mL. It'll take 150 mL of this potassium hydroxide for us to reach the equivalence point. Now before any of this strong base is added, so before the titration has even started, what we will have initially is just the strong acid. And remember, if we have a strong acid, we're going to say its concentration is equal to the concentration of H\(^+\) ions.

So, if we take a look here, we have 0.250 Molar HCl. We don't have to involve the volume yet; that volume comes into play once we start the titration process. So once we start adding the strong base, that's when we'll utilize the volume. Because we only have strong acid here by itself, just focus on this molarity. So since it's a strong acid, it dissociates 100%, creating 100% of H\(^+\) and Cl\(^-\).

Now, we're going to say here the concentration of the strong acid is equal to the concentration of each of those ions. So we have exactly 0.250 molar H\(^+\). And if we know the molarity or concentration of H\(^+\) ions, then we can calculate pH, because remember pH equals the negative log of H\(^+\). So, we take this and we plug it in. When we take the negative log of that, that gives me a pH of 0.602.

So before we've added any strong base, this is our initial pH. Now as we start to slowly add the strong base to our strong acid, we should expect the pH to gradually increase, and that's what we're going to see here in terms of the steps involved in this titration. Alright. So now that we've seen what happens before any strong base is added, let's move on to the next video and see what process we take as this potassium hydroxide is slowly being added.

Strong Acid-Strong Base Titration

So remember, we're going to need 150 ml of potassium hydroxide in order to reach the equivalence point. So as we begin to add potassium hydroxide, until we get to this number, we'll be dealing with calculations before the equivalence point. So if we take a look here, it says, once our acid and base begin to mix, we use an ICF chart to determine the pH. So an ICF chart, remember, stands for Initial, Change, Final. If we take a look here, we have the titration of 120 ml's of 0.250 molar HCl with 100 ml's of 0.200 molar KOH.

We haven't quite reached our equivalence volume yet; we're only at 100. In an ICF chart, the units have to be in moles. Remember, moles equals liters times molarity. That's just a rearrangement of the molarity equation. When we multiply both sides here by liters, we can isolate our moles.

So we're going to have to divide these ml's by 1000; that'll give us liters of each, so that gives us 0.120 liters. 'Of' means multiply, so then it would multiply with the molarity of each compound. This would give me 0.100 liters multiplied by the 0.200 molar. So doing that gives me the moles of HCl, which came out as 0.030 moles. Plugging this in gave me 0.020 moles of KOH.

Now, we're going to say that in this reaction, remember, acid is a proton donor, so the HCl donated an H⁺ to OH⁻, together they formed water here as a product, and then we're going to say next that the K⁺ left behind and the Cl⁻ left behind combine together to form this conjugate base. Now since we are dealing with a titration before the equivalence point, we'll have some strong acid remaining at the end. That's because there's not enough strong base to completely neutralize and overtake the strong acid. So if we go back to the ICF chart, we're going to look at the reactant side. We have these two mole amounts.

We're going to say that the smaller moles, which is our limiting amount, will deduct from the larger moles. So subtract 0.020 from 0.030. At the end, we'll have no strong base remaining, and then we'll have some strong acid left. Through the law of conservation of mass, remember, matter cannot be created nor destroyed, it just changes form. So in reality, we're not exactly losing these compounds, they're just being reformed into this conjugate base.

So on this side, on the reactant side, we have a loss of 0.02, on this side here, we'll have an addition of 0.020. So bring that down. Now, in the titration of a strong acid and a strong base, we have the creation of a neutral salt. So this neutral salt, if we don't take into account things such as activity coefficients, we can ignore it. Okay?

Activity coefficients, we'll talk about later on when we try to connect it to pH and pOH. But for now, this neutral salt will not influence our pH, so we can ignore it. So what's important here is we have a strong acid left at the end. Now this strong acid, because we have it, we can determine its concentration. So we're going to say here that the strong acid, we're going to take the moles left, we're going to divide them by the total liters.

So the moles left of our strong acid are 0.010 moles. The total volume would be this 0.120 liters plus this 0.100 liters. Altogether that gives me 0.045 molar HCl. And since I have the concentration of a strong acid, when I take the negative log of it, that will give me my pH. Because remember, what was said above, that the concentration of a strong acid is equal to the concentration of H⁺.

So that's going to equal 1.35. So we can see that we've added some strong base, and it's caused my pH to increase. We started out around 0.602 initially when we only had the strong acid by itself. Adding the base is going to cause the pH to increase over time. At this point, now we're at 1.35.

Now that we've seen the calculation setup that we use when we're dealing with before the equivalence point, move on to the next video and see what happens when we get to the equivalence point between this strong acid and strong base.

Strong Acid-Strong Base Titration

So recall, it's going to take 150 mounds of potassium hydroxide in order to reach the equivalence point. Now, in this question, we've finally reached that total. So we're going to say at the equivalence point of a strong acid-strong base titration, the solution is neutral and the pH will equal 7 at 25 degrees Celsius. So, remember, when we reach the equivalence point between a strong acid and a strong base, no matter which one is the analyte and which one is the titrant, our pH should be equal to 7 at 25 degrees Celsius. They completely neutralize each other and we'll be left with a conjugate base, that is a neutral salt.

That neutral salt, if we don't take into account things such as activity coefficients, will have no impact on the pH. So if we take a look here, we divide these by 1,000, that's going to give us liters of each, multiplying them by their molarities will give us moles. So doing that, we see that we have identical moles of KOH and HCl, which makes sense because at the equivalence point, we have equal moles of our acid and our base. When they subtract from one another, we'll see that we have 0 left of both. We have the addition of our conjugate base here on this side, which again, since it's a neutral salt, it doesn't have an impact on the pH.

This further justifies why our solution is neutral at the end. So again, whether we're doing a strong acid strong base titration or strong base strong acid titration, as long as both species are strong, when we get to the equivalence point, we're going to have a pH that will be neutral. Okay? So we're going to have a pH equal to 7 if the temperature is also at 25 degrees Celsius. Now that we've seen this, let's move on to the last video and see what happens when we go beyond the equivalence point.

Strong Acid-Strong Base Titration

So here, we've gone beyond the equivalence volume. Remember we needed 150 ml of KOH. So after the equivalence point of a strong after strong base titration, we will have excess strong base remaining. We have excess strong base. If we take a look here, we divide these ml's by 1000 to get liters and then multiply them by their molarities, that'll give us the moles of each compound.

That gives us 0.036 moles of potassium hydroxide and 0.030 moles of hydrochloric acid. Looking at the reactant side, we subtract the smaller mole total from the larger one. As a result of this, hydrochloric acid is 0 at the end, and then we'll have 0.006 molesKOH. What we're gonna do is realize that we're gonna add this same amount to the other side. Since it's a neutral salt, we will ignore it.

Now because we're after the equivalence point, we're gonna have strong base present. That's our excess base left at the end. If I can find its new concentration, that can give me the concentration of OH^-. We're gonna take the moles that we have left of KOH and divide it by the total volume, which is 0.120 liters+0.180 liters. So that gives me 0.020 for my KOH, the molarity of my KOH.

Since I know the molarity of my strong base, taking the negative log of that will give me my pOH. So plug that in, that gives me 1.70. And if I know pOH, then I know pH because pH equals 14 minus pOH. So we have 12.3 as our final pH. So we can see that as we've added strong base to our initial strong acid, we've seen that the pH has jumped up, starting at a lowly point 6.02 all the way up to this final pH of 12.3, and that's what's going on here.

Our strong base is driving up our pH. Now that we've seen all the steps at different points of this titration, move on to the next video and let's start doing some practice and example questions.

Strong Acid-Strong Base Calculations 1

So here the example says, calculate the pH of the solution resulting from the titration of 50 ml of 0.10 molar hydroiodic acid with 20 ml of 0.30 molar sodium hydroxide. So we have our strong acid here that is being titrated by our strong base. We're going to write down our compound, so HI + NaOH. We know that HI will protonate the OH^- to give us H₂O, so it gives an H⁺ to the OH^-. Then we're going to have Na⁺ remaining and I^- remaining, and they combine together to give us our conjugate base.

Because we have an acid and a base titrating one another, we set up our ICF chart. So initial, change, final. Water is a liquid, liquids and solids are not found in either ICE charts or ICF charts. Remember, in an ICF chart, we need the units to be in moles, so divide the ml's by 1,000 to get liters, then multiply them by their molarities, that'll give us the moles of HI and NaOH. When we do that, we're gonna get 0.0050 moles and 0.0060 moles.

Initially, our conjugate base is 0 since we're not given any information on it. On the reactant side, the smaller moles, which is our limiting amount, will deduct from the larger moles. So subtract 0.0050, 0.0050. At the end, we're gonna have 0 left of our strong acid. That means that we've gone beyond the equivalence point where there's going to be an excess of strong base remaining.

So this is 0.0010 moles remaining. Here we have our conjugate base. It's a neutral salt, so we don't care about it. We care about the strong species because it has a greater impact on our final pH. Since we have a strong base remaining, we just need to find its concentration.

So we're gonna take the moles left of it and divide it by the total volume, which would be 0.070 liters. Alright. So then when we plug all that into our calculator, we'll get a molarity of 0.014 molar of NaOH. Since we know the concentration of our base, we can use that to help us find pOH, so negative log of OH^-, so plug that concentration in. So that equals 1.85.

Since we know the pOH, we can find the pH. pH would equal 14 minus that pOH total. So that comes out to be 12.15. So in this case, we had excess strong base titrant left, so we're dealing with after the equivalence point. So find the concentration of that strong base and use it to find pOH, and then from there pH.

Now that we've seen this example question, click on to the next video and let's continue our discussion on this type of titration.

Strong Acid-Strong Base Calculations 1

So here our practice question states, calculate the pH of the solution resulting from the titration of 90 ml of 0.40 molar chloric acid with 50 ml of 0.50 molar potassium hydroxide. So again, we have a strong acid being titrated by a strong base. So we're going to write those down as our reactants: HClO₃ plus KOH. The H⁺ from the acid will be given to the OH^- from the base, giving us water. The K⁺ and ClO₃^- left behind will combine together to give us our conjugate base.

Since we have an acid and base titrating one another, we have an ICF chart, so we have initial, change, final. Water is ignored. Now remember, we need moles as our units within this ICF chart, so we're going to divide the ml's by 1,000 to get liters, and then multiply them by the molarities to get the moles of our chloric acid and our potassium hydroxide. When we do that, we should get 0.036 moles of our strong acid, and 0.025 moles of our strong base, and 0 of our conjugate base. On the reactant side, the smaller moles, which is our limiting amount, will subtract from the larger moles.

So at the end, we'll have 0 left of my strong base, but 0.011 moles of our strong acid. Since we still have some strong acid left, that means there wasn't enough strong base to completely neutralize it. This is dealing with titration before the equivalence point. Here, we're adding this again as we normally do, but it's a neutral salt. The strong acid is what we care about here.

We're going to say that the concentration of our strong acid, we take the moles left divided by the total volume, so that would be 0.090 liters from the strong acid and base. Together, that gives me 0.079 molar of this strong acid. Now because I have a strong acid, its concentration is equal to the concentration of H⁺, and with that, I can determine the pH. So plug that in, and when I do that, I get 1.10 as my final pH. So we can see that the titration between two strong species is pretty simple as long as you adhere to the ICF chart principles.

Just make sure you see how much, moles are left of either one to help us determine if we'll figure out pOH or pH at the end.

Here’s what students ask on this topic:

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

The equivalence point in a strong acid-strong base titration occurs when equal moles of the acid and base have reacted completely. At this point, the solution is neutral, and the pH is 7 at 25°C. This is because the strong acid and strong base neutralize each other, forming water and a neutral salt. The equation used to determine the equivalence point is:

$m\mathrm{acid}\times V\mathrm{acid}=m\mathrm{base}\times V\mathrm{base}$

where $m$ is molarity and $V$ is volume.

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

Before the equivalence point, the pH is determined by the concentration of the excess reactant. Use an ICF (Initial, Change, Final) chart to find the moles of the remaining strong acid or base. For example, if you have excess strong base, calculate its concentration:

$\mathrm{Concentration}=\frac{\mathrm{moles}}{\mathrm{total\; volume}}$

Then, find the pOH:

$\mathrm{pOH}=-\log \left(\mathrm{OH}-\right)$

Finally, convert pOH to pH:

$\mathrm{pH}=14-\mathrm{pOH}$

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

After the equivalence point, the pH is determined by the concentration of the excess strong acid or base. For example, if excess strong acid is present, calculate its concentration:

$\mathrm{Concentration}=\frac{\mathrm{moles}}{\mathrm{total\; volume}}$

Then, find the pH directly:

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

If excess strong base is present, calculate its concentration and find the pOH:

$\mathrm{pOH}=-\log \left(\mathrm{OH}-\right)$

Then, convert pOH to pH:

$\mathrm{pH}=14-\mathrm{pOH}$

Why is an ICF chart used in strong acid-strong base titrations?

An ICF chart (Initial, Change, Final) is used in strong acid-strong base titrations to track the moles of reactants and products throughout the titration process. Unlike ICE charts, which are used for weak acids and bases, ICF charts are suitable for strong acids and bases because they dissociate completely. The ICF chart helps determine the remaining moles of acid or base at any point in the titration, which is crucial for calculating the pH or pOH of the solution.

How do you determine the equivalence volume in a strong acid-strong base titration?

The equivalence volume in a strong acid-strong base titration is determined using the equation:

$m\mathrm{acid}\times V\mathrm{acid}=m\mathrm{base}\times V\mathrm{base}$

where $m$ is molarity and $V$ is volume. By rearranging the equation to solve for the unknown volume, you can find the volume of titrant required to reach the equivalence point. For example, if you have 120 mL of 0.250 M HCl and 0.200 M KOH, the equivalence volume of KOH is:

$\frac{0.250\times 120}{0.200}=150mL$