So now we're going to take a look at weak acid-strong base titrations. Now we're going to say here that our weak species, which is our acid, will behave as the analyte, and our strong base, which we'll slowly add to it, represents our titrant. Now we're going to say whenever you titrate a weak species, in this case, the acid, with a strong species, in this case, the strong base, the fact that we have an acid and a base mixing together means that we're going to have to use the ICF chart. Now when we say ICF, ICF stands for Initial, Change, Final. Now we're going to use the following roadmap, to determine the pH at different points in our titration between a weak acid and a strong base. Now, before we even begin all of this, we first look at the equivalence volume. Now remember, we calculate the equivalence volume, \( v_e \), in order to determine the volume of titrant required to reach the equivalence point. In this case, the titrant again is our strong base. Remember, the equivalence point is where your moles of acid equals moles of base. And remember that moles itself equals liters times molarity. So if we take a look here, it says the titration of 300 ml of 0.100 molar nitrous acid with 0.30 molar potassium hydroxide. So nitrous acid represents our weak acid, potassium hydroxide represents our strong base. We're going to say here that \( m_{\text{acid}} \times v_{\text{acid}} = m_{\text{base}} \times v_{\text{base}} \). So the molarity of my acid is 0.100 molar, though its volume is 300 milliliters, the molarity of my strong base is 0.30 molar, and we're looking for the volume of our base, our equivalence volume in this case. Divide both sides by 0.30 molar, here you'll see that molarities cancel out and we'll be left with milliliters as our volume. So when we work that out, we'll get 100 ml's of KOH is required in order to reach the equivalence point. Now let's look at the next step, this is before any strong base has been added:

Before any strong base has been added, we just have a weak acid initially. Here we have the titration of 300 ml of 0.100 molar nitrous acid with 0 ml of 0.300 molar KOH. Here, there is no strong base being added, so again, we just have a weak acid by itself. Remember, when we have a weak acid or a weak base by itself, we need to utilize an ICE chart in order to determine our pH. Now, an ICF chart represents Initial, Change, Final, but an ICE chart represents Initial, Change, Equilibrium. Remember, in these types of ICE charts, our units will be molarity. Because no strong base is being added, we don't need this volume initially. All we require is the initial molarity of my weak species. We plug that in. Remember, weak acids and weak bases react with water within an ICE chart. So this nitric acid would react with the water. Remember, an acid is a proton donor. So \( \text{HNO}_2 \) donates an \( \text{H}^+ \) to give us \( \text{NO}_2^- \) at the end, which is our nitrite ion, and then \( \text{H}_2\text{O} \) accepts an \( \text{H}^+ \) to become \( \text{H}_3\text{O}^+ \). Our change here, we lose reactants in order to create products. So our reactant here will be minus x, our products here that are being formed are plus x. Initially no information is given on them, so initially there's 0. We bring down everything for our equilibrium line, so 0.100 minus x, and then plus x and plus x comes down. Now at this point, we would say that since we're dealing with a weak acid, we have to use our acid dissociation constant. Remember, weak acids use \( K_a \), weak bases, which we'll see later, deal with \( K_b \). Now \( K_a \) is the equilibrium constant for the weak acid. It equals products over reactants, so we know that'll be x times x, which is \( x^2 \), divided by our initial concentration, minus x. If we can isolate our x variable, that x variable will give us \( \text{H}_3\text{O}^+ \), which is the same thing as \( \text{H}^+ \). If we know the concentrations of \( \text{H}^+ \), we can take the negative log of that to find pH. Now there are going to be situations where we can ignore this minus x here. In those situations where we can ignore that minus x, we can avoid the quadratic formula. In cases where we cannot ignore that minus x, we keep it in, and we set up all the math that we need to do, and from there, we use a quadratic formula. Now in order to determine if we can keep that minus x or not, we do what I call a 5% approximation. So basically, we have our initial concentration of our weak acid divided by its \( K_a \) value. If this ratio gives us a value greater than 500, then we can ignore the minus x. Now in this case, the initial concentration of weak acid is 0.100, and the \( K_a \) of our nitrous acid from your book is approximately \( 7.1 \times 10^{-4} \). When you plug that in, you'll see that it gives you 140.8. Here it's not greater than 500, so we'd have to keep this minus x and use the quadratic formula in order to isolate x. So if we work this out we'd have, \( K_a \) so I'm going to take this expression here. So \( K_a \) is \( 7.1 \times 10^{-4} \) equals \( x^2 \) divided by \( 0.100 - x \). So you would have if you work this out, you'd have \( 7.1 \times 10^{-4} \times (0.100 - x) = x^2 \). Then, we're going to say here, you're going to distribute. When you work that out, you're going to get \( 7.1 \times 10^{-5} - 7.1 \times 10^{-4}x = x^2 \). This x has the highest power, so it's the lead term. When you rearrange this, it gives you \( x^2 + 7.1 \times 10^{-4}x - 7.1 \times 10^{-5} \). Then we're going to say here, the quadratic formula remember is negative b, plus or minus, square root of b squared minus 4ac over 2a. Okay. You would plug into that formula. So bring it down, it'd be negative \( 7.1 \times 10^{-4} \), plus or minus \( 7.1 \times 10^{-4} \) squared minus 4 times 1 times, don't forget the negative sign for c, negative \( 7.1 \times 10^{-5} \) divided by 2 times 1. Now because of this plus-minus point here, you'd get 2 x values, one that's positive and one that's negative. But here's the thing, at equilibrium, you're only allowed to have a positive value, so we would disregard the negative x and go with the positive x answer. That would give us x equals 0.00808 molar. Again, in this situation, x would be equal to \( \text{H}_3\text{O}^+ \), which is equal to \( \text{H}^+ \). Taking the negative log of that number would give you my pH, which is 2.09. So that would be the long version of this reaction. So again, at this point, we don't have any strong base being added. We just have a weak acid by itself, so we must utilize an ICE chart in order to isolate our pH. Now, click on the next video, and let's start slowly adding our strong base and see what happens in our titration.