We've been looking at the titration curve for the titration of a weak acid, acetic acid, with a strong base, sodium hydroxide. And in part A, we found the pH before we'd added any base at all. So we found this point on our titration.
and we also found in part B, the pH after you add 100 milliliters of base, so we found the pH at this point. In part C, our goal is to find the pH after the addition of 200 mLs of 0.05, 0.05 molar solution of sodium hydroxide. So how many moles of hydroxide ions are we adding here?
So if the concentration of sodium hydroxide is 0.05 molar, that's the same concentration for hydroxide. So for hydroxide, we have a concentration of 0.05 molar, and molarity is moles over liters, right? So moles over liters, we have 200 milliliters, and if we move our decimal place, one, two, three. that's 0.2 liters, so this is 0.2 liters here. So we solve for moles, so 0.05 times 0.2 is going to give us 0.01, 0.01 moles of hydroxide ions.
So that's how many moles of hydroxide ions we're adding to our original acid solution. How many moles of acid did we start with? Well, we had 50 milliliters of a 0.2 molar solution.
of acetic acid. So the concentration of acetic acid, the concentration of acetic acid was equal to 0.2 molar, and that's equal to moles over liters. So 50 milliliters would be.05 liters. So this would be.05 liters.
Once again, solve for moles. So.2 times.05 gives us.01. zero moles of acetic acids. So notice, we have the same number of moles of acetic acid. of acid as we do of base.
And the base is going to neutralize the acid. Let's go ahead and write the reaction. Let's write the neutralization reaction.
If we have, we start with some acetic acid, and to our acetic solution we add our sodium hydroxide. So we're adding some sodium hydroxide here. The hydroxide ions are going to take the acetic proton. So a hydroxide ion takes this acetic proton, proton right here, H plus and OH minus give us H2O.
If you take away the acidic proton from acetic acid, you're left with acetate, CH3COO minus. And we're starting with.01 moles of acetic acid. So let's color coordinate here. So we're starting with.01 moles, over here we'll put moles. So.01 moles of acetic acid.
And that's the same. number of moles of base, right,.01. So we have.01 moles of base. Notice our mole ratio is one to one.
So we have one to one here. So all of the base is going to react. It's going to completely neutralize the acid that we originally had present.
So all of our base reacts and we end up with zero here and all of our acid has been completely neutralized. All right, we lose all. all of this, so we lose all of that, and so we've neutralized all of our acid too. So this is the equivalence point for this titration.
And if we're losing acetic acid, we're converting acetic acid into acetate. So if we think about starting with zero moles of acetate, and we lose.01 moles of acetic acid, that turns into acetate. So we have to write plus.01 over here. so we're making that, so we end up with.01 moles of our acetate anion. Alright, next, if we have moles of our acetate anion, we have a concentration, so if we find the total volume of our solution, we can find the concentration of acetate anion, so let's do that next.
We'll go back up here, what is the total volume now? We started with 50 milliliters and we added 200 more, so. So 50 and 200 give us 250 milliliters.
So that's our total volume now. And 250 milliliters would be, move our decimal place, 3.25 liters. So next, we find our concentration of acetate.
So what is our concentration? of acetate now. It would be moles over liters, so.01.
So we have.01. moles over.25 liters. Over.25 liters. So we can go ahead and find the concentration.
You could do this in your head, or I'll just show you on the calculator..01 divided by.25 is going to give us a concentration of.04. So the concentration is.04 molar. So we have acetate anions in solution at the equivalence point. and acetate reacts with water. So let's go ahead and show that reaction.
So we have acetate, which can react with water, and this reaction will eventually come to an equilibrium here. So the acetate anion acts as a base and takes a proton from water. So if acetate picks up a proton, it turns into acetic acid, so CH3COOH.
If we take a proton away from water, we're left with hydroxide. so OH minus. So now, let's think about our initial concentration of acetate, so our initial concentration of acetate was.04, so we write here initial concentration is.04 molar, and we're assuming that we don't have any products yet, so we write zero here. Next, we think about our change. Well, a certain concentration of acetate is going to react, so we're going to lose.04 molar.
a certain concentration which we call x. Whatever we lose for acetate, we gain for acetic acid. So if it's minus x for acetate, it must be plus x over here for acetic acid, and therefore also plus x for hydroxide. Alright, so at equilibrium, at equilibrium, our concentration would be.04 minus x for acetate, and then we'd have x over here and x over here.
So if we write it an equilibrium expression for this. Acetate is acting as a base. So for our equilibrium expression, we would write Kb. And remember, that's concentration of products over reactants. So that would be x times x, so x times x, all over.04 minus x, leaving water out.
So this is over.04 minus x. And we did all this in weak base equilibrium. so make sure that you've watched that video before you watch this one, because I have to go a little bit faster here. Alright, so next we need to find the Kb value. We can get the Kb by, because we know the Ka.
In the last video, the Ka for acetic acid was 1.8 times 10 to the negative five. And we know Ka times Kb for a conjugate acid-base pair is equal to 1.0 times 10 to the negative 14. Again, this is from an earlier video. So if we plug If we plug in Ka into here, we can solve for Kb.
And I won't do it to save time on the calculator, I'll just give you the answer, that Kb is equal to 5.6 times 10 to the negative 10. So we plug this in to our equilibrium expression, so we get 5.6 times 10 to the negative 10 is equal to x squared over, here we make the assumption, like we did in all the earlier videos, that x is a very small number, very small. concentration. So 0.04 minus x is approximately the same thing as 0.04. So we write 0.04 in here.
And next we need to solve for x. So let's take out the calculator. So we have 5.6 times 10 to the negative 10. We need to multiply that by 0.04. And then we need to take the square root of our...
answer, and so we get x is equal to 4.7 times 10 to the negative 6. So x is equal to 4.7 times 10 to the negative 6. What does x represent? Right, we go up here and we notice that x represents the concentration of hydroxide ions. So this, this is equal to, this is equal to the concentration of hydroxide. So 4.7 times 10 to the negative six molar.
Alright, our goal was to find the pH. So at this point it makes sense to find the pOH. pOH is equal to the negative log of the concentration of hydroxide ions. So we plug that into here and we solve for the pOH. Negative log of 4.7 times 10 to the negative six gives us a pOH of...
of 5.33, so the pOH is equal to 5.33. And one more step, pH plus pOH is equal to 14. So if we plug in our pOH into here, pH is equal to 14 minus 5.33, which is 8.67. So we're at the equivalence point, but this is a titration of a a weak acid with a strong base.
And so we have a basic salt solution at the equivalence point, right? So our pH is in the basic range, right? It's above seven. So let's find that point on our titration curve.
So we've added here 200 mLs of our base, and the pH was 8.67, so we've added 200 mLs of our base, and the equivalence point should be somewhere in... there, right about there, about 8.67. That's our equivalence point for a titration of a weak acid with a strong base for this particular example. Finally, we're on to part D, which asks us, what is the pH after the addition of 300 milliliters of a.05 molar solution of sodium hydroxide?
So once again, we need to find the moles of hydroxide ions that we are adding. the concentration would be equal to.05. So.05 molar is our concentration of hydroxide ions, and that's equal to moles over liters.
So 300 milliliters would be.3 liters. liters, so we have 0.3 liters here. Multiply 0.05 by 0.3 and you get moles. So 0.05 times 0.3 is equal to 0.015 moles of hydroxide ions. In part C, we saw that we needed 0.01 moles of hydroxide ions to completely neutralize the acid that we originally had present.
So we're going to we're going to use up.01 moles of hydroxide. That's how much was necessary to neutralize our acid. So how many moles of hydroxide are left over after the neutralization? Well that would just be.005. So.005 moles of hydroxide ions are left over after all of the acid has been neutralized.
So our goal is to find the pH. So we could find the pOH if we found the concentration of hydroxide, so what is the concentration of hydroxide ions now after the neutralization has occurred? So concentration is moles over liters, so it's.005 divided by, what's our total volume? We started with 50, and we have now added 300 milliliters more, so 300 plus 50 is 350 milliliters, or.305. 35.35 liters. So what is our concentration of hydroxide ions?
So this is.005 divided by.35. So our concentration of hydroxide ions is.014. So let's write.014 molar here.
Once we know that, we can calculate the pOH. So the pOH is the negative log of the concentration of. hydroxide ion, so it's a negative log of.014.
So we can do that on our calculator, negative log of.014, and we get a pOH of 1.85. Alright, so our pOH, let's get a little bit more room here, pOH is equal to 1.85. And finally, to get the pH, we know that pH plus pOH is once again equal to four.
14, so we plug in the pOH into here, and the pH would be equal to 14 minus 1.85, so 14 minus 1.85 gives us 12.15. So we're now past the equivalence point. Alright, so we're past the equivalence point here, so our pH is 12.15 after we've added 300 milliliters of our base. So let's find the point. on our titration curve.
We've added 300 mLs of our base, so we're right here. So we go up, we go up to our titration curve, and that would be right here. So we just found the pH is a little bit over 12. So approximately, we got 12.15.
So that's the pH right here on our titration curve.