We are now at part F where we have 20 millilitres of the sodium hydroxide being added to the butanoic acid. We are at this point at the equivalence point. Now how do I know that? If I look up here at the amounts that are given in the problem, I see that there is 20 millilitres of the butanoic acid and the concentration is 0.249. Since the concentration of the sodium hydroxide is 0.249, then it would require the same volume. Other sodium hydroxide to reach the equivalence point. Let's go ahead and write the reaction. If you did not recognized that it was at the equivalence point you would eventually recognize it as you works through the problem because of what the definition of the equivalence point is. So let's see how that works. We have our asset, reaction was our base to give us the conjugate base and water as we've been having for each of the other parts we have done for B and C. We're going to do an ICF table. We're going to plug moles into the table. Now the number of moles of the hydroxide would be the molarity of the hydroxide which is zero. 0.249 moles per liter times the volume which is 0.0200 liters, and that would give me 4.98 times 10 to the minus 3 moles. Okay, that's an equal sign there. That is the most of the hydroxide which is what we've been using for the in every point along the way but since it's the same molarity and the same volume, certainly it would be the same mole. So we'd have 4.98 times 10 to -3 here. 4.98 times 10 to the -3 under the hydroxide. Non of the conjugate base A- which is a butane net weight ion. And then we've got the water that we don't need to have numbers fr. We'll subtract the smallest. Well they're both the same number and we have minus 0.498 times 10 to the -3 in both of those occasions and we're going to have plus 4.98 x 10 to the -3 in this location. When we put the 0's in, we'll have 4.98 x 10 to -3 as the most of the A- left over. Now, how do we know we're at the equivalence point? Well, at the equivalence point, by definition, the moles of the acid and the moles of the base will be equal according to the coefficients of the balanced equation. And since it's one to one up there we should have the same quantity of the acid and the base being added. Also, whenever you're at the equivalence point, consumed your reactives and have just had converted over to your products. Okay, so now what do we have? We have sitting right here, a weak base. And as a weak base it will undergo equilibrium with water as any weak base would do. So, let's write that reaction. A-plus water. Since it's a weak acid with water it is equilibrium so we do reversible arrows. We exchange the proton, so the A- will accept the proton, become HA. The water is donated the proton, so it becomes And as an equilibrium, we do an ICE table instead of an ICF table. In our ICE table, we're going to put molarity. Instead of mols. Mols is what is sitting right here. So, if I want to know the concentration of A-, I'm going to need the mols of A-. Divided by the volume of the solution. Now we made a 20 millilitres of both solutions together, so that's 40 millilitres or 0.4000 liters and this is going to give me a value for this concentration of 0.12 Four, five. Now we really need only three significant figures, but I'm going to carry that extra one along as I work this problem and round to three at the very end. So I will put the 0.12 45 here. Again I don't need numbers for water. I put 0 in these spots I'll consume some and produce here we're not working with real numbers because it's in equilibrium. We have to use this value of x in order to configure out the concentration. A goals is find a pH is the solution so if we could determine a hydroxide concentration or X we will be set. And now we're going to keep in mind that is A- is a base. So we're going to need the KB For the base form. Well, we were given the Ka, so if we take Kw which is 1.0 x 10 to the -14 divided by 1.54 x 10 to the -5, we will have the Kb of that solution which equals 6.49 x 10 to the -10 And that is going to be equal to products over reactions raised to the power of coefficients to x times x or x squared divided by 0.1245 minus x. Now if we look at the value of our KB it is very small. And if it's very small then we can assume that x is going to be very small compared to 0.1245 and he can ignore that term. So this is going to be equal to x squared divided by 0.124 45. Now if we multiply both sides by the 0.1245 and then take the square root of that value we will obtain x. And x is 8.99 X 10 To the -6. And certainly that was a good assumption that x is much, much smaller than .1245 because when I take .1245 and subtract this number from it, I still have .1245. That's my x value. And I know that that x value according to my ICE table was the hydroxide concentration. So now I know the hydroxide concentration. From there I can get the pOH. The pOH is the negative log of that number. Eight point nine nine times ten to the minus six. And the value of that is equal to five point four six. The ph would then be 14 minus that number, 14 minus 5.046 and that will give me a ph of 8.954 so, that's a PH at the equivalence point. Let's think if that number makes sense. This is a PH that is greater than seven. Would you expect it to be greater than seven? Any time you have a weak acid, any strong base, you will have produced a basic solution, and you would expect a PH above seven. Look at our table. Just one last time here. We see here what's being represented in our calculation at 20 millilitres, we are up here with a pH at the equivalence point, whenever there's a strong base reacting with a weak equivalence point. It will be somewhere greater than seven. Back to the numbers just one last time. And that's the conclusion of this problem.
