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2015 AP Chemistry free response 3a

Net-ionic equation for a weak base titrated with HCl. From 2015 AP Chemistry free response 3a. 

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Video transcript

- [Voiceover] Potassium sorbate, and they give us its formula right over here, has a molar mass of 150 grams per mole. They put this decimal here to show us that these are actually three significant figures. Even the zero is a significant digit here. Is commonly added to diet soft drinks as a preservative. A stock solution of potassium sorbate, dissolvent's an aqueous solution here, of known concentration must be prepared. A student titrates 45 millileters of the stock solution with one point two five molar hydrochloric acid using both an indicator and a pH meter. The value of K A for sorbic acid is one point seven times ten to the negative fifth. All right, so let's tackle this piece by piece. Write the net ionic equation for the reaction between potassium sorbate and hydrochloric acid. All right, so first off, let's write the ionic equation, and then we'll, I'll write the net ionic equation, and hopefully you'll see the difference. So ionic, ionic equation. And the way we think about ionic equations is we think about well, if these are dissolved in water, it's an aqueous solution, these are going to disassociate into their, into ions. And so we would write that out on both of the, on both the reactant and the product side. So the potassium sorbate, we can write that as, it's gonna be a potassium ion dissolved in an aqueous solution, plus the C six H seven O two, this is also going to be an ion dissolved in the aqueous solution, plus the hydrochloric acid will dissolve, so you have the hydrogen proton dissolved in the aqueous solution plus the chloride ion, or anion I guess we could say it. so that's going to be in our aqueous solution, and then they react. What happens? Well, you're going to have the C six H seven O two react with the hydrogen proton to get to sorbic acid. So you're gonna have sorbic acid, H C six H seven O two, that's the sorbic acid. It's going to be in an aqueous solution. So I took care so far of that and that, and then you're going to have, and then you're going to have your potassium ions, your potassium ions and your chloride ions. It's going to be just like that. So this right over here is the ionic equation, not the net ionic equation. I have the ions on the reaction, on the reactant side, and then on the product side right over here. And did I, yep, I included everything. Now you do the net ionic reaction, you can imagine what's going to happen here. I have potassium ions on the left. I have potassium ions on the right. I could net them out on both sides. So let's net them out on both sides. I have chloride ions on the left. I have chloride ions on the right. I could net them out on both sides. And then I can write the net ionic equation. So the net ionic equation, net ionic equation, is going to be well, I have my C six H seven O two ion dissolved in an aqueous solution. You combine that with the hydrogen proton dissolved in the aqueous solution, and it's going to, it's going to give us sorbic acid: H C six H seven O two in our aqueous solution. So there you have it. That is the net ionic equation.