10. Acid-Base Titrations
Strong Acid-Strong Base Titrations
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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 gonna write down our compound, so HI+ NaOH. We know that HI will protonate the Oh minus to give us H2O, so it gives an H+ to the Oh minus. Then we're going to have NaI positive remaining and I minus 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 a 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 giving any information on it. On the reactant side, the smaller moles, which is our limiting amount, will deduct or subtract 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 gonna be an excess of strong base remaining. So this is point 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 0.0 so 0.050 liters plus 0.020 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 p o h, so negative log of o h minus, so plug that concentration in. So that equals 1.85. Since we know the p o h, we can find the pH. PH would equal 14 minuteus that p o h 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 of this type of titration.
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