So here the example says calculate the ph of the solution resulting from the thai tradition of 50 mls of 500.10 molar hydrochloric acid with 20 ml of 200.30 moller sodium hydroxide. So we have our strong acid here. That is being tight traded by a strong base. We're gonna write down our compound. So H. I. Plus an A. O. H. We know that H. I will protein it the O. H. Minus to give us a church to O. So it gives an H. Plus to the O. H. Minus. Then we're gonna have N ai positive remaining and I minus remaining. And they combined together to give us our conjugate base because we have an acid and a base thai trading one another. We set up our I. C. F. Chart. So initial change. Final water is a liquid liquid and solid are not found in either ice charts or I. C. F. Charts. Remember in an I. C. F. Chart, we need the units to be in moles. So divide the mls by 1000 to get leaders, then multiply them by their polarities. That will give us the moles of H. I and N A. O. H. When we do that, we're gonna get 00.50 moles And .0060 moles. Initially our conjugate base is zero. Since we're not giving any information on it on the react inside the smaller molds, which is our limiting amount will deduct or subtract from the larger moles. So subtract 0.50 point 0050 at the end. We're gonna have zero left of our strong acid. That means that we've gone beyond the equivalence point where there's going to be an excess of strong based remaining. So this is .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 molds left of it and divided by the total volume, Which would be zero. So point 050 liters plus 500.20 liters. Alright, so then when we plug all that into our calculator, we'll get a polarity of 0.14 Moeller of N A. O. H. 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 minus that P. O. H total. So that comes out to be 12.15. So in this case we had excess strong based titrate left. So we're dealing with after the equivalence point. So find the concentration of that strong base and use it to find P. O. H. And then from there ph now that we've seen this example, question click onto the next video and let's continue our discussion of this type of filtration.
