So remember it's going to require on our part, 100 mls of potassium hydroxide in order to reach the equivalence point in this question here, we have only 50 mls of K O. H. So that means we're dealing with calculations before the equivalence point. Now, remember once our acid and base begin to mix, we transition away from the ice chart and we move to the I. C. F. Chart to determine the ph and remember in an I. C. F chart, the units have to be in moles, moles itself equals leaders, times more clarity. We divide these mls here by 1000 to change them into leaders and then multiply them by their polarities. That would give me the moles of each of these species. So here are moles of nitrous acid would give me 0.30 moles. And then here this would give me the moles of K O H 00.15 moles. We don't have any of our nitrate ion nitrite ion. So we started initially zero, like in ice chart. I cf charts ignore water because it's a liquid. It's not really affecting our ph value. So we can ignore it. Now we look at the react inside, we know on the react inside the smaller moles, which is our limiting amount will deduct from the larger moles. So our smaller moles are these .015 moles. They subtract subtract from themselves because they're being used to neutralize the acid. We subtract them from the acid as well. At the end, we'll have left zero of my strong base and I'll have some of this weak acid remaining based on the law of conservation of mass or matter. Um We know that matter can't be created or destroyed. It just changes forms. So whatever I'm losing on the reacting side, I'm actually gaining as this conjugate base on the product side. So at the end of this what we have is we have weak acid and we have conjugate based remaining. Remember weak acid conjugate base means we have a buffer. And if we have a buffer then we utilize the Henderson Hasselbach equation in order to calculate our ph so if we look here remember our Henderson household back equation as P. H equals P. K. A. Plus log of the conjugate base amount divided by the weak acid amount. We said earlier that the so negative log on P. K. Equals negative log of K. A. And we said earlier that the K. Of nitrous acid is 7.1 times 10 to the negative four according to your book. And then it's log of the conjugate base amount divided by their weak acid amount. When you plug that in. That gives you 3.15 as your ph. So you're seeing as we're slowly adding strong base we see that our ph is increasing. And remember before we reached the equivalent volume of the strong based hatred. That means we will still have a buffer and we can utilize the Henderson Hasselbach equation to calculate our ph this particular example, we see that the end we have equal amounts of my weak acid and conjugate base. This happens we're more at the half equivalence point. And it's important to remember that at the half equivalence point p H equals P K. That's because log of 0.15 divided by 0.15 is really long. Of. One, log of one Is equal to zero. So at the half equivalence point that the half of the volume we needed to get to the equivalence point P H equals P K. A. Now that we've seen these types of calculations, let's move on to what happens when we're at the equivalence point between a weak acid and a strong base.