So here, let's take a look at this question. It says consider the nutrition of 75 mls of 750.300 moller of H three C 303. It's given at having a K. A. Of 4.1 times 10 to the minus three. Um and it's being tight traded with 30 ml of 300.450 Mohler K. O. H. Here we're told to calculate the ph since it's K. Is less than one, we know that this represents a weak acid and we know that potassium hydroxide is a strong base. We're gonna write them here as our reactant. Remember the acid is going to be a proton donor? It's gonna donate a proton two or H plus two O. H minus two. Give us water. And what we have left over here is our conjugate bases. KH 2 C30203. Since we have an acid and a base tight trading one another. We utilize an I. C. F. Chart in this I C. F. Truck. We're gonna ignore water which is a liquid. Remember in an I. C. F. Chart, we use moles as our units moles equals leaders, times more clarity. So divide these mls by a 1000 Multiplied by their polarities gives us the moles of our weak acid which is .00225 moles. Here, K.O. H. is .00135 moles. And we're not giving any initial information on our conjugate base. So initially it's zero. Now remember looking at the react inside the smaller moles which is our limiting amount will subtract from the larger moles. So subtract 0.135. Subtract 0.135. So when we do that We're gonna have zero left of this for our weak acid. We're gonna have .00090 moles. And remember conservation of mass. This side would increase .00135. Now look and see at the end of this hydration. We have left weak acid, we have left its conjugate base. Remember we casted conjugate base means that we have a buffer and with a buffer that means we have to utilize the Henderson Hasselbach equation. So we're gonna stay here that ph equals P. K. A. Which is the negative log of K. A plus log of the conjugate base amount. So there goes my conjugate base amount oops divided by our weak acid amount. Okay, so that equals 2.56. Now remember because we still have a buffer in terms of this hydration, that means we're dealing with calculations before the equivalence point. And because we still have a buffer we can utilize the Henderson Hasselbach equation to find our ph now that we've seen this one, click onto the next video and take a look at the practice question below
