Consider the titration of 50.0 mL of 1.00 M H3PO4 with 1.00 M KOH. Calculate the pH after the addition of each of the following volumes of base. (a) 25.0 mL

Video transcript

Hello. In this problem, we are told that iteration of 75 mL of a 1.25 molar solution of citric acid. The 37.5 mL of a 1.25 molar sodium hydroxide solution was done grass to calculate the ph if the P. K. Values for the 1st, 2nd and 3rd association or 3.134 point 76 and 6.40. Respectfully, citric acid has formula C six H 807. It is a try product acid, which means it has three acidic protons. So we will just denote this then as H three a. So in our first step, then we're gonna write the equations for the dissociation of the three acidic protons. And for simplicity, we'll drop this study mine. So we have citric acid. Then we'll react with um hydroxide to form water and the base of our acid. This thing goes on to react with additional base, remove the second acidic proton and then third association we removed the last acidic proton, the addition of our base. So if we were to look at a situation plot, then this is a try product acid would have ph is a function of the hydroxide iron concentration. And we would have pre equivalence points. And so the ph that's first halfway point would be equal to the P. K. A one halfway to the second equivalent point would be P. K. A. Two and halfway to the third equivalent point would be P. K. Three. So in our next step is find our initial moles of our citric acid and sitting hydroxide. So we have 75 ml of citric acid volume from milliliters to liters. And then we'll make use of the concentration to go from Volume to moles at 1.25 moles citric acid per leader. Sorry milliliters of acid and our leaders of acid cancel. This works out to 0. rules of our acid. And we added 37.5 ml of our base. Sitting hydroxide will convert our volume from milliliters to liters And frustration. Certain hydroxide is the same as that of our acid. So 1.25 moles the leader so our volume base canceled in milliliters and leaders. This works out to 0.046 7. Mark five. Um In our next step then we will determine moles of our acid and it's going to be based after we add percent 0.5 million. Sitting outside threading the generic form of our waiting for the first association. We have initial the change and then final. So before we add any base we initially have 0.09375 moles of our acid. We have 0.046875 moles of our base and none of the conjugate base. And since water won't impact ph will just ignore it the next then we have the change. Then we're gonna assume all of the hydroxide reacts. We'll have no base left and we'll have 0.0 or six. Basin 5 are acid left over and will form an equal amount of our conjugate base. And the last step we're going to make use of the Henderson Hasselbach equation. Find ph to recall the Henderson Hasselbach equation as ph equal to the P. K. A. Blog of our thanks to our acid and this problem are acid in our base are equal. And so the ratio of Our face to our acid concentration would be equal to one and the log of one is zero. So we would have pH is equal to P. K. A. And this is P K one which we were told is 3.13. So this is the ph Of our solution. After the addition of 37.5 ml of sitting hydroxide. When we look at our filtration plot, then it is this point right here where we have added a sufficient amount based to react with half of the acid to remove half of the first acidic proton. So we're to the half point of the first equivalence point, that's where our ph is equal to P K. One, which is what we found in this problem. And this corresponds then to answer B. Thanks for watching hope this out