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. (d) 100.0 mL

Video transcript

Hello everyone today. We have the following problem 75 ml of 1.50 moller of arsenic acid is tight traded with 1.50 moller of sodium hydroxide. What is the Ph for the Thai tradition? After 150 ml of base is added. And then we have the following K values for arsenic acid. So the first thing that we have to do is we have to find the number of moles of our acid. In this case we have to find a number of moles for our arsenic acid. And we do that by starting with the volume. So we have 75 ml. We're gonna convert that into liters by using the conversion factor that one leader is equal to 1000 militia leaders. And then we're then going to multiply by our polarity of arsenic acid. Which we were given as 1.50 moles per liter parities and units of moles per one liter. Which is why we were able to convert it in this fashion when our units cancel, We are left with 0.1125 moles of our arsenic acid. So next we need to find the moles of our base which in this case is sodium hydroxide. As before we're going to start with our volume. We have 150 ml of bass that was added. We're gonna convert this into leaders by using the conversion factor that one leader is equal to militia leaders. And we're going to multiply the polarity of our base, which is the same as that of our acid, which is 1.50 moles per one liter As before. Our units will cancel. And we will be left with 0.225 moles of sodium hydroxide. So notice that we have this ratio of our acid to our base which essentially is one more Of our acid per two moles of our base. And this is going to be the second equivalent point. So now that we have this information we can go ahead and attempt to solve for our ph and we can use the formula that the ph of a try protic acid at the second equivalent point is as follows, the ph is equal to the P. K. Of our second acid plus the P. K. A. Of our third acid divided by two. So how do we get these P. K. Values? Well what we're gonna do is we're going to solve before the P. K. A. Of two first in the form of for PK is simple, it is negative log of the K. A. So we're gonna do negative log of the second K. And so that's gonna look like negative log of 1.7 times 10 to the negative seventh. And we're gonna get this P. K. Eagling six 0.77. So you have 6.77 here we're gonna sell for the P. K. For our third titillation. Which is going to be the same formula except we're gonna be using K. A. The third one. So we're gonna say negative log of 5.1 times 10 to the negative 12th. That's going to give us 11.3. So once we combine these We're going to get a final ph of about 9.0. And with that we have solved the problem overall. I hope this helped and until next time.