Two 25.0-mL samples, one 0.100 M HCl and the other 0.100 M HF, are titrated with 0.200 M KOH. c. Which titration curve has the lower initial pH?

Video transcript

Hello everyone today we are being asked to solve the following problem. It says a 0.150 molar solution of N A. O. H. Is used to titrate to separate asset solutions. The first one is a 25 mL of a 250.124 moller of hydrogen iodide and the second one is a 20 millimeter 0.184 molar of acetic acid. Were then asked to find which of the two to titrate in curves will have a higher ph at the start. So the very first thing we wanna do is we want to calculate the initial ph for each solution. So the first thing we'll do, we'll start with the 25 ml one. We know that H. I. Is a strong acid. So therefore the concentration of H. I. We equal to the concentration of H plus ions. We then must find the ph of this by doing so we can say the ph is equal to negative log of that H plus concentration, Plugging in our values for our H. Plus. We have 0.124 from the polarity. And when we put this in our calculator we will get that the ph is equal to 0.907. To find the ph of the second asset solution. We must set up an ice chart. So we must set up our initial compound. We must show it reacting with water and then we must show the resulting products so we'll have ch three C O minus from that deep rotation. And then we're just going to say we're going to add that. We're going to also produce H 30. Plus or H plus ions Where they're gonna write i. c. and E. on the side on the left hand side here and I stands for initial concentration. Our initial initial concentration of our acetic acid is 0.184. Water is a liquid and we're not involved involving liquids with ice charts. So we'll just put a giant X. Throughout all the columns there for our acetate. We're going to have an initial concentration of zero of course. And the same with our H 30. Plus for our acetic acid. For the change, that's what C. Stands for. We're going to say we're going to have a minus X. Change because we're reacting that. And so we're going to do the opposite side. We're going to have the opposite sign. So we're gonna be adding plus X. Or plus some concentration that we did note as X. E. Is going to be basically the rows of I plus C. Or the columns. So therefore this E. For the first one for acetic acid will be 10.184 minus X. For our acetate. On the right here we'll just have X. H 30 plus. We'll have X. As well. The K. A value for acetic acid is 4. five times 10 to the -5. And while it is not shown here, it should be provided for you when we do that. we're going to equal this K. A. We're going to say 4.75 times 10 to the negative five is equal to X squared over 0. minus X. Whenever we solve for R. K. A. We have to equal it too. Whatever our products are over our reactant. In this case our products are the acetate and the H. 30 plus ions. When we have to excess and when we have an X times X. That becomes X squared over our reactant which is that 300.184 minus X. We can then denote that since X is going to be such a small value, it is going to be considered negligible. So we're going to put we're just going to cross out that X. We're just going to be left with our K. Equal to X squared over 0.184. And so solving for X we eventually get X is equal to 2.96 times 10 to the negative third molar. And so the ph from that we're gonna do the same thing we did with our H. I. We're going to see P. H. Is equal to the negative log of our H plus ions. So plugging in our H plus ions will get negative log of 2.96 times 10 to the negative third. And then we plug this into our calculator. We'll get a ph of 2.53. We can see that acetic acid has a much higher ph or much higher initial ph, and therefore our acetic acid titrate in curve will have a higher ph at the start. I hope this helped, and until next time.