Pyridinium bromide 1C5H5NHBr2 is a strong electrolyte that dissociates completely into C5H5NH+ and Br-. An aqueous solution of pyridinium bromide has a pH of 2.95. (b) Using Appendix D, calculate the Ka for pyridinium bromide.

Video transcript

Hi everyone. This problem reads try ethel ammonium chloride completely dissociates into its ions because it is a strong electrolyte. A solution of tri ethyl ammonium chloride has a ph of 2.79. What is the K. A. Of trifle ammonium chloride? If tri ethyl amine has a KB of 6.5 times 10 to the negative five. So our goal here is to solve for the value of K. A. And we're given K B. So we need to think of the equation that relates K A. And K B. And that is our ion product constant. KD W is equal to K A. Times K B. And in this case we want to solve for K A. So we need to isolate K A by dividing both sides by K B. Okay, so we get K A. Is equal to K W over. K B. K W is a constant that we should know and have memorized and that value is 1.0 times 10 to the negative 14. And KB was given in the problem, we're told that it is 6.5 times 10 to the negative five. So we have all of the information that we need to solve for K. A. So let's go ahead and do that. So once we divide this out we get our K A. Is equal to 1.538 times 10 to the negative 10 as our final answer. Okay, so that is it for this problem? I hope this was helpful