A 100.0-mL buffer solution is 0.100 M in NH3 and 0.125 M in NH4Br. What mass of HCl can this buffer neutralize before the pH falls below 9.00?

Video transcript

Hello. In this problem we are told that 0.115 moller hydroxy amine and 0. molar hydroxide unemployed are mixed to produce 100 50 mL of buffer solution is to determine the mass of hydrogen bromide. The buffer can neutralize before the ph of the solution falls below 5.50. Were given the base association constant for hydroxy amine in our first step, then let's find the concentration of our base that of our acid. Using the Henderson Hasselbach equation. So the Henderson Hasselbach equation says that the ph is equal to the P K. A. Plus the log of the base concentration, which for us is the hydroxyl amine to that of the acid, which is the hydroxyl monument. So we can find then the K A gall. It's equal to then I'm product constant water divided by the KB nine product constant water is one times 10 minus 14. And Our Base Association constant for domain is 1. times 10 - -8. This works out to 9.09 Times 10 -7. You can then find the P K. A. PK is the negative log of R K A value. This works out to 6.041. Now we'll find the concentration of our face to that of our acid. So we moved the p K. To the other side and take the inverse laudable sides. So we get 10 then to the ph minus the p K Ph is 5.50 RPK. 6.041. And this works out to 0.287. So the concentration of the drops will mean then Is equal 2.287 times of concentration of hydroxyl monument. So now in the next step, let's set up a table. Find the hydrogen ion concentration that results from adding hydrogen bromine. So beginning with our reaction. So we have our base then that will react with added acid will form the acid of our base. We have initial change and then final Initially. Then the amount of our base is 0.115 moller. We're trying to determine the amount of acid that is added And the amount of our conjugate acid is 0.135. We assume that all of the acid reacts so it's b minus y minus Y plus Y, combining the initial and the change we get the final. You can then plug these in for our concentrations in the equation above. So we have then the concentration of our base is equal to 0.115 -6. This is equal to 0.287 times the concentration of our conjugate acid simplifying things and we get 0.115 minus Y. Is equal to 0. Plus. Y. Everything with the Y variable to one side. So we end up with 0.7626 Is equal to 1.287. Why? Why then? Is equal to 0.7626 divided by 1.287. This then works out to 0.0 Mueller. And this is then the concentration of our high Drinan's. And the next step then let's find the mass of hydrogen bromide. So we have a 150 of our solution will convert middle leaders to leaders to make use then of the concentration of our hydro nines and everyone role of hydrogen ions. We had one mole hydrant bromide and then we'll make use of the smaller mass to get from most mass. One mole of hydrogen bromine And has a massive .912g. So our units of milliliters cancels, leaders cancels molds of hygiene in cancels molds of hydrogen bromide cancels. And we're left with grams. This works out then to 0. g hydrant bromide. This then corresponds to answer the Thanks for watching. Hope this out