Consider the titration of a 25.0-mL sample of 0.175 M CH3NH2
wit...

Question

Consider the titration of a 25.0-mL sample of 0.175 M CH3NH2 
with 0.150 M HBr. Determine each quantity. 
a. the initial pH

Accepted Answer

Hey everyone, we're asked to consider the tight rations of a 50 millimeter of 0.25 moller aniline solution with 0.125 molar hydrochloric acid. What is the ph of the solution before adding any acid? So here we know that an allen is a week base. So let's go ahead and write out our reaction. We have our aniline and this is going to react with water since this is before adding any acid. When these two react, our water is going to donate a proton to our an allen. And we're going to get the pro rated version of our an allen and we have our hydroxide ion. Now let's go ahead and create our ice chart, Creating our ice chart. We were told that we had 0.250 moller of our aniline. Initially, since water is a liquid, this won't be included. And initially we had zero of our products formed. Our change is going to be a minus X on our reactant side and a plus X on our product side. Since we're losing reactant and gaining products at equilibrium in our react inside, we have 0.250 minus X and an X. And an X in our product side. Now when we look up the KB of our aniline, this comes up to 3.9 times 10 to the negative 10. And we also know that our KB is equal to our products over our reactant. In this case it will be our X times r X. All over 0.250 minus X. Now looking at the X. In our denominator, we can go ahead and check if this is negligible to do. So We're going to take 0. and divide this by our KB of 3.9 times 10 to the negative 10. Now, if we get a value greater than 500 then our X is negligible. And in this case we do so we can go ahead and disregard our X. And our denominator, Solving for X. We get 3.9 times 10 to the negative 10 and this is going to be equal to X squared over 0.250, Multiplying both sides by 0.250. We end up with X squared equals 9.75 times 10 to the negative 11. Taking the square root of both sides. We end up with an X. Of 9.8742 times 10 to the negative six. And this is also going to be the concentration of our hydroxide ion. Now to determine our ph we can go ahead and calculate for our P. O. H. First. So our p. O. H. Is going to be the negative log of the concentration of our hydroxide ion. In this case it's 9.8742 times 10 to the negative six. This gets us to a p. O. H. Of 5.0. Now as we've learned, we know that our ph plus our p. O. H is equal to 14, solving for r p. H. We get p h equals 14 minus 5.0. So our ph comes up to 9. and this is going to be our final answer. Now, I hope this made sense and let us know if you have any questions.

Consider the titration of a 25.0-mL sample of 0.175 M CH3NH2 with 0.150 M HBr. Determine each quantity. a. the initial pH

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