What is the pH at the equivalence point for the titration of 0.26M CH3NH2(aq) with 0.26M HClO4(aq)? For CH3NH2, Kb = 3.6 x 10^-4.
I arrived to the answer of pH=5.57 but its wrong.
Can you tell me why it is wrong? Thanks!
It's tough to tell you what is wrong when you don't show your work. The most probable error is that you didn't account for the change in the concn of the salt. That will not be 0.26 M but 0.13 M because it is diluted with an equal amount of titrant. My calculation gives 5.72
How did you get 0.13M?
That will not be 0.26 M but 0.13 M because it is diluted with an equal amount of titrant
There is 0.26M of both CH3NH2 and HClO4. So that means it will create 0.26M of CH3NH3+ and ClO4- right?
Then I put it in the ICE table: CH3NH3+ + H20 - > CH3NH2 + H30+
CH3NH3+ at Equlibrium would be 0.26M-x while the both products will be x
I put it in the Keq equation with Ka (I used Kw/Kb to find Ka) and thus i got 5.57. I don't understand how you got 5.72. Explain please Thank youu
You obtained 5.57 because you used 0.26 M. I obtained 5.72 because I used 0.26/2 = 0.13 M. Let's just take some convenient value for the volume of CH3NH2. Let's take 100 mL of 0.26 M. At the equivalence point we will have added 0.1 x 0.26 M = 0.026 moles CH3NH2. If we titrate with 0.26 M HClO4, we will add 100 mL of HClO4 and that gives us 0.1 x 0.26 = 0.026 moles HClO4. At the equivalence point, then we have 0.026 moles of the salt in how much solution? That is M(salt) = moles/L = 0.026 moles/0.200 L = 0.13 M.
Without you showing your work I guessed at what you had done; I have seen that mistake a thousand times and it't a common mistake students make.