PH of solution (not enough infromation?)

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This discussion focuses on calculating the pH of solutions involving ethylammonium chloride (C2H5NH3Cl) and aluminum nitrate (Al(NO3)3). For the 0.25M ethylammonium chloride solution, the Kb value of ethylamine (C2H5NH2) is 5.6 x 10^-4, leading to a calculated concentration of [C2H5NH3+] at 0.25M and [H+] at 2.1 x 10^-6M. For the 0.050M Al(NO3)3 solution, the Ka value for Al(H2O)6^3+ is 1.4 x 10^-5, which is essential for determining the pH through equilibrium constant equations.

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Both are confusing me. I don't know what information to use (K values)
I'd really appreciate some help!


Calculate the concentration of all species present in a 0.25M solution of ethylammonium chloride (C2H5NH3CL)
(Not sure you if you may need this, but C2H5NH2 has a Kb value of 5.6*10^-4... it is not given in the question, it's in the appendixes of my book, which are used a lot in these problems)



Calculate the pH of a 0.050M Al(NO3)3 solution. Ka value for Al(H2O)6 3+ is 1.4*10^-5.



EDIT: hmmm after reading analyzing it a little bit, I think I'm supposed to find the Kb of C2H5NH2 for the first one.. in other words
1ee-14/5.6ee-4
right?
And that would lead to the answer being
[C2H5NH3 +] = 0.25M (% dissociation is insignificant)
[Cl -] = 0.25M
[H +] = [C2H5NH2] = 2.1ee-6M
Does this look right?
 
Last edited:
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Hi, both of these questions involve setting up an equilibrium constant equation. So you can start by showing us this setup for both problems.
 

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