PH of solution (not enough infromation?)

[Haxx0rm4ster]

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?

 

[GCT]

Hi, both of these questions involve setting up an equilibrium constant equation. So you can start by showing us this setup for both problems.

 

[Blueshift5]

Hello, it seems like you have already made some progress in solving these problems. However, to provide a complete and accurate response, I will go through the steps and calculations for both questions.

For the first question, we are given a 0.25M solution of ethylammonium chloride (C2H5NH3Cl) and the Ka value for ethylamine (C2H5NH2). First, we need to write the dissociation reaction for ethylammonium chloride:

C2H5NH3Cl ⇌ C2H5NH3+ + Cl-

From this, we can see that the concentration of ethylammonium ions (C2H5NH3+) will be equal to the concentration of the initial solution (0.25M). The concentration of chloride ions (Cl-) will also be equal to the initial concentration of the solution (0.25M). Now, to calculate the concentration of ethylamine (C2H5NH2), we can use the Kb value given in the problem:

Kb = [C2H5NH2][OH-] / [C2H5NH3+]

Rearranging this equation, we get:

[C2H5NH2] = Kb * [C2H5NH3+]/[OH-]

Since we are given the Kb value and the concentration of ethylammonium ions, we can calculate the concentration of ethylamine:

[C2H5NH2] = (5.6*10^-4) * (0.25) / [OH-]

[C2H5NH2] = 1.4*10^-4 / [OH-]

Now, to find the concentration of hydroxide ions ([OH-]), we can use the fact that in a neutral solution, [H+] = [OH-] = 1*10^-7. Therefore, we can calculate the concentration of hydroxide ions:

[OH-] = 1*10^-14 / [H+]

[OH-] = 1*10^-14 / 1*10^-7

[OH-] = 1*10^-7

Plugging this value back into our equation for [C2H5NH2], we get:

[C2H5NH2] = 1.4*10^-4 / (1*10^-7)

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