Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: yankeekd25 on April 01, 2009, 08:36:00 AM
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An aqueous solution contains 0.353 M ethylamine (C2H5NH2).
How many mL of 0.328 M hydrobromic acid would have to be added to 225 mL of this solution in order to prepare a buffer with a pH of 10.600.
Kb= 4.3 x 10^-4
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Show your attempt!
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An aqueous solution contains 0.353 M ethylamine (C2H5NH2).
How many mL of 0.328 M hydrobromic acid would have to be added to 225 mL of this solution in order to prepare a buffer with a pH of 10.600.
Kb= 4.3 x 10^-4
Ethylamine + H+ --> EthylamineH+ + H2O (** don't know how the reaction exactly happens)
0.079425 moles-0.0738 moles .0738 moles created
0.005625/ .0738
But now I have no idea what to do with the pH. I was thinking about trying to backwards solve the H-H equation. Would that work?
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You can actually use the Henderson-Hasselbalch Equation (http://en.wikipedia.org/wiki/Henderson-Hasselbalch_equation) to solve this very quickly.
pH = pKa + lg{(Conj. Base/Acid)}
Remember, though, that what your question gave was Ka and not Kb
And if it helps,
CH3CH2NH2 + H2O <-> CH3CH2NH3+ + OH-
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An aqueous solution contains 0.353 M ethylamine (C2H5NH2).
How many mL of 0.328 M hydrobromic acid would have to be added to 225 mL of this solution in order to prepare a buffer with a pH of 10.600.
Kb= 4.3 x 10^-4
10.6= 10.63 + log
Looking good so far?
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Don't forget that some of the ethylamine is converted into its conjugate acid form, i.e. it should be lg{x / (0.353 - x)} and not lg{x/0.353}, otherwise, looks good.
PS. I didn't check whether pKb you gave was correct.
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Okay, so once you solve for x, what is the next step please?
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Stoichiometry. Write reaction equation.
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I cannot figure this problem out, or how to find mL of the base need. It's a 1:1 ratio I believe, so how do I go from here?
10.6= 10.63 + log x /[.353-x]
-0.03 = log x / (.353-x)
.933 = x / .353 - x
.933 (.353-x) = x
.329 - .933x = x
.329 = 1.933x
x= .170
Do I need to do anything with the original volume of the acid?
Perhaps, C1V1= C2V2??
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Write reaction equation.
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Write reaction equation.
CH3CH2NH3+ + OH- <-> CH3CH2NH2 + H2O
Now what?
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This is not the reaction taking place.
You should look at the (correct) reaction equation to find out how the stoichiometry is related to the amount of acid and conjugate base.
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Write reaction equation.
CH3CH2NH3+ + OH- <-> CH3CH2NH2 + H2O
Now what?
CH3CH2NH3+ + H+ <-> CH3CH2NH4+ + H2O
I believe that is correct now.
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No, it is still wrong. Charge is not balanced. Seems to me like you have no idea how amines are protonated.
Do you know what happens to ammonia in water? When you add acid? Same with amines.
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NH3 + H2O --> NH4+ + OH-
NH3 + H+ --> NH4 + H2O
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NH3 + H+ --> NH4 + H2O
Perhaps its nitpicking, but charges are not balanced.
This way I will never know if what you write is wrong because you are making typos, or because you don't understand what you are doing. Hard to help in such situation.
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NH3 + H2O --> NH4+ + OH-
NH3 + H+ --> NH4 + H2O
Just made a typo on the last rxn. It's NH4+ + H2O right?
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Yes, and that's almost the same as the necessary reaction of amine.
How many moles of conjugated acid do you need in the final solution?
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Yes, and that's almost the same as the necessary reaction of amine.
How many moles of conjugated acid do you need in the final solution?
Would it be .328 M x .225 L= 0.0738 ?