[yankeekd25]

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

[AWK]

Show your attempt!

[yankeekd25]

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?

[lancenti]

You can actually use the Henderson-Hasselbalch Equation to solve this very quickly.

pH = pK_a + lg{(Conj. Base/Acid)}

Remember, though, that what your question gave was K_a and not K_b

And if it helps,

CH₃CH₂NH₂ + H₂O <-> CH₃CH₂NH₃⁺ + OH^-

[yankeekd25]

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

• /[.353]

Looking good so far?

[lancenti]

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 pK_b you gave was correct.

[yankeekd25]

Okay, so once you solve for x, what is the next step please?

[Borek]

Stoichiometry. Write reaction equation.

[yankeekd25]

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??

[Borek]

Write reaction equation.

[yankeekd25]

Write reaction equation.

CH3CH2NH3+ + OH- <-> CH3CH2NH2 + H2O

Now what?

[Borek]

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.

[yankeekd25]

Write reaction equation.

CH3CH2NH3+ + OH- <-> CH3CH2NH2 + H2O

Now what?

CH3CH2NH3+ + H+ <-> CH3CH2NH4+ + H2O

I believe that is correct now.

[Borek]

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.

[yankeekd25]

NH3 + H2O --> NH4+ + OH-


NH3 + H+ --> NH4 + H2O