Very hard chemistry problem. >:(

[DivideByZero]

Ok. First of all this is a homework problem so please provide your steps on how you solved this.

 

Calculate the [math]K_{a}[/math] for the following acids using the given information.

 

(a) 0.220 M solution of H3AsO4, pH = 1.50

 

 

 

so i know the concentration of H+ ions would be 10^-1.5 = 0.0316 M

 

I think the dilution of the H3AsO4 is:

H3AsO4 --> 3H + AsO4

 

now what?

[big314mp]

You are going to be calculating K(a1), or the K of the first proton.

 

So your reaction is:

 

H3AsO4 -> H+ + H2AsO4-

 

Now if you have a reaction of:

 

A + B -> C + D

 

Then K = ([A]*)/([C]*[D])

 

Where [A] = concentration of A

[Cap'n Refsmmat]

From what I recall of chemistry, you can use the concentration of H+ you found and the equation you found ([ce]H3AsO4 -> 3H+ + AsO4-[/ce]) to determine the final concentration of [ce]H3AsO4[/ce] and [ce]AsO4-[/ce]. From those you should be able to find [math]K_a[/math].

 

Oh, yeah, and what big314mp said about the first proton... I had forgotten about that. Sorry.

 

big314mp: Shouldn't that be

 

[math]K_a = \frac{[A^a][b^b]}{[C^c][D^d]}[/math]

 

where a, b, c and d are the coefficients of A, B, C, and D, respectively?

[DivideByZero]

woah woah woah!!

 

I though it was

 

[math]

K_a = \frac{[C^c][D^d]}{[A^a][b^b]}

[/math]

 

if so, then it should be

 

[math]

K_a = \frac{[H+]^2}{[H_3 AsO_4]}

[/math]

[Cap'n Refsmmat]

Yikes, you're right about that. However, are you sure it's [math][H+]^2[/math]? As big314mp points out, you only want the first proton.

[DivideByZero]

I though it should be [H+]^2

because the concentration of the H+ and H3AsO4 should be equal in a weak acid. But even after I compute the answer, I keep getting it wrong.

[Cap'n Refsmmat]

because the concentration of the H+ and H3AsO4 should be equal in a weak acid.

 

Why do you say that? It depends on the [math]K_a[/math] of the acid.

[big314mp]

Check you denominator. Are you doing:

 

[H3AsO4] - [H+]

 

Because, once the H3AsO4 deprotonates, it no longer counts as a reactant. So however much deprotonates needs to be subtracted from the initial concentration of H3AsO4.

[hermanntrude]

i feel an ICE table coming on

[DivideByZero]

Check you denominator. Are you doing:

 

[H3AsO4] - [H+]

 

Because, once the H3AsO4 deprotonates, it no longer counts as a reactant. So however much deprotonates needs to be subtracted from the initial concentration of H3AsO4.

 

YESS!!!!! I LOVE YOU MAN!!!!!!!!!!!!!!!!!!!!!! HEARTS ALL AROUND!!!!

 

 

thank you everyone!!!!

[hermanntrude]

I'd strongly reccomend that this type of question be answered using an ICE table. It IS possible without but it's much clearer in a table format