Gen chem que

[predentgal]

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Acetic acid has a Ka of 1.8 x 10-5. What is the molarity of an acetic acid solution if it is 1.3% ionized?
4.8 x 10-40.33 M0.0014 M0.11 M
Answer is .11M

How do you calculate this?

 

[sfoksn]

So HAc -> H+ + Ac-

Ka = ([H+][Ac-])/(x), putting x as the molarity.

1.8x10^-5 = (1.3x10^-2*x)^2/x just ICE table

(1.3x10^2*x)^2 = 1.69x10-4 x^2

(1.69x10^-4*x^2)/x = 1.69x10^-4*x, which must equal to 1.8x10^-5

divide 1.8x10^-5 by 1.69x10^-4 and answer is 0.11M

 

[engr2dent]

So HAc -> H+ + Ac-

Ka = ([H+][Ac-])/(x), putting x as the molarity.

1.8x10^-5 = (1.3x10^-2*x)^2/x just ICE table

(1.3x10^2*x)^2 = 1.69x10-4 x^2

(1.69x10^-4*x^2)/x = 1.69x10^-4*x, which must equal to 1.8x10^-5

divide 1.8x10^-5 by 1.69x10^-4 and answer is 0.11M

Why is [Ac-] equal to x and not equal to the same value as [H+] (1.3x10^-2) in your calculation? Shouldn't the concentration of [Ac-] and [H+] be the same when it ionizes? Thanks.

 

[sfoksn]

Why is [Ac-] equal to x and not equal to the same value as [H+] (1.3x10^-2) in your calculation? Shouldn't the concentration of [Ac-] and [H+] be the same when it ionizes? Thanks.

Yes, it is, that's why it's squared.

I just simplified (1.3x10^-2*x)(1.3x10^-2*x) as (1.3x10^-2*x)^2

 

[engr2dent]

Yes, it is, that's why it's squared.

I just simplified (1.3x10^-2*x)(1.3x10^-2*x) as (1.3x10^-2*x)^2

Apologies in advance for me being an idiot. I think I see what you're doing but can you confirm? What I've been doing is this (which is wrong)...

Ka = ([H+][Ac-])/(x)

1.8x10^-5 = (1.3x10^-2)^2/x

(1.3x10^2)^2 = 1.69x10-4^2

(1.69x10^-4)/x = 1.8x10^-5

divide 1.69x10^-4 by 1.8x10^-5 and get 9.39 which is obviously wrong.

The reason for your "x" in
(1.3x10^-2*x)(1.3x10^-2*x) as (1.3x10^-2*x)^2 is because this is the molarity times the percentage dissociated to find [H+] and [Ac-], correct? Thanks!

 

[predentgal]

Thankss guys... my mistake was that instead of doing (1.3 x 10^2x)(1.2 x 10^2x) I assumed that you were suppost to use 1.3 x 10( x)^2

 

[sfoksn]

Yes, that's correct

 

[luv8724]

I cant seem to find these kind of questions in KBB... should I be using the Destroyer after I have been thru KBB once?