Equilibrium Constant

[Mask667]

Hello everyone,

I can't figure out how to do this problem, it is very easy, but I just don't know how.

What is the concentration of H+ in a 2.0 M aqueous solution of acetic acid, CH3COOH? (Ka of acetic acid = 1.8 X 10^-5 at 25 deg. C)

This is what I did and it is completely wrong:

1.8 X 10^-5 = [H+]/[2M]

The answer is 6.0 x 10^-3

Please show me how to do this, with the numbers plugged in and everything.

Thank you,

Mask667

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[Leh]

Hello everyone,

I can't figure out how to do this problem, it is very easy, but I just don't know how.

What is the concentration of H+ in a 2.0 M aqueous solution of acetic acid, CH3COOH? (Ka of acetic acid = 1.8 X 10^-5 at 25 deg. C)

This is what I did and it is completely wrong:

1.8 X 10^-5 = [H+]/[2M]

The answer is 6.0 x 10^-3

Please show me how to do this, with the numbers plugged in and everything.

Thank you,

Mask667

okay...in this problem we have Ka= ([ch3coo-]*[H+])/[ch3cooh]
given 1.8 X 10^-5 = [H+][ch3coo-]/[2M] -------->
3.6 X 10 ^-6= [H+][ch3coo-]
since this is a monoprotic acid (ch3coo- to H+ ratio is 1:1) x^2=3.6 X 10^-6
square root of both sides gives 6X10^-3 which is equal to x and our H+ concentration.
gyea

 

[Lily-VA]

okay...in this problem we have Ka= ([ch3coo-]*[H+])/[ch3cooh]
given 1.8 X 10^-5 = [H+][ch3coo-]/[2M] -------->
3.6 X 10 ^-6= [H+][ch3coo-]
since this is a monoprotic acid (ch3coo- to H+ ratio is 1:1) x^2=3.6 X 10^-6
square root of both sides gives 6X10^-3 which is equal to x and our H+ concentration.
gyea

Good job, Leh.
Everything looks good...only one calculation^^ I want to point out.

(1.8 x 10 ^ -5 ) x 2 = x^2
therefore, 3.6 x 10^-5 =x^2 changing it to..36x10^-6 = X^2
square root of both sides give X=6 x 10^-3

Good question and answer..I am copying this down to study!

 

[Mask667]

aaaahhh, I got it. So if CH3OO- had a concentration of let say 3 (it wouldn't but hypothetically) then you would simply divide by 3, right?

Thanx,

Mask667