## vera_ewing one year ago .

1. Photon336

it's definitely not D. let me see. BTW do you have some kind of time limit for this?

2. Photon336

@Rushwr what do you think?

3. Rushwr

4. Rushwr

what if we consider hydrolysis !

5. Photon336

I think i have it $1.76x10^-5 = \frac{ [x][x] }{ [0.28-x] }$

6. Photon336

Vera i got 2.65 ill show you how i got it

7. Photon336

$\sqrt{ }(1.76x10^-5)*0.28 = 0.00221M = [H^+]$

8. Photon336

pH = -Log[H+] = -Log[0.00221] = 2.65

9. Rushwr

@Photon336 I agree I got the same answer

10. Photon336

you have to set up an ice table.. you know that CH3COOH --> CH3COO- + H+ $\frac{ [CHCOO-][H^+] }{ [CH3COOH]}$ $\frac{ [+x][+x] }{ [CH3COOH-x }$ concentration of CH3COOH will go down by x and the concentration of your acid and the conjugate base CH3COO- will both be x. you can assume that because this is a weak acid we know we will have more of CH3COO- so we can ignore the -x here for CH3COO-

11. Photon336

re-arranges to give $\frac{ x ^{2} }{ 0.28 } = 1.76x10^-5$