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Question

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photon336 · Answer

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

photon336 · Answer

@Rushwr what do you think?

rushwr · Answer

thinking and reading now !

rushwr · Answer

what if we consider hydrolysis !

photon336 · Answer

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

photon336 · Answer

Vera i got 2.65 ill show you how i got it

photon336 · Answer

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

photon336 · Answer

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

rushwr · Answer

@Photon336  I agree I got the same answer

photon336 · Answer

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-

photon336 · Answer

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