calculate [ClCH2COOH] for chloroacetic acid. Given:0.100 M soln of chloroacetic acid is 11.0% ionized. I have already found the [H+] which is 1.10*10^-2 and this is correct. I keep coming up with 1.0*10^-12... any ideas on how to do this problem?

Question

calculate [ClCH2COOH] for chloroacetic acid. Given:0.100 M soln of chloroacetic acid is 11.0% ionized.  I have already found the [H+] which is 1.10*10^-2 and this is correct.  I keep coming up with 1.0*10^-12... any ideas on how to do this problem?

aaronq · Answer

do you know the Ka?

aaronq · Answer

regardless you know 11% of 0.1M is ionized so 0.011M the conc of dissociated acid.. the rest is the undissociated acid

anonymous · Answer

ka is dissociated acid?

aaronq · Answer

no no, i was gonna say if you know the Ka you can find it by setting up an equation 

Ka=[A-][H3O+]/[HA]   where [A-]=x and [HA]=0.1-x

anonymous · Answer

got ya.....       i found [H+] to be 1.10*10^-2M which is correct... maybe that helps?

aaronq · Answer

yeah, 1.1x10^-2 =0.011 which is 11% of 0.1M you merely had to find 11% then subtract by the original amount to find the concentration of undissociated acid left

anonymous · Answer

so... 0.100-0.011?

aaronq · Answer

yep

anonymous · Answer

finally got it right! thanks so much

aaronq · Answer

no prob !