[tarik3001]

A solution of monochloroacetic acid, ClCH2COOH, is prepared by dissolving 5.80 g of this acid in deionized water and then diluting to a total volume of 500.00 mL in a volumetric flask.  At the molar concentration of this solution, the acid is 12.0% ionized to H+ (aq) and ClCH2COO- (aq).  What is the total concentration of ClCH2COOH in this solution?  What are [H+], [ClCH2COO-], and [ClCH2COOH] in this solution?  What is the pH of this solution?

I did...
5.80g ClCH2COOH * (1mol/94.5g ClCH2COOH) = .0614M
.06154M * .500L = .0307M ClCH2COOH in solution
.0307 * .12 = .0036 M of H+ and ClCH2COO-
ph= -log[H3O+] = -log[.0307] = 1.51

Thanks!

[opti384]

-log[H3O+] = -log[.0307] = 1.51

Is [H3O+] =  .0307?

[tarik3001]

I don't know..i haven't done a calculation like this since gen chem..

[opti384]

You got the [H^+] but when you calculated the pH you just plugged in the concentration of ClCH2COOH

[tarik3001]

how do I find the [H3O+]?

[opti384]

H+ and H3O+ can be seemed as basically same things. H3O+ is formed when H+ attracts the water molecule.

[tarik3001]

Ok, so I now get  -log[.003684] = 2.434?

Also does everything else look good?

Thanks

[Borek]

If 12% of ClCH2COOH dissociated, concentration of ClCH2COOH is not 0.0307M.

[tarik3001]

so that should also get multiplied by .12 and that would make all of the concentrations .0036 then?

[Borek]

No. 12% dissociated, how much was left?

[tarik3001]

I see so 88%...maybe I should be reading these questions a little bit more carefully...Since 12% ionized to H+ and ClCH2COO- does that mean that I should also be using 88% to calculate those? Because that would mean that all of the concentrations would be the same..