Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: tarik3001 on October 19, 2010, 09:30:45 PM
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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!
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-log[H3O+] = -log[.0307] = 1.51
Is [H3O+] = .0307?
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I don't know..i haven't done a calculation like this since gen chem..
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You got the [H^+] but when you calculated the pH you just plugged in the concentration of ClCH2COOH
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how do I find the [H3O+]?
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H+ and H3O+ can be seemed as basically same things. H3O+ is formed when H+ attracts the water molecule.
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Ok, so I now get -log[.003684] = 2.434?
Also does everything else look good?
Thanks
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If 12% of ClCH2COOH dissociated, concentration of ClCH2COOH is not 0.0307M.
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so that should also get multiplied by .12 and that would make all of the concentrations .0036 then?
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No. 12% dissociated, how much was left?
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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..