# Homework Help: C2H3O2- concentration given pH of 30mL NaC2H3O2 solution

1. Feb 1, 2012

### Phyzwizz

I'm working on a lab for my chem class and we have to calculate all of the different ion concentrations within some titrations that we did. Here I have to calculate the concentration before anything is added. Na+ + C2H3O2- + H2O → HC2H3O2 + OH- + Na+
the pH is 8.4458 for a 30.0 mL NaC2H3O2 solution. I tried finding the concentration of the Na+ and C2H3O2- ions by: 14.00 - 8.4458 = pOH = 5.55452 then 10-5.55452 = 2.79 X 10-6 which is the OH- concentration. In the Kb? equation this would be x so: 5.3 X 10-6 = (2.79 X 10-6)2/(x-(2.79 X 10-6). Calculating all this through I get 4.25 X 10-6. With the Liters being 30.0 mL I take this concentration and divide it by .0300 L and I get 1.42 X10-4. This just seems incorrect because the Molarity was supposed to be somewhere near 1.0M, what did I do wrong?

2. Feb 2, 2012

### Staff: Mentor

You really know pH with that high accuracy? Wow. Most pH meters I know will show 8.45, and I would not believe the last digit.

You mean - if you will take different volume, pH will have different value?

Probably typo. 5.5542, if anything. Not that it changes the result by much.

OK

5.3x10-6 is not Kb of the acetate ion, so anything you did past this point can't be correct. Still, you managed to write something that I can't left uncommented:

So you took concentration and you divided it by volume to get... what? Concentration again?

3. Feb 3, 2012

### Phyzwizz

yeah I don't know what I did to get that Kb value I know its (1.0 X 10-14)/(1.8 X 10-5) = 5.6 X 10-10.

Looking at it now I'm unsure why I would have gone any farther than the 4.25 X 10-6 this answer would definitely be different with the correct Kb value but wouldn't this give me the ion concentrations of both C2H3O2- and Na+ since they are 1:1?

4. Feb 4, 2012

### Staff: Mentor

Yes.

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