C2H3O2- concentration given pH of 30mL NaC2H3O2 solution

[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 K_b? 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?

 

[Borek]

the pH is 8.4458

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.

for a 30.0 mL NaC2H3O2 solution

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

14.00 - 8.4458 = pOH = 5.55452

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

2.79 X 10^-6 which is the OH^- concentration

OK

In the K_b? equation this would be x so: 5.3 X 10^-6

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:

With the Liters being 30.0 mL I take this concentration and divide it by .0300 L and I get 1.42 X10^-4.

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

 

[Phyzwizz]

yeah I don't know what I did to get that K_b 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 K_b value but wouldn't this give me the ion concentrations of both C₂H₃O₂^- and Na⁺ since they are 1:1?

 

[Borek]

Yes.