Inorganic Chemistry Question Help!

[Hopefulone111]

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Sodium benzoate (C6H5CO2Na) is used as a food preservative. Calculate the pH and the concentrations of all species present (Na+, C6H5CO2-, C6H5CO2H ,H3O+, and OH-) in 0.050 M sodium benzoate; Ka for benzoic acid (C6H5CO2H) is 6.5e-5

I got pH pf 8.4
Na+ 0.050 M
C6H5CO2- 0.050M
C6H5CO2H 2.7e-6M
H3O+ 4e-9 M
OH- 2.7e-6 M

 

[syoung]

Do you have a solvent or is it in water only?

NaC6H5CO2 + H2O -> Na+ + HC6H5CO2 + OH-
And
HC6H5CO2 + H2O -> H3O+ + C6H5CO2-
Is that what you are working with?

 

[Buttafuoco]

I get a pH of 8.4 also, so I'm thinking your numbers are good to go.

 

[Hopefulone111]

I wasnt sure of the equation...I worked with
C6H5CO2- + H2O -> OH- + C6H5CO2H

 

[Buttafuoco]

yeah exactly, convert the ka to kb via kakb = kw, then set up kb = X^2/.05

 

[pfaction]

Buttafuoco, can you post your entire thing? I'm getting a pH of 8.6ish???

KaKb = 1x10^-14
1^-14 / 6.5^-5 so I got 1.5^-10

Kb = x^2/.05 * 1.5^-10^-10 ; X = 2.5^-6 [ close to what is listed for OH]
Therefore, pOH = 5.4 or something
14-5.4 is 8.6

My question is to you, how did you know to use KB and not the KA given?

 

[syoung]

Ka is for the acid H-C6H5CO2, while Kb is for its conjugate base.
You start with the conjugate base grabbing a H+
So you get C6H5CO2- + H2O -> OH- + HC6H5CO2

So your ice chart has to have Kb = x^2 / (conc of base)
then convert the Ka to Kb and solve for X, and then solve for pOH, and then solve for pH

 

[Buttafuoco]

You're told the ka for benzoic acid, and then you're told that you have a bunch of benzoate in solution, the conjugate base of benzoic acid. Since you have the conjugate base, you want to use the kb of that base for the reaction:

benzoate + H20 ---> Benzoic acid + OH-

so kb = x^2/0.5M where x = the concentration of OH- ions and benzoic acid.

so you do the math and find the concentration of OH- slash benzoic acid, and from there the easiest thing is take the -log of OH- to get the pOH and then subtract that from 14 to get the pH. You can get the concentration of hydronum ions by dividing kw by hydroxide concentration or take 10^-pH.

As TS implied in his answers, so little benzoate picks up protons that the amount that it's reduced by is negligible. This should be the case with any MCAT question because no calculators- you need to be able to solve questions like this without dicking around with the quadratic equation. That goes for common ion effect problems too.

 

[Class1P]

you shouldn't really need an ice chart for this.

I just converted the Ka to a PKa so 6.5E-5 is about 4.3

Then subtracted that from 14 to get the PKb = 9.7

then used the equation from TBR.

0.5*(9.7) - 0.5* log of (0.05)

which is about 4.8 + 0.7 = 5.5 = pOH

5.5 - 14= 8.5

I am sure there was a shortcut but I got my head turned around.

The ion concentrations are easy and the OH and H+ concentrations you can get from the pH