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# Thread: Chemistry help needed! (calculating the Ka or Kb value of salt in solution) *help?*

1. ## Chemistry help needed! (calculating the Ka or Kb value of salt in solution) *help?*

heres the question

2. Ka = [10^-9.25]^2/1

3. Originally Posted by Pirate4lyfe
Ka = [10^-9.25]^2/1
smart dude

4. sorry how do u do it?

5. Acetate breaks into equal concentrations of CH3COO- ion and H+ ion. Ka is: [H+ ion][Acetate anion]/[Starting concentration minus the concentrations of the ions]
You calculate the [H+] like this: 10^(-9.25). This value is basically insignifcant so you can leave one as a denomiator. So Ka= [10^(-9.25)]^2

6. bum[p

7. Originally Posted by mikay456
Acetate breaks into equal concentrations of CH3COO- ion and H+ ion. Ka is: [H+ ion][Acetate anion]/[Starting concentration minus the concentrations of the ions]
You calculate the [H+] like this: 10^(-9.25). This value is basically insignifcant so you can leave one as a denomiator. So Ka= [10^(-9.25)]^2
Nope, sorry. What are your sources?

8. Originally Posted by liftw8s
Nope, sorry. What are your sources?
What I learned in AP chem.

9. pH = -log Ka

10. You have to make an ICE chart and find the concentrations at equilibrium I think, so to find the concentration of H+ ions u need to do 10^(-9.25)=5.6234x10^-10. and then Ka=(concentration of products]/{concentration of reactants]. I cant make the ICE chart right now cuz I cant find my formula sheet.

11. Originally Posted by mikay456
Acetate breaks into equal concentrations of CH3COO- ion and H+ ion. Ka is: [H+ ion][Acetate anion]/[Starting concentration minus the concentrations of the ions]
You calculate the [H+] like this: 10^(-9.25). This value is basically insignifcant so you can leave one as a denomiator. So Ka= [10^(-9.25)]^2
That's wrong. Acetic acid falls apart into acetate and a proton.

So overall I believe this would be the Kb because it is acetate (C2 H3 O2) not acetic acid. So pOH = 14 - pH. Then calculate from there.

12. god this is pissing me off its easy if I knew how to set up the reaction.

13. Originally Posted by OhioUstud
That's wrong. Acetic acid falls apart into acetate and a proton.

So overall I believe this would be the Kb because it is acetate (C2 H3 O2) not acetic acid. So pOH = 14 - pH. Then calculate from there.
Okay, so it is acetic acid that, my bad. But why is Kb involved?

Edit: forget I asked, aceate is the conjugate weak base, forgot that.

14. cuz Kb is used for the concentration of OH ions because weak bases in water react incompletely to produce OH- ions

15. K final answer... You were given sodium acetate so you must calculate the Kb of acetate since it is a weak base.

pOH = 4.75

Kb = [10^(-4.75)]^2.

16. k if u find out how to right the equation this is what u do.
do an ICE chart, then using the equilibrium values that u got go Kb= [porducts]/ [reactants].

17. Originally Posted by Rockwilder
k if u find out how to right the equation this is what u do.
do an ICE chart, then using the equilibrium values that u got go Kb= [porducts]/ [reactants].
ICE chart isn't needed in this case because the solution is already at equilibrium. You only do ICE charts when you're mixing solutions or changing the pressure or molarities.

18. okay lol, this problem was pissing me off.

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