## IsTim 3 years ago The hydroxide ion concentration in a 0.157 mol/L solution of sodium propanoate, NaC3H5O2 (aq) is found to be 1.1*10^-5 mol/L. Calculate the base ionization constant for the propanoate ion.

1. IsTim

I am using the Formula Ka*Kb=Kw. Rearranged then into Kb=Kw/Ka. Kw is a constant; 1.0*10^-14. Originally attempted using formula Kw=[H30}{OH^-}. Answer was 9.09*10^-10. Actual answer is 7.7*10^-10.

2. IsTim

@saifoo.khan @Hero @ParthKohli

3. IsTim

Finally, got some people.

4. Hero

What world am I in? Chemistry? Really? Sorry I'm not the guy.

5. IsTim

@.Sam. @Callisto

6. A.Avinash_Goutham

that's the salt of a strong base and a weak acid.....

7. IsTim

Sodium propanoate?

base ionization constant means PoH?

9. IsTim

I don't think so, but it's mentioned in this part of the textbook.

10. A.Avinash_Goutham

so ther's some other formula to calculate that i suppose.......

yes thats the PoH and 1st one is pH(Hydrogen Concentration)

12. IsTim

So far, I attempted to use Kw=[H3O][OH], kakb=kw and kb=[HB][OH]/[B]

please have a look on http://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/Calculating_pHandpOH.htm

14. A.Avinash_Goutham

2ph=14+pka+logc @25 degree c

15. A.Avinash_Goutham

use that.....u l get it....

16. IsTim

Sorry, which formula did oyu use? (Best use the equation editor or draw it out to get the message across)

17. A.Avinash_Goutham

help ur self......

18. IsTim

Wait, is this question related any way to pH Also, that was a rather rude way to brush me off.

yes

20. huneya
21. IsTim

Ok. Thank you all. I was not suppose to do questions related with pH yet. Thank you for clearing that up.

22. huneya

ur welcome