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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.

Chemistry
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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.
Finally, got some people.

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Other answers:

What world am I in? Chemistry? Really? Sorry I'm not the guy.
@.Sam. @Callisto
that's the salt of a strong base and a weak acid.....
Sodium propanoate?
base ionization constant means PoH?
I don't think so, but it's mentioned in this part of the textbook.
so ther's some other formula to calculate that i suppose.......
yes thats the PoH and 1st one is pH(Hydrogen Concentration)
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
2ph=14+pka+logc @25 degree c
use that.....u l get it....
Sorry, which formula did oyu use? (Best use the equation editor or draw it out to get the message across)
help ur self......
Wait, is this question related any way to pH Also, that was a rather rude way to brush me off.
yes
http://herh.ccrsb.ca/staff/FarrellL/chem12/acids/acid-base-review-answers.pdf
Ok. Thank you all. I was not suppose to do questions related with pH yet. Thank you for clearing that up.
ur welcome
You are welcome friend!

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