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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.
 one year ago
 one year 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.
 one year ago
 one year ago

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IsTimBest ResponseYou've already chosen the best response.0
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.
 one year ago

IsTimBest ResponseYou've already chosen the best response.0
@saifoo.khan @Hero @ParthKohli
 one year ago

IsTimBest ResponseYou've already chosen the best response.0
Finally, got some people.
 one year ago

HeroBest ResponseYou've already chosen the best response.0
What world am I in? Chemistry? Really? Sorry I'm not the guy.
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
that's the salt of a strong base and a weak acid.....
 one year ago

muhammad9t5Best ResponseYou've already chosen the best response.0
base ionization constant means PoH?
 one year ago

IsTimBest ResponseYou've already chosen the best response.0
I don't think so, but it's mentioned in this part of the textbook.
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
so ther's some other formula to calculate that i suppose.......
 one year ago

muhammad9t5Best ResponseYou've already chosen the best response.0
yes thats the PoH and 1st one is pH(Hydrogen Concentration)
 one year ago

IsTimBest ResponseYou've already chosen the best response.0
So far, I attempted to use Kw=[H3O][OH], kakb=kw and kb=[HB][OH]/[B]
 one year ago

muhammad9t5Best ResponseYou've already chosen the best response.0
please have a look on http://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/Calculating_pHandpOH.htm
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
2ph=14+pka+logc @25 degree c
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
use that.....u l get it....
 one year ago

IsTimBest ResponseYou've already chosen the best response.0
Sorry, which formula did oyu use? (Best use the equation editor or draw it out to get the message across)
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
help ur self......
 one year ago

IsTimBest ResponseYou've already chosen the best response.0
Wait, is this question related any way to pH Also, that was a rather rude way to brush me off.
 one year ago

huneyaBest ResponseYou've already chosen the best response.1
http://herh.ccrsb.ca/staff/FarrellL/chem12/acids/acidbasereviewanswers.pdf
 one year ago

IsTimBest ResponseYou've already chosen the best response.0
Ok. Thank you all. I was not suppose to do questions related with pH yet. Thank you for clearing that up.
 one year ago

muhammad9t5Best ResponseYou've already chosen the best response.0
You are welcome friend!
 one year ago
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