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IsTim
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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.
 2 years ago
 2 years ago
IsTim Group Title
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.
 2 years ago
 2 years ago

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IsTim Group TitleBest 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.
 2 years ago

IsTim Group TitleBest ResponseYou've already chosen the best response.0
@saifoo.khan @Hero @ParthKohli
 2 years ago

IsTim Group TitleBest ResponseYou've already chosen the best response.0
Finally, got some people.
 2 years ago

Hero Group TitleBest ResponseYou've already chosen the best response.0
What world am I in? Chemistry? Really? Sorry I'm not the guy.
 2 years ago

IsTim Group TitleBest ResponseYou've already chosen the best response.0
@.Sam. @Callisto
 2 years ago

A.Avinash_Goutham Group TitleBest ResponseYou've already chosen the best response.0
that's the salt of a strong base and a weak acid.....
 2 years ago

IsTim Group TitleBest ResponseYou've already chosen the best response.0
Sodium propanoate?
 2 years ago

muhammad9t5 Group TitleBest ResponseYou've already chosen the best response.0
base ionization constant means PoH?
 2 years ago

IsTim Group TitleBest ResponseYou've already chosen the best response.0
I don't think so, but it's mentioned in this part of the textbook.
 2 years ago

A.Avinash_Goutham Group TitleBest ResponseYou've already chosen the best response.0
so ther's some other formula to calculate that i suppose.......
 2 years ago

muhammad9t5 Group TitleBest ResponseYou've already chosen the best response.0
yes thats the PoH and 1st one is pH(Hydrogen Concentration)
 2 years ago

IsTim Group TitleBest ResponseYou've already chosen the best response.0
So far, I attempted to use Kw=[H3O][OH], kakb=kw and kb=[HB][OH]/[B]
 2 years ago

muhammad9t5 Group TitleBest ResponseYou've already chosen the best response.0
please have a look on http://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/Calculating_pHandpOH.htm
 2 years ago

A.Avinash_Goutham Group TitleBest ResponseYou've already chosen the best response.0
2ph=14+pka+logc @25 degree c
 2 years ago

A.Avinash_Goutham Group TitleBest ResponseYou've already chosen the best response.0
use that.....u l get it....
 2 years ago

IsTim Group TitleBest 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)
 2 years ago

A.Avinash_Goutham Group TitleBest ResponseYou've already chosen the best response.0
help ur self......
 2 years ago

IsTim Group TitleBest 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.
 2 years ago

huneya Group TitleBest ResponseYou've already chosen the best response.1
http://herh.ccrsb.ca/staff/FarrellL/chem12/acids/acidbasereviewanswers.pdf
 2 years ago

IsTim Group TitleBest 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.
 2 years ago

muhammad9t5 Group TitleBest ResponseYou've already chosen the best response.0
You are welcome friend!
 2 years ago
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