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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.
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.

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IsTim
 3 years ago
Best ResponseYou've already chosen the best response.0I 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.

IsTim
 3 years ago
Best ResponseYou've already chosen the best response.0@saifoo.khan @Hero @ParthKohli

IsTim
 3 years ago
Best ResponseYou've already chosen the best response.0Finally, got some people.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that's the salt of a strong base and a weak acid.....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0base ionization constant means PoH?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so ther's some other formula to calculate that i suppose.......

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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.02ph=14+pka+logc @25 degree c

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0use that.....u l get it....

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0http://herh.ccrsb.ca/staff/FarrellL/chem12/acids/acidbasereviewanswers.pdf

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You are welcome friend!
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