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.

Question

istim · Answer

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.

istim · Answer

@saifoo.khan @Hero @ParthKohli

istim · Answer

Finally, got some people.

hero · Answer

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

istim · Answer

@.Sam. @Callisto

anonymous · Answer

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

istim · Answer

Sodium propanoate?

anonymous · Answer

base ionization constant means PoH?

istim · Answer

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

anonymous · Answer

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

anonymous · Answer

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

istim · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

istim · Answer

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

anonymous · Answer

help ur self......

istim · Answer

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

anonymous · Answer

yes

anonymous · Answer

http://herh.ccrsb.ca/staff/FarrellL/chem12/acids/acid-base-review-answers.pdf

istim · Answer

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

anonymous · Answer

ur welcome

anonymous · Answer

You are welcome friend!