[reddog690]

What is the original NH3 molarity if an equilibrium mixture with water is 4.2 % ionized?

first I used the equation NH3 + H2O ::equil::NH4+ +OH- and expressed percent ionization (the equivalent of percent protonation?) as

%ion= (conc. of ammonium ion/initial conc. of NH3)= 0.042

I am confused how to finish this problem.  I looked for help from previous posts and found this link http://www.titrations.info/acid-base-titration-indicators

but it did not help me.

Thanks for any help offered.

[Borek]

Start with Kb. Express concentrations of both ionized and not ionized base in terms of its total concentration and percent ionization. Substitute, solve for C.

[reddog690]

I did look at Kb=[NH4+][OH-]/[NH3]= 1.76 X 10-5, but not sure what to substitute.    I seem to have no trouble using Ka to find answers for pH questions, but having conceptual difficulty in making a similar connection to ionization of bases.
Thank you again for assistance.

[Borek]

Assume analytical concentration is C (that's your unknown). If so, ionized form concentration is 0.042C. Can you express unionized form in a smilar way?

[reddog690]

OK, if C is the analytical concentration (the original conc of ammonia?), then C= conc.ionized plus conc un-ionized, and conc. un-ionized= (1-0.042)=0.958c.  I end up with an expression like the one below:
Kb={(0.042c)[OH-]/0.958c}= 1.76 x 10-5, and solving for C I get 9.56 X 10-3M (NH3) using the assumption that and OH- present is a result of the formation of NH4+.

If I did this correctly, what is the best way to check my answer?

It seems logical that the ionized NH3 (NH4+)= 4.02 x 10-4 since 4.02 x 10-4/9.56x10-3= 0.042 but I would think that other ratios would also fit this criteria.

I am still a little unsure of my approach (and answer).

Thanks for looking at this again, I appreciate your *delete me*

[Borek]

If I did this correctly, what is the best way to check my answer?

Start with ammonia solution of a given concentration, and calculate percent ionization.

It seems logical that the ionized NH3 (NH4+)= 4.02 x 10-4 since 4.02 x 10-4/9.56x10-3= 0.042 but I would think that other ratios would also fit this criteria.

For other concentrations degree of dissociation would be different, so percent dissociation would be different as well.

You may try to calculate degree of dissociation for different concentrations - you may use BATE from my signature for that, free trial will do, you don' have to buy the program to play with it. You will see that percent dissociation changes with concentration.and is never identical for two different concentrations.

Compare http://en.wikipedia.org/wiki/Ostwald_dilution_law