tough pka questions

[Smooth Operater]

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a solution is composed of equal concentrations of aqueous ammmonia (pKa = 4.74) and ammonium chloride (pka = 9.26). What would be the pH of the resulting solution?

the answer is 9.26, but I am not sure how to arrive to this answer. If you know, plz let me know. THANKS!

 

[doc toothache]

This is a common ion effect problem.

from Ka, [OH-]=Ka[NH3]/[NH4+] since the concentrations of NH3 and NH4+ are equal, [OH-]=ka=1.8 x 10^5
then pOH=4.74, and pH=14.0-4.74=9.26

 

[mr.dds]

how do you know the log of 1.8x10^5 is equal to 4.74?

 

[doc toothache]

pKa=-log Ka;
pKa= -log 1.8 x 10^5;
log of 10^5=5; log of 1.8 (from log tables)=0.2553 or ~0.26
pKa=5- 0.26=4.74

 

[haavoc]

how do you know the log of 1.8x10^5 is equal to 4.74?

You can either know the log tables or approximate the answer by doing the following:

log (m * 10 ^ n) --> n - log(m) --> log(m) ranges from 0 to 1 based on m being from 1 to 10, therefore, the closer m is to 1, the final answer will be closer to n, if it is closer to 10, then the answer will be closer to n - 1.

They will most likely not give you answers which vary closely, so with this method you could usually approximate your answer.

 

[Smooth Operater]

hello, thanks for the help tooth doc, but i think there is a problem

for [OH-]=Ka[NH3]/[NH4+], should it be Kb instead of Ka since you are trying to find [OH-] not [H3O+]. Can you explain how you set up this equation? thanks!

 

[doc toothache]

The correct numbers for ammonia are pKa=9.26 (not 4.74) and pKb=4.74.
(Since the constant given is the correct one (for pKb) , my guess is that the Ka notation was used as a generic notation for acid-base dissociation constant rather than the true value of the acid dissociation constant, pKa, of NH3).
Aqueous ammonia reaction is:

NH3 + H2O = NH4+ + OH-

Kb=[NH4+][OH-]/[NH3]; rearanging and solving for [OH-]

[OH-]=Kb[NH3]/[NH4+]
The rest of the calculations remain the same.