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Concentration of ammonia in a solution

by sveioen
Tags: ammonia, concentration, solution
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sveioen
#1
Nov5-07, 12:08 AM
P: 14
Hello all,

I had chemistry a long time ago, but now I am very rusty at it so I am hoping you can get me started with this problem I have;

A particular solution of ammonia (Kb = 1.8 x 10-5) has a pH of 8.3.
What is the concentration of ammonia in this solution?

Is it the concentration of NH3 I have to find? I know I can find [OH-] since I know the pH, but what does the final equation look like? Something like [tex]Kb=[OH-][NH4+]/[NH3][/tex]?

Thank you for any help!
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Antineutron
#2
Nov5-07, 12:28 AM
P: 76
NH4 and OH- is going to have same amount of equilibrium concentration gained from the NH3. So If you know the pH, then how do you find the pOH, and what is the concentration of OH-? Multiply both sides by NH3 and divide bothsides by Kb. What happens?
sveioen
#3
Nov5-07, 12:37 AM
P: 14
Ok, so pOH = 14 - pH = 14 - 8,3 = 5,7. Concentration of OH- and NH4 is therefore [tex]1,995 \times 10^{-6}[/tex]? And then [tex] [NH3] = \frac{[OH^-][NH4^+]}{K_b}[/tex]?

Antineutron
#4
Nov5-07, 12:48 AM
P: 76
Concentration of ammonia in a solution

So plug the values and see what you get, I hope this answer agree with the true answer, does it?? If not tell me.
sveioen
#5
Nov5-07, 12:54 AM
P: 14
I got [tex]2,21 \times 10^{-7}[/tex], which seems reasonable I guess. Maybe a bit low?!
Antineutron
#6
Nov5-07, 12:59 AM
P: 76
You don't have the answer? It should be reasonable right? because its the equilibrum concentration right?
sveioen
#7
Nov5-07, 01:09 AM
P: 14
Nope dont have answer (yet) :(, but it seems kinda right.. Probably is equilibrum concentration..
lightarrow
#8
Nov8-07, 06:08 AM
P: 1,521
Quote Quote by sveioen View Post
I got [tex]2,21 \times 10^{-7}[/tex], which seems reasonable I guess. Maybe a bit low?!
I got [tex]2,21 \times 10^{-6}[/tex] instead.


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