Calculating pH?
Thanks!
Answer:
___NH3 + H2O = NH4+ + OH-
I __.500_________0_____0
C___-x__________+x____+x
E__.50-x_________+x____+x
1.8*10^-5 = [NH4+][OH-]/[NH3]
1.8*10^-5 = (x^2)/(.50 - x)
x is negligible compared to .5 so just forget it and rewrite
1.8*10^-5 = (x^2)/(.50)
solve for x
x = .003
-log(.003) in calculator
=
2.523 = pOH
14 - 2.523
=
11.477 = pH
*edit*
Sorry, my ICE table didn't post very well. Crazy FunQA.com forum setup..
I hope you can still understand it
Aqueous Ammonia is NH4OH
1.8 x 10^-5 = ([NH4][OH])/[NH4OH]
It is assumed that [NH4]=[OH] and that NH4OH is constant
Therefore
[OH]^2 = 1.8 * 10^-5 * 0.50
=9 * 10^-6
[OH] = 3 * 10 ^ -3
pOH = - log 3 * 10 ^ -3 =2.52
ph = 14-pOH = 14- 2.52 = 11.48
Here is the equilibria
NH3 + H2O <> NH4+ + OH- (HYDROLYSIS)
NH4OH---NH4+ + OH- (IONIZATION)
Let x represent the conc of NH4OH and OH- at equilibrium
[NH4OH]=[OH-} = x
NH4+ = C(NH4+) - x = 0.50 - x
Since Conc(NH4+) >> Kb then neglect x compared to the concentration of the NH4+ ion therfore you have
[(x)(x)]/0.50 = 1.8E-5
and x=4.5 x 10-6
therefore [OH-] = Kw/Kb .. (1E-14)/(4.5x10-6) = 2.0E-9
now pH = -log(2.0E-9)=8.7
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