Chemistry pH question? Given Kb value and Molarity?

L

Lema

New member
Kb for NH3 is 1.8*10^-5. What is the pH of a 0.35 M aqueous solution of NH4Cl at 25 degrees C?

I set up the equation ...
NH3 + H20 <-> NH4 + OH-
0.35M 0 0
-x +x +x
0.35-x x x

1.8*10^-5 = x^2/0.35-x
x=0.002501
[OH-]=0.002501
pOH=2.6
14-2.6=11.4=pH
This is wrong though. Can someone explain what I'm doing wrong and how to fix it.
 
The problem tells you that the acqueous solution is only made of NH4Cl wich is a salt deriving from a weak base (NH3 or better NH4OH because it is acqueous) and a strong acid (HCl). This salt react with water in a reaction called hydrolysis:

NH4Cl --> NH4(+) + Cl(-)

NH4(+) + H2O <--> NH4OH + H(+)

Since H(+) ions are released we expect that pH will be acidic:

[H+] = sqrt (Kw/Kb * Cs) where Cs is salt concentraion (0.35M)
[H+] = sqrt (10^-14/1.810^-5 * 0.35)
[H+] = sqrt (1.94*10^-10)
[H+] = 1.39 * 10^-5

pH = -log[H+]
pH = -log(1.39*10^-5)
pH = 4.85

Hope this is correct, contact me for any question :-)
 
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