Determine the ph of a 0.188 m nh3 solution at 25°c. the kb of nh3 is 1.76 × 10-5.

Determine the pH of a 0.188 M NH3 solution at
25degrees Celcius.The Kb of NH3 is 1.76 X
10-5.
3 b 3 -5.

NH3 dissolved in water dissociates slightly to produce :
NH3 + H2O ? NH4+ + OH-
Use the Kb equation to determine the [OH-]
Kb = [NH4+] [OH-] /* [NH3]
Because [NH4+] = [OH-] and dissociation is very slight we can
write:
(1.76*10^-5) = [OH-]² / 0.188
[OH-]² = (1.76*10^-5) * 0.188
[OH-]² = 3.3088*10^-6
[OH-] = 1.819*10^-3
In order to calculate pH you require [H+]
Equation:
[H+] [OH-] = 10^-14
[H+] = 10^-14 / [OH-]
[H+] = 10^-14 / (1.819*10^-3)
[H+] = 5.50*10^-12
pH = -log [H+]
pH = -log (5.5*10^-12)pH = 11.26
Hope this will help you. Write me in comment for any
question regarding this.Thanks

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