Determine the ph of a 0.188 m nh3 solution at 25°c. the kb of nh3 is 1.76 × 10-5.
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Ответ:
NH3 + H2O → NH4+ + OH-
We use the Kb expression to determine the [OH-] concentration,
Kb = [NH4+] [OH-] /* [NH3]
We can write NH4+ as OH- since they are of equal ratio.
(1.76*10^-5) = [OH-]² / 0.188
[OH-]² = 3.3088*10^-6
[OH-] = 1.819*10^-3
We calculate for H+ concentration as follows:
[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
Ответ:
lol
Explanation:
no