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September 2019
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Estime o pH de uma solução 0,15 molL -1 NH4Cl(aq). Kb (NH3 ) = 1,8 x 10-5 mol.L-1 Resp.: pH = 5,04
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jacquehersheysp7x5pf
NH4Cl + H2O ----> NH4OH + H+ + Cl-
HCl ----> H+ + Cl-
__________________________________--> cortar os termos comuns
NH4+ + H2O ----> NH4OH + H+
NH4+ + H2O -----> NH4OH + H+
0,15 - x x x
Kh= Kw / Kb = [NH4OH] [H+] / [NH4+] -->1x10^-14 / 1,8x10^-5 = x² / (0,15-x)
x = 9,12x10^-6 --- [H+] = 9,12x10^-6
pH = - log [H+] ----> pH = - log 9,12x10^-6 ---> pH = 5,03
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HCl ----> H+ + Cl-
__________________________________--> cortar os termos comuns
NH4+ + H2O ----> NH4OH + H+
NH4+ + H2O -----> NH4OH + H+
0,15 - x x x
Kh= Kw / Kb = [NH4OH] [H+] / [NH4+] -->1x10^-14 / 1,8x10^-5 = x² / (0,15-x)
x = 9,12x10^-6 --- [H+] = 9,12x10^-6
pH = - log [H+] ----> pH = - log 9,12x10^-6 ---> pH = 5,03