Articles
Register
Sign In
Search
LFSS
@LFSS
September 2019
1
105
Report
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
Please enter comments
Please enter your name.
Please enter the correct email address.
Agree to
terms and service
You must agree before submitting.
Send
Lista de comentários
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
3 votes
Thanks 0
More Questions From This User
See All
LFSS
December 2019 | 0 Respostas
Responda
LFSS
December 2019 | 0 Respostas
Responda
LFSS
September 2019 | 0 Respostas
Responda
LFSS
May 2019 | 0 Respostas
Responda
Recomendar perguntas
Deividyfreitas
May 2020 | 0 Respostas
BlackShot
May 2020 | 0 Respostas
Vanessakellen
May 2020 | 0 Respostas
Guiduarter
May 2020 | 0 Respostas
Mrzaine
May 2020 | 0 Respostas
O QUE SERIA AUTONOMIA?
Grazifer
May 2020 | 0 Respostas
Joazinho
May 2020 | 0 Respostas
a palavra rapidez formou se de qual derivacao
Celiana
May 2020 | 0 Respostas
Joazinho
May 2020 | 0 Respostas
Anatercia
May 2020 | 0 Respostas
×
Report "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.... Pergunta de ideia de LFSS"
Your name
Email
Reason
-Select Reason-
Pornographic
Defamatory
Illegal/Unlawful
Spam
Other Terms Of Service Violation
File a copyright complaint
Description
Helpful Links
Sobre nós
Política de Privacidade
Termos e Condições
direito autoral
Contate-Nos
Helpful Social
Get monthly updates
Submit
Copyright © 2024 ELIBRARY.TIPS - All rights reserved.
Lista de comentários
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