Jawaban:
Penjelasan:
pH sebelum penambahan HCl
[H+] = Ka x (mol asam.valensi/mol garam.valensi)
= 1,8. 10^-5 x (Ma. Va. 1/Mg. Vg. 1) (valensinya = 1)
= 1,8. 10^-5 x (0,1. 100/0,1. 100)
= 1,8. 10^-5
pH = - log [H+]
= - log 1,8. 10^-5
= 5 - log 1,8
------------------------------------------------------------------------------------------
pH setelah penambahan HCl
mol HCl = M.V = 0,1. 10 = 1 mmol
mol CH3COOH dan CH3COOH dapat dari M. V diatas
M 10 1 10
R 1 1 1 1
S 9 - 11 1
= 1,8. 10^-5 x (11/9)
= 2,2. 10^-5
= - log 2,2. 10^-5
= 5 - log 2,2 (A)
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Jawaban:
Penjelasan:
pH sebelum penambahan HCl
[H+] = Ka x (mol asam.valensi/mol garam.valensi)
= 1,8. 10^-5 x (Ma. Va. 1/Mg. Vg. 1) (valensinya = 1)
= 1,8. 10^-5 x (0,1. 100/0,1. 100)
= 1,8. 10^-5
pH = - log [H+]
= - log 1,8. 10^-5
= 5 - log 1,8
------------------------------------------------------------------------------------------
pH setelah penambahan HCl
mol HCl = M.V = 0,1. 10 = 1 mmol
mol CH3COOH dan CH3COOH dapat dari M. V diatas
M 10 1 10
R 1 1 1 1
S 9 - 11 1
[H+] = Ka x (mol asam.valensi/mol garam.valensi)
= 1,8. 10^-5 x (11/9)
= 2,2. 10^-5
pH = - log [H+]
= - log 2,2. 10^-5
= 5 - log 2,2 (A)