Jawabanmu

2014-05-09T14:22:23+07:00
POH = 14 - pH = 14 - (11+log 4) = 3 - log 4 = - log 4x10^-3
[OH-] = 4x10^-3 M
Kb = [OH-]^2 / [LOH] = (4x10^-3)^2 / 0,02 = 8x10^-4

n LOH = M x V = 0,02 x 0,1 = 0,002 mol
n HBr = M x V = 0,01 x 0,2 = 0,002 mol

LOH . + . HBr . -> . LBr . + . H2O
m 0,002 .... 0,002 ..... - ........... -
r . 0,002 ..... 0,002 .... 0,002 ... 0,002
--------------------------------------...
s .. 0 ........... 0 .......... 0,002 ... 0,002

kemudian gunakan rumus hidrolisis garam

M LBr = n /V = 0,002 / 0,3 = 6,67x10^-3 M

..... LBr ..... -> ..... L^+ ..... + ..... Br^- ....
6,67x10^-3 M . 6,67x10^-3 M . 6,67x10^-3 M

[H+] = √(Kw/Kb x [L+]) = √(10^-14/8x10^-4 x 6,67x10^-3) = √8,3x10^-14 = 2,9x10^-7

pH = - log 2,9x10^-7 = 7 - log 2,9

semoga membantu :)
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