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e-modul KIMIA 2022

9. D. 100 mL
Pembahasan:
pH = 9
pOH = 14 – 9
= 5
[OH ] = 10
-
-5
NH4OH + HCl  NH4Cl
m = x 10 mmol -
r = -10 mmol -10 mmol +10 mmol
s = x – 10 mmol – 10 mmol

[OH ] = Kb x
-

;10
10 = 10 x
-5
-5
10
x – 10 mmol = 10 mmol
x = 20 mmol
n = M x V
20 mmol = 0,2 M x Vml
20
V =
0,2
V = 100 mL

10. C. 11 dan 10 – log 5
Pembahasan:
 pH awal
[OH ] = Kb x M
-
= 10 x 0,1
;5
= 10
;6
= 10
-3
pOH = - log [OH ]
-
= - log 10
-3
= 3

pH = 14 – pOH
= 14 – 3
= 11
 pH setelah penambahan NH4Cl
xxs

[OH ] = Kb x
-

20
= 10 x
-5
10
= 2 x 10 -5

e-modul Kesetimbangan Ion dan pH Larutan Penyangga Halaman 61