Tentukan pH larutan berikut :
A. = 0,1 molor H2504
B. =0,1 molor NH40H, 4Ka 10-⁵![]()
A. = 0,1 molor H2504
B. =0,1 molor NH40H, 4Ka 10-⁵
Bagian A
Diketahui
- Ma = 0,1 mol = 10^-1 mol
- H2SO4 => 2H^+ + SO4^2-
- a = 2
Ditanya
- pH
JAWAB
Hitung konsentrasi [H^+] terlebih dahulu
[H^+] = a . Ma
[H^+] = 2 . 10^-1
[H^+] = 2 × 10^-1
Maka, pH :
pH = - log [H^+]
pH = - log [2 × 10^-1]
pH = 1 - log 2
Bagian B
Diketahui
- Mb = 0,1 mol = 10^-1 mol
- NH4OH => NH4^+ + OH^-
- Kb = 4 × 10^-5
Ditanya
- pH
JAWAB
Hitung konsentrasi [OH^-] terlebih dahulu
[OH^-] = √Kb . Mb
[OH^-] = √(4 × 10^-5) . (10^-1)
[OH^-] = √4 × 10^-6
[OH^-] = 2 × 10^-3
Nilai pOH :
pOH = - log [OH-]
pOH = - log [2 × 10^-3]
pOH = 3 - log 2
Maka, nilai pH :
pH = 14 - pOH
pH = 14 - (3 - log 2)
pH = 14 - 3 + log 2
pH = 11 + log 2
[answer.2.content]
