PH dari larutan garam (NH4)2SO4 0,2M adalah… (kb NH3= 1,8×10-5 mol/L)

(NH4)2SO4 → 2NH4+ + SO4^2-

[H+] = √Kw / Kb . n M
[H+] = √10^-14 / 18 x 10^-6 . 2 . 2×10^-1
[H+] = √10^-8 . 10^-1 . 4/18
[H+] = √10^-9 . 2 (1/9)
[H+] = √10^-8 . 0,2 (1/9)
[H+] = 1/9√0,2 x 10^-4

pH = -log[H+]
pH = -log 1/9√0,2 x 10^-4
pH = 4 – log√0,2/9
pH = 4 – (-0,03)
pH = 4 + 0,03
pH = 4,03