terdapat 4 L larutan nh4cl 0,01 m jika kh= 10^9 ph larutan garam tsb adalh

M = n/v
M = 0,01 / 4L
M = 0,0025
M = 25 x 10^-4

[H+] = √Kh . M
[H+] = √10^-9 . 25 x 10^-4
[H+] = √25 x 10^-12 x 10^-1
[H+] = 5√0,1 x 10^-6

pH = -log [H+]
pH = -log5√0,1 x 10^-6
pH = 6 - log5√0,1
pH = 5,8

→ Atau

pH = 1/2 (14 - Kb - log[G])
pH = 1/2 (14 - (-log10^-5) - log25 x 10^-4)
pH = 1/2 (14 - 5 + 4 - log25)
pH = 1/2 (13 - log25)
pH = 6,5 - log25/2
pH = 6,5 - 0,69
pH = 5,8