Question

PH 500 ml larutan yang memgandung 0,2 mol NH4Cl (Kb NH3 = 10^-5) adalah?

Answer 1

[G] = n/v
[G] = 0,2mol/0,5Liter = 0,4 M
[H+] = √Kw/Kb.[G]
[H+] = √10^-14.10^-5.4x10^-1
[H+] = √4x10^-10
[H+] = 2x10^-5
pH = 5 - log 2