Question

Besarnya [OH-] dalam 100 mL larutan yang mengandung 3,2 gram amonium nitrat adalah . . . M. (Kb NH4OH = 10^-5; Ar : N = 14, O = 16, H = 1)

Answer 1

NH4NO3
V = 100 mL
m = 3,2 gram
Kb NH4OH = 10^-5

[OH^-] = ... ?

Mr NH4NO3
Mr = 2×Ar N + 4×Ar H + 3×Ar O
Mr = 2×14 + 4×1 + 3×16
Mr = 28 + 4 + 48
Mr = 80

Konsentrasi NH4NO3
M = gram/Mr × 1000/mL
M = 3,2/80 × 1000/100
M = 0,04 × 10
M = 0,4

NH4NO3 bersifat asam, jadi
[H^+] = √( Kw/Kb × M )
[H^+] = √( 10^-14/10^-5 × 0,4 )
[H^+] = √( 10^-9 × 0,4 )
[H^+] = √( 10^-10 × 4 )
[H^+] = 10^-5 × 2
[H^+] = 2 × 10^-5

[H^+][OH^-] = 10^-14
[OH^-] = 10^-14 / [H^+]
[OH^-] = 10^-14 / (2 × 10^-5)
[OH^-] = 10^-9 × 0,5
[OH^-] = 10^-10 × 5
[OH^-] = 5 × 10^-10