Question

500 ml larutan basa lemah LOH mempunyai PH =10 +log 5,joka kb LOH adalah 5x10^-5.Tentukan massa LOH pada larutan (Mr=35)

Answer 1

LOH
V = 500 mL
pH = 10 + log 5
Kb LOH = 5 × 10^-5
Mr = 35

massa LOH ... ?

*Cari Molaritas LOH dari pH-nya

pH = 10 + log 5
pOH = 14 - pH
pOH = 14 - (10 + log 5)
pOH = 14 - 10 - log 5
pOH = 4 - log 5

......... pOH = 4 - log 5
......... pOH = - log [OH^-]
... 4 - log 5 = - log [OH^-]
.. 5 × 10^-4 = [OH^-]
...... [OH^-] = 5 × 10^-4

.......... [OH^-] = √Kb × M
......5 × 10^-4 = √5 × 10^-5 × M
(5 × 10^-4)^2 = (√5 × 10^-5 × M )^2
.. 25 × 10^-8 = 5 × 10^-5 × M
..................M = (25 × 10^-8) / (5 × 10^-5)
................. M = 5 × 10^-3
................. M = 0,005

*Cari massa LOH
M = gram/Mr × 1000/mL
0,005 = gram/35 × 1000/500
0,005 = gram/35 × 2
0,005 × 35 = gram × 2
0,175 = gram × 2
0,175/2 = gram
gram = 0,0875 gram

Jadi, massa LOH pada larutan adalah 0,0875 gram