Question

Diketahui 0,8 gam NaOH dilarutkan dalam air hingga volume 500 mL. pH dari larutan tersebut adalah… (Ar Na= 23, O= 16, H= 1)
2 – log 4
4 – log 2
8 – log 2
10 + log 2
12 + log 4

Answer 1

V = 500mL = 0,5 L
nNaOH = m/Mr = 0,8/40 = 0,02
Mnaoh = n/v = 0,02/0,5 = 4 x 10^-2

konsentrasi OH = 4 x 10^-2
pOH = - log [OH]
= - log 4 x 10^-2
=2 - log 4 (sifat log)

pH = 14 - (2 - log 4)
=12 + log 4