Berapa gram NaOH terlarut dalam 750ml larutan NaOH yang nilai phnya 12 + log6?

Diketahui :
v = 750 mL = 0,75 L
pH = 12 + log 6

Ditanya :
m ?

Jawab :
pOH = Kw - pH
pOH = 14 - (12 + log 6)
pOH = 14 - 12 - log 6
pOH = 2 - log 6
[OH⁻] = 6.10⁻² M

NaOH → Na⁺ + OH⁻
[NaOH] = [OH⁻] = 6.10⁻² M

n NaOH = M NaOH × v
n NaOH = 6.10⁻² M × 0,75 L
n NaOH = 0,045 mol

Mr NaOH = 40 g/mol

m NaOH = Mr NaOH × n NaOH
m NaOH = 40 g/mol × 0,045 mol
m NaOH = 1,8 g