larutan 0,4 gram NaOH (Mr=40) dalam 1 liter air mempunyai harga PH A) 2 - Log4 B) 2 + Log4 C) 12 D) 12 - Log4 E) 12 + Log4

mol NaOH = massa/mr 0.4/40 = 0.01 mol
M NaOH = mol/v = 0.01/1 = 10^-2 M

OH- = jumlah OH x M = 1 x 10^-2 = 10^-2
pOH = -log OH- = 2
pH = 14 - 2 = 12

Mapel: Kimia
Materi: Asam Basa
Kelas : XI

Jawab

M= gr/Mr × 1000/V
M= 0,4/40 × 1000/1000
M= 0,01

NaOH → Na+ OH-

[OH-] = M×Vb
[OH-] = 0.01×1
[OH-] = 10-²

pOH= -log[OH-]
pOH = -log10-²
pOH= 2

pH= 14-pOH
pH= 14-2
pH= 12

Jawabnya = [C]