ini gimana caranya? mohon bantuannya ☺

[OH-] = akar (Kw/Ka × M × x)
= akar (10^-14 / 2 × 10^-5 × 25 × 10^-3 × 2)
= akar (10^-9 × 25 × 10^-3)
= akar (25 × 10^-12)
= 5 × 10^-6 M

pOH = -log [OH-]
= -log (5 × 10^-6)
= 6 - log 5

pH = 14 - (6 - log 5)
= 8 + log 5
Jawaban: D