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= √ ( 10⁻¹⁴ / 10⁻⁵) x 9 x 10⁻¹
= √ 9 x 10⁻¹⁰
= 3 x 10⁻⁵
pOH = 5 - log 3
= 5 - 0,5
= 4,5
pH = 14 - 4.5
= 9,5 atau 9 + log 3 <------------------------
= √10^-14/10^-5 . 9.10^-1
= √9x10^-15/10^-5
= √9x10^-10
= 3x10^-5 M
pOH = -log [OH⁻]
= -log 3x10^-5
=5-log 3
pH = 14 - pOH
= 14 - (5-log 3)
= 14 - 5 + log 3
= 9 + log 3
= 9 + 0,48
= 9,48