Hitung Ph larutannya Ba(OH)2 0.0005M
[ASAM BASA | XI SMA]
Ba(OH)₂ 0,0005 M
[OH⁻] = M x valensi
[OH⁻] = 5 x 10⁻⁴ x 2
[OH⁻] = 10⁻³
pOH = -log [OH⁻]
pOH = -log 10⁻³
pOH = 3
pH = 14 – pOH
pH = 14 – 3
pH = 11
Hitung Ph larutannya Ba(OH)2 0.0005M
[ASAM BASA | XI SMA]
Ba(OH)₂ 0,0005 M
[OH⁻] = M x valensi
[OH⁻] = 5 x 10⁻⁴ x 2
[OH⁻] = 10⁻³
pOH = -log [OH⁻]
pOH = -log 10⁻³
pOH = 3
pH = 14 – pOH
pH = 14 – 3
pH = 11