Ba(OH)₂
Massa Ba(OH)₂ = 17,1 gr
Mr Ba(OH)₂ = 137 + 2. (16 +1) = 137 + 2. 17 = 137 + 34 = 171
mol Ba(OH)₂ = Massa / Mr = 17,1 gr / 171 = 0,1 mol
Molaritas Ba(OH)₂ = Mol / Volume = 0,1 / 4 = 0,025 M = 2,5 × 10⁻²M
Molaritas OH⁻ = 2 × 2,5 × 10⁻²M = 5 × 10⁻²M
pOH = - log OH⁻
pOH = - log [5 × 10⁻²]
pOH = 2 - log 5
pH = 14 - pOH
pH = 14 - (2 - log 5)
pH = 12 + log 5
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Ba(OH)₂
Massa Ba(OH)₂ = 17,1 gr
Mr Ba(OH)₂ = 137 + 2. (16 +1) = 137 + 2. 17 = 137 + 34 = 171
mol Ba(OH)₂ = Massa / Mr = 17,1 gr / 171 = 0,1 mol
Molaritas Ba(OH)₂ = Mol / Volume = 0,1 / 4 = 0,025 M = 2,5 × 10⁻²M
Molaritas OH⁻ = 2 × 2,5 × 10⁻²M = 5 × 10⁻²M
pOH = - log OH⁻
pOH = - log [5 × 10⁻²]
pOH = 2 - log 5
pH = 14 - pOH
pH = 14 - (2 - log 5)
pH = 12 + log 5