c Ba(OH)₂ = 0,01 mol/dm³
Ba(OH)₂ ---> Ba²⁺ + 2OH⁻
c OH ⁻ = 2 * c Ba(OH)₂ = 2 * 0,01 mol/dm³ = 0,02 mol/dm³
pOH = - log c OH⁻ = - log 0,02 = 1,7
pH + pOH = 14
pH = 14 - pOH = 14 - 1,7 = 12,3
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c Ba(OH)₂ = 0,01 mol/dm³
Ba(OH)₂ ---> Ba²⁺ + 2OH⁻
c OH ⁻ = 2 * c Ba(OH)₂ = 2 * 0,01 mol/dm³ = 0,02 mol/dm³
pOH = - log c OH⁻ = - log 0,02 = 1,7
pH + pOH = 14
pH = 14 - pOH = 14 - 1,7 = 12,3