solutia 1: KOH : Vs1 = V cM1 = 3ₓ10⁻⁴ n1KOH = 3ₓVₓ10⁻⁴ moli = n1HO⁻
solutia 2 : KOH : Vs2 = 9,7V cM2 = 10⁻³ n2KOH = 9,7Vₓ10⁻³ moli = n2HO⁻
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solutia finala : Vs = 10,7V nHO⁻ = 100ₓVₓ10⁻⁴ = Vₓ10⁻²
[HO⁻] = Vₓ10⁻²/(10,7V) = 9,345ₓ10⁻⁴
pOH = 4 - lg9,345 = 4 - 0,97 = 3,03 pH = 14 - 3,03 = 10,97