initial : solutia de HCl : Vs1 cM1 = 0,1mol/l nHCl = 0,1Vs1 = nH⁺
solutia de NaOH : Vs2 cM2 = 0,05mol/l nNaOH = 0,05Vs2= nHO⁻
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reactioneaza : 0,05Vs2 HO⁻ + 0,05Vs2H⁺
final : Vs = 0,1 l pH = 2 ([H⁺]exces = 10⁻²) nH⁺ exces = 0,1ₓ10⁻² = 10⁻³ moli
nH⁺ exces = nH⁺initial - nH⁺ reactionat =10⁻¹Vs1 - 5ₓ10⁻²Vs2
5ₓ10⁻²(2Vs1- Vs2) = 10⁻³ 100Vs1 - 50Vs2 = 1
Vs1 + Vs2 = 0,1
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150Vs1 = 6 Vs1 = 0,04 l = 40ml Vs2 = 60ml
