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n NaOH = M NaOH x V NaOh = 0,1 x 20 = 2 mmol
HCl + NaOH ⇒ NaCl + H₂O
M 5 2
R -2 -2 +2 +2
S 3 - 2 2
[H⁺] =
pH = -log [H⁺] = -log (4,28 x 10⁻²) ≈ 1,4