Respuesta :
pH = -log[H₃O⁺]
[H₃O⁺] = 10^-pH
[H₃O⁺] = 10^-2,32
[H₃O⁺] =0,0047863 ≈ 4,79×10×10⁻³ mol/dm³
B) -- 4,79×10⁻³ M
[H₃O⁺] = 10^-pH
[H₃O⁺] = 10^-2,32
[H₃O⁺] =0,0047863 ≈ 4,79×10×10⁻³ mol/dm³
B) -- 4,79×10⁻³ M