1. Calcule el pH de las siguientes soluciones: HCl 10 ^– 5 N, KOH 3 x 10^-6 N, HCl 3 x 10^-4 N y el pOH del NaOH 5 x 10^-3N porfa ayuda con esto se lo agradecere bastante.
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HCl
pH = - log [H⁺]
pH = -log (10⁻⁵)
pH = -(-5) = 5
KOH = 3 x 10⁻⁶
pOH = -log [OH⁻]
pOH = - Log(3 x 10⁻⁶)
pOH = -(-5.52) = 5.52
pOH + pH = 14
pH = 14 - 5.52 = 8.48
HCl 3 x 10⁻⁴
pH = - log [H⁺]
pH = -log ( 3 x 10⁻⁴)
pH = -(-3.52) = 3.52
NaOH = 5 X 10⁻³
pOH = - log [OH⁻]
pOH = -log (5 x 10⁻³)
pOH = -(-2.30) = 2.30
pH + pOH = 14
pH = 14 - 2.30 = 11.7
pH = - log [H⁺]
pH = -log (10⁻⁵)
pH = -(-5) = 5
KOH = 3 x 10⁻⁶
pOH = -log [OH⁻]
pOH = - Log(3 x 10⁻⁶)
pOH = -(-5.52) = 5.52
pOH + pH = 14
pH = 14 - 5.52 = 8.48
HCl 3 x 10⁻⁴
pH = - log [H⁺]
pH = -log ( 3 x 10⁻⁴)
pH = -(-3.52) = 3.52
NaOH = 5 X 10⁻³
pOH = - log [OH⁻]
pOH = -log (5 x 10⁻³)
pOH = -(-2.30) = 2.30
pH + pOH = 14
pH = 14 - 2.30 = 11.7
Edisson96:
muchas gracias :) amigo, lo nececitaba de verdad
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