SOLUCIONESBUFFER

Una solución Reguladora, Buffer , Tampón o Amortiguadora es:

Un sistema que tiende a mantener elpH casi constante cuando seagregan pequeñas cantidades deácidos (H+) ó bases (OH-).

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COMPONENTESBuffer ácido: Buffer básico:

Formado por un ácidodébil y su sal.Ejemplo: CH3COOH/CH3COONa

Formado por unabase débil y su sal.Ejemplo:NH3/NH4Cl

La Ecuación de Henderson HasselbachpH= pKa + Log [Sal]

[Ácido]

Donde: pKa = -log KaY para las bases:

pOH= pKb + Log [Sal][Base]

Donde: pKb = -log Kb

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1. Calcule el pH de unasolución Buffer formada por0.25 moles de CH3COOH(ácido acético) y 0.4 molesde CH3COONa (acetato desodio) disueltos en 500 mlde solución. Teniendo unaKa = 1.8 x 10-5

Tenemos:0.25 moles de CH3COOH0.40 moles de CH3COONa500 ml de soluciónKa= 1.8 x 10-5

Calcular : [CH3COOH]= 0.25 moles= 0.5M0.5 L

[CH3COONa]=0.40 moleS=0.8M0.5L

[H+]= Ka [ácido][sal]

[H+]= 1.8 x 10-5 [0.5M] = 1.125 x 10 -5[0.8M]

pH = -log 1.125 X 10-5 = 4.94

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Venus has a beautiful name and is the second planet from the Sun. It’s terribly hot and its atmosphere is extremely poisonous

by VENUS by SATURN

Con la ecuación de Henderson-Hasselbach

pH = pKa + log [sal][ácido]

pKa=-log KapKa = -log ( 1.8 x 10-5) = pKa =4.74pH= 4.74 + log (0.8M)

(0.5M)pH= 4.74+0.20= 4.94

Tenemos:0.2 moles de CH3NH20.3 moles de CH3NH3Cl1 Lt de soluciónKb= 4.4 x 10-4[OH-]= Kb [base]

[sal][OH-]= 4.4 x 10-4 [0.2M] = 2.93 x 10 -4

[0.3M]pOH = -log 2.93 X 10-4 = 3.53

Recordar: pH+ pOH= 14pH= 14 - 3.53= 10.47

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Con la ecuación de Henderson-Hasselbach

pOH = pKb + log [sal][base]

pKb=-log KbpKb = -log ( 4.4 x 10-4) = pKb =3.36pOH= 3.36 + log (0.3M)

(0.2M)pOH= 3.36 + 0.176= 3.53pH = 14 – 3.53 = 10.47

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