EQUILIBRIO QUÍMICO

PREGUNTAS

1. la Ka del ácido benzoico es 6.5×10-5 .calcule el pH de una solución 0.10 M.

HC7H5O2 + H2O H3O + C7H5O2

Ka = [disociada] / [asociada]

6.5×10-5 = [H3O] [C7H5O2] / [HC7H5O2]

6.5×10-5 = x2 / 0.10 – x

6.5×10-5 (0.10 – x) = x2

x2 + 6.5×10-5 x - 6.5×10-6 = 0

x= -b±√(b^2 )-4(a)(c)/2(a)

x = -6.5×10-5 ± √4.225×10-9 + 2.6×10-5 / 2

x = -6.5×10-5 ± √2.6×10-5 / 2

x = -6.5×10-5 ± 5.09 / 2

x = 2.51×10-3

pH = -Log [2.51×10-3]

pH = 2.6

2. calcule el pH de una solución 0.36M de CH3COONa.

CH3COONa + H2O OH- + CH3COOH

Ka = 5.5 ×10-10

Ka = [disociada] / [asociada]

5.5×10-10 = [OH] [CH3COOH] / [CH3COONa]

5.5×10-10 = x2 / 0.36 – x

5.5×10-10 (0.36 – x) = x2

x2 + 5.5×10-10 x – 1.98 ×10-10 = 0

x= -b±√(b^2 )-4(a)(c)/2(a)

x = -5.5×10-10 ± √3.025×10-19 + 7.92×10-10 / 2

x = -5.5×10-10 ± √7.92×10-10 / 2

x = -5.5×10-10 ± 2.81×10-5 / 2

x = 1.40×10-5

pOH = -Log [1.40×10-5]

pOH = 4.85

pH + pOH = 14

pH =14-pOH

pH = 9.15

3. Calcule el pH de un sistema amortiguador formado por NH3 0.15 M/NH4Cl 0.35.

El ácido NH4= 0.35M

La base NH3= 0.15M Kb= 5.9×10-10

Kw (constante del agua) = 1×10-14

Kw = Ka × Kb

Ka = 1×10-14 / 5.9×10-10

Ka = 1.69×10-5

(H+) = Ka × (Ca / Cb)

(H+) = 1.69×10-5 × (0.35 / 0.15)

(H+) = 3.94×10-5 M

pH = -log [3.94×10-5]

pH

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