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Andrei [34K]
3 years ago
8

El pH de una solución acuosa 10^-14 M de ácido acético, a 25°C, es igual a, teniendo en cuenta que Ke = 1.76x10^-5

Chemistry
2 answers:
frez [133]3 years ago
5 0

Answer:

pH=14

Explanation:

Hola!

En este caso, consideramos que la disociación de ácido acético ocurre:

CH_3COOH\rightarrow CH_3COO^-+H^+

Así, mediante la solución del equilibrio ácido, podemos calcular la concentración de iones hidronio que posteriormente sirven para calcular el pH de la solución, por tal razón, debemos calcular el equilibrio dada la constante de equilibrio y por medio de la ley de acción de masas en términos del cambio x como cualquier problema de equilibrio:

Ke=\frac{CH_3COO^-][H^+]}{[CH_3COOH]}\\\\1.76x10^{-5}=\frac{x*x}{1x10^{-14}M-x}

Resolviendo para x, tenemos x=0.999x10^{-14}

Así, la concentración de hidrógeno es igual a x, por lo que el pH:

pH=-log([H^+])=-log(0.999x10^{-14})\\\\pH=14

Dicho valor tiene sentido desde que la concentración de hidrógeno es casi despreciable, por lo que se puede asumir que tiende a ser básica.

Saludos!

ycow [4]3 years ago
5 0

Answer:

pH = 4.75

Explanation:

English Translation

The pH of a 10 ^ -14 M aqueous solution of acetic acid, at 25 ° C, is equal to, taking into account that Ka = 1.76x10 ^ -5

Solution

For Acetic acid to ionized partially, we have

CH₃COOH ⇌ CH₃COO⁻ + H⁺

If x is the amount that ionizes,

CH₃COOH ⇌ CH₃COO⁻ + H⁺

(10⁻¹⁴ - x) ⇌ (x) | (x)

Ka = [CH₃COO⁻] [H⁺]/[CH₃COOH]

(1.76 × 10⁻⁵) = (x)(x)/(10⁻¹⁴ - x)

x² = (10⁻¹⁴ - x) × (1.76 × 10⁻⁵)

x² + 0.0000176x - (1.76 × 10⁻¹⁹) = 0

Solving the quadratic equation

x = (1.00 × 10⁻¹⁴) or -(1.76 × 10⁻⁵)

Since concentration cannot be negative,

x = (1.00 × 10⁻¹⁴)

But from the Henderson - Hasselbalch equation, the pH of a weak acid in the presence of its conjugate base is given as

pH = pKa + log {[conjugate base]/[weak acid]}

[Conjugate base] = [CH₃COO⁻] = 10⁻¹⁴

[Weak acid] = [CH₃COOH] = 10⁻¹⁴

log {[conjugate base]/[weak acid]} = log 1 = 0

pH = pKa

pH = pKa = - log (1.76 × 10⁻⁵) = 4.75

Hope this Helps!!!

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