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TiliK225 [7]
3 years ago
12

39. Consider the following molecules:

Chemistry
1 answer:
sweet [91]3 years ago
5 0
I) cc14
ii) h20
iii) i’m not sure

not fully sure but hope this helps!
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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
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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!

5 0
3 years ago
Read 2 more answers
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