Un estudiante preparo 200 ml de solución de acetato de potasio (CH3COOH ; Masa molar = 98 g/mol ; Ka CH3COOH = 1,8 x 10-5), disolviendo 3,5 g de ácido, la que denomino "Solución A". Posteriormente preparo otra solución de menor concentración del mismo soluto que denomino "solución B" y para ello tomo un volumen de 5,5 ml de la solución A y agrego agua hasta 70 ml de solución. Señale el pH de solución A y B y, la concentración molar de CH3COO- en la solución b.

Answers

Answer 1

Answer:

mirar respuesta abajo

Explanation:

Muy bien. Antes de responder lo que piden en el problema, vamos a calcular la concentración inicial de la solución A:

1) Concentración de la solución A:

En este caso, se sabe que se usaron 3,5 g de la sal, se puede calcular los moles usando el peso molecular y la masa:

n = m/PM  (1)

Aplicando tenemos:

n = 3,5 / 98 = 0,0357 moles

Conociendo los moles, podemos calcular la concentración:

M = n/V (2)

Aplicando la formula tenemos:

M = 0.0357 / 0.200 = 0.1785 M

Esta es la concentración del ácido etiquetado como "Solución A".

Ahora podemos ver la concentración de la solución B, para luego calcular las concentraciones molares de los iones en solución y sus respectivos pH.

2) Concentración del ácido en la solución B:

Con la concentración de "A", se puede determinar la concentración de la solución B. Aqui podemos esperar que sea un valor mas bajo, puesto que es una dilución la que estamos haciendo. Por lo tanto.

Si se toma 5.5 mL de la solución A, entonces:

n = 0.1785 * 0.0055 = 0.00098 moles

Con esto, se calcula la nueva concentración:

M = 0.00098 / 0.070 = 0.014 M

Esta es la concentración de la solución B. Ahora para calcular pH y concentraciones de los iones en equilibrio, hay que plantear la reacción acido base en equilibrio. Como es el mismo compuesto, usaremos la misma ecuación.

3) pH de las soluciones A y B:

Planteamos la reacción de equilibrio:

CH₃COO⁻ + H₂O <------> CH₃COOH + OH⁻     Kb

Calculando el Kb, sería asi:

Kb = Kw/Ka

Kb = 1x10⁻¹⁴ / 1.8x10⁻⁵ = 5.56x10⁻¹⁰

Ahora reescribimos la ecuación y hacemos una tabla de equilibrio:

       CH₃COO⁻ + H₂O <------> CH₃COOH + OH⁻     Kb = 5.56x10⁻¹⁰

i)        0.1785                                    x              x

eq)    0.1785-x                                  x              x

Kb = [OH⁻] [CH₃COOH] / [CH₃COO⁻]

Reemplazando nos queda:

5.56 * 10⁻¹⁰ = x² / (0.1785-x)      

Y como Kb es muy pequeño, se asume que el valor de x será también pequeño, asi que podemos redondear la resta a simplemente 0.1785, quedando tan solo:

5.56 * 10⁻¹⁰ = x² / (0.1785)

x² = 5.56*10⁻¹⁰ * 0.1785

x = √9.9246*10⁻¹¹

x = 9.96*10⁻⁶ M

Esta es la concentración de [OH⁻] y [CH₃COO⁻] en la solución A.

Aplicando lo mismo para la solución B (Cambiando solo el dato de concentración) nos queda:

x = √5.56*10⁻¹⁰ * 0.014

x = 2.79x10⁻⁶ M = [OH⁻] = [CH₃COO⁻]

Finalmente, para calcular pH, se calcula primero el pOH y luego el pH:

pH = 14 - pOH

pOH = -log[OH⁻]

Para la solución A:

pOH = -log(9.96*10⁻⁶) = 5

pH = 14 - 5

pH = 9

En el caso de la solución B:

pOH = -log(2.79*10⁻⁶) = 5.55

pH = 14 - 5.55

pH = 8.45

Espero te ayude


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