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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Edg. 2020

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Answer:

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now explanation because that is answer

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Answers

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Have a good day!

Which electron configuration represents the electrons in an atom of sulfur in an excited state? 2 – 8 – 6
2 – 7 – 7
2 – 8 – 7
2 – 7 – 8

Answers

The relations of the quantum numbers allow to find that the correct answer for the configuration of the excited state is:

  2 - 7 - 7

The electronic configuration of the elements is the distribution of electrons and different levels and sublevels fulfilling the relationships between quantum numbers.  

The principal quantum number (n) goes from 0 to infinity The orbital quantum number (l) goes from 0 to n-1, in general it is represented by lera s, p, d, f The magnetic quantum number ([tex]m_l[/tex]) ranges from -l to l The spin quantum number ([tex]m_s[/tex]) can have two values ​​+ ½ and - ½

In the base configuration Sulfur of the periodic table is 2 - 8 - 6

This is at the lowest energy configuration, when the atom acquires energy, an electron from its shell must be promoted to the next level.

                       

That is, an electron from level n = 2 that is full is promoted to level n = 3 since in this shell there is still room for two electrons, the configuration of the excited state is:

                  2- 7 - 7

Let's examine the different alternatives:

1) 2 - 8 -6

False, this is the configuration of the base state

2) 2 - 7 - 7

True. An electron is promoted to the next level due to the excitation of the atom

3) 2 - 8 -7

False, in this configuration the atom becomes an ion

4) 2 - 7 - 8

False, there is an electron transfer between two levels, but there is an extra electron that turns that atom into an ion

In conclusion using the relationships of quantum numbers we can find that the correct answer for the excited state configuration is:

2 - 7 - 7

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