Determine el PH y el % de disociación de una solución de ácido débil, sabiendo que se disuelven 20 gramos del ácido (masa molar= 55 grs) en un volumen total de 2100 ml. Ka=1,65 x 10 -8

pH = 4.27. Porcentaje de disociación: 0.03%

Explanation:

El pH de un ácido débil, HX, se obtiene haciendo uso de su equilibrio:

HX(aq) ⇄ H⁺(aq) + X⁻(aq)

Donde la constante de equilibrio, Ka, es

Ka = 1.65x10⁻⁸ = [H⁺] [X⁻] / [HX]

Como los iones H⁺ y X⁻ vienen del mismo equilibrio podemos decir:

[H⁺] = [X⁻]

[HX] es:

20g * (1mol/55g) = 0.3636moles / 2.100L = 0.1732M

Reemplazando es Ka:

1.65x10⁻⁸ = [H⁺] [H⁺] / [0.1732M]

2.858x10⁻⁹ = [H⁺]²

5.35x10⁻⁵M = [H⁺]

pH = -log[H⁺]

El porcentaje de disociacion es [X⁻] / [HX] inicial * 100

Reemplazando

5.35x10⁻⁵M / 0.1732M * 100

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.

answermeridional ;