Answer:
The dissociation constant of phenol from given information is
.
Explanation:
The measured pH of the solution = 5.153

Initially c
At eq'm c-x x x
The expression of dissociation constant is given as:
![K_a=\frac{[C_6H_5O^-][H^+]}{[C_6H_5OOH]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BC_6H_5O%5E-%5D%5BH%5E%2B%5D%7D%7B%5BC_6H_5OOH%5D%7D)
Concentration of phenoxide ions and hydrogen ions are equal to x.
![pH=-\log[x]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5Bx%5D)
![5.153=-\log[x]](https://tex.z-dn.net/?f=5.153%3D-%5Clog%5Bx%5D)



The dissociation constant of phenol from given information is
.
When we have:
Zn(OH)2 → Zn2+ 2OH- with Ksp = 3 x 10 ^-16
and:
Zn2+ + 4OH- → Zn(OH)4 2- with Kf = 2 x 10^15
by mixing those equations together:
Zn(OH)2 + 2OH- → Zn(OH)4 2- with K = Kf *Ksp = 3 x 10^-16 * 2x10^15 =0.6
by using ICE table:
Zn(OH)2 + 2OH- → Zn(OH)4 2-
initial 2m 0
change -2X +X
Equ 2-2X X
when we assume that the solubility is X
and when K = [Zn(OH)4 2-] / [OH-]^2
0.6 = X / (2-2X)^2 by solving this equation for X
∴ X = 0.53 m
∴ the solubility of Zn(OH)2 = 0.53 M
That wouod be the ionosphere!
The answer would be B - homogeneous mixture