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
PH + pOH = 14
12.52 + pOH = 14
pOH = 14 - 12.52
pOH = 1.48
[OH⁻] = 10^ -pOH
[OH⁻] = 10 ^- 1.48
[OH⁻] = 0.033 M
Hello there.
<span>
Some elementary particles are positively or negatively
</span><span>
b. charged particles.
</span>
They are pretty much the cause and effect. The independent variable that is being changed in the experiment, so it is the cause. The dependant variable is the result of change the independant variable, so the effect.