Part 1)
when the balanced equation for this reaction is:
and by using ICE table
Mg(OH)2(s) ↔ Mg2+(aq) + 2OH-(aq)
initial 0 0
change +X +2X
Equ X 2X
When KSp expression = [Mg2+][OH-]
when we have KSp = 5.61 x 10^-11
and when we assumed [Mg2+] = X
and [ OH-] = (2X)^2
when we assume X is the value of molar solubility of Mg(OH)2
so, by substitution:
5.61 x 10^-11 = 4X^3
∴ X = 2.4 x 10^-4 M
∴ molar solubility of Mg(OH)2 = X = 2.4 x 10^-4 M
Part 2)
the molar solubility of Mg(OH)2 in 0.16 m NaOH we assumed it = X
by using the ICE table:
Mg(OH)2(s) → Mg2+(aq) + 2OH-
initial 0 0.16m
change +X +2X
equ X (0.16+2x)
when Ksp = [mg2+][OH-]^2
5.61 x 10^-11 = X * (0.16+2X)^2 by solving for X
∴ X = 1.3 x 10^-5 M
∴ the molar solubility = X = 1.3 x 10^-5 M
I’m pretty sure it’s none of the above cause I’m googling it a bunch and and it say you use the dideoxy method
Answer & Explanation:
Using the numbers in the configuration 1s² 2s² 2p^6 3s^1, we can find out it has 11 electrons (by adding 2,2,6,1), meaning the element is sodium (Na). Sodium is in the outermost column of the periodic table, meaning it has one valence electron, making it highly reactive because it can very easily give one electron in a reaction. Atoms will prefer to have a full eight valence electrons, meaning an atom with seven valence electrons is also incredibly reactive since it only needs to gain one more in a reaction to gain eight. If sodium, an element with one valence electron, reacts with an element with seven (i.e chlorine), they both will be very quick to chemically bond to form a full valence shell.
