It condenses very long strings of numbers while retaining the general accuracy of the figure.
Answer is: the equilibrium concentrations fluorine anion are 0.004 M and lead cation are 0.002 M.<span>
Chemical reaction: PbF</span>₂(aq) → Pb²⁺(aq) + 2F⁻(aq).<span>
Ksp = 3,2·10</span>⁻⁸.
[Pb²⁺] = x.
[F⁻] = 2[Pb²⁺] = 2x<span>
Ksp = [Pb²</span>⁺] ·
[F⁻]².
Ksp = x · 4x².
3,2·10⁻⁸ = 4x³.
x = ∛3,2·10⁻⁸ ÷ 4.
x = [Pb²⁺] = 0,002M = 2·10⁻³ M.
[F⁻] = 2 · 0,002M = 0,004 M = 4·10⁻³ M.
I don't know terribly much about radioactive decay, but yes, it WILL decay. If it's half life is 25 days, it will be completely consumed in 50 days. By 100 days, it should be long gone. As far as I know, the reason for this is (besides the simple math which is self-explanatory) the Thorium has so many protons/neutrons, it's unstable and thus undergoes radioactive decay as it cannot maintain stable form.
I hope this helps! :)
that were all round and smooth
