Answer
is: molar solubility of barium carbonate is 4.47·10⁻⁵ M.
Chemical
reaction:BaCO₃(s) →
Ba²⁺(aq) + CO₃²⁻(aq).
Ksp
= 2.0·10⁻⁹.
s(BaCO₃) = ?M.
s(BaCO₃) = s(Ba²⁺)
=s(CO₃²⁻) = x.
Ksp
= s(Ba²⁺) · s(CO₃²⁻).
Ksp
= x · x.
2.0·10⁻⁹ =
x².
x
= √2.0·10⁻⁹ = 4.47·10⁻⁵
M.
