Question

What is the equilibrium Ba²⁺ concentration when 2.96 L of a 0.134 M barium acetate solution are mixed with 3.11 L of a 0.495 M potassium fluoride solution?

[Ba²⁺] = M

Answer 1

moles of Ba+2 = 2.96 x 0.134 = 0.3966

moles of KF = 3.11 x 0.495 = 1.539

Ba+2   + 2 F-    ----------------> BaF2

1            2                                 1

0.3966    1.539

here limiting reagent is Ba+2.

remaing F- = 1.539 - ( 2 x 0.3966 ) = 0.7458

concentration of F- = 0.7458 / 2.96 + 3.11 = 0.1229 M

Ksp = [Ba+2][F-]^2

1.84×10^–7 = [Ba+2] (0.1229)^2

[Ba+2] = 1.22 x 10^-5 M