Question

What is the equilibrium Ba²⁺ concentration when 1.66 L of a 0.138 M barium sulfide solution are mixed with 1.63 L of a 0.253 M potassium carbonate solution?

[Ba²⁺] = M

Answer 1

concentration of BaS = 1.66 x 0.138 / 1.66 + 1.63

                                   = 0.0696 M

concentration of K2CO3 = 1.63 x 0.253 / 1.66 + 1.63

                                       = 0.125 M

BaS (aq) + K2CO3 (aq)   -------------> BaCO3 (s) + K2S (aq)

   1                   1

0.0696          0.125

here BaS is limiting reagent .

remaining CO32- = 0.125 - 0.0696 = 0.0554 M

Ksp = [Ba2+][CO32-]

8.1 x 10^-9 = [Ba2+] (0.0554)

[Ba2+] = 1.46 x 10^-7 M