Determine the molar solubility for BaCO2 by construing an ICE table, writing the solubility constant expression, and solving for molar solubility.

Question

Assume that the solution already contains 1.4x10^-4 M CO3^-2

Answer 1

There really is no need to construct an ICE table, but we will do so since the questions requests it.

BaCO₃(s) <===> Ba²⁺(aq) + CO₃^2-(aq)

......1............................0................1.4x10^-2 ..........Initial

-x..........................+x..............+x......................Change

1-x.........................x..............1.4x10^-2 + x........Equilibrium

Upon looking up the Ksp for BaCO₃, it was found to be 8.1x10^-9. You may have a different value, and if so, simply substitute your value instead.

Ksp = 8.1x10^-9 = [Ba²⁺][CO₃^2-]

8.1x10^-9 = (x)(1.4x10^-2 + x) ... assume x is very small compared to 1.4x10^-2 M and ignore it

8.1x10^-9 = (x)(1.4x10^-2)

x = [Ba²⁺] = 5.8x10^-7 M (note: this is very small so above assumption was valid)

Molar solubility of BaCO₃ under these conditions = 5.8x10^-7 M