Question

The solubility of BaCO3(s) in water at a certain temperature is 4.4 10–5 mol/L. Calculate the value of Ksp for BaCO3(s) at this temperature.

Answer 1

Answer : The value of $K_{sp}$ for $BaCO_3$ is $19.36\times 10^{-10}mole^2/L^2$.

Solution : Given,

Solubility of $BaCO_3$ in water = $4.4\times 10^{-5}mole/L$

The barium carbonate is insoluble in water, that means when we are adding water then the result is the formation of an equilibrium reaction between the dissolved ions and undissolved solid.

The equilibrium equation is,

                            $BaCO_3\rightleftharpoons Ba^{2+}+CO^{2-}_3$

Initially                   -                   0        0

At equilibrium       -                   s         s

The Solubility product will be equal to,

$K_{sp}=[Ba^{2+}][CO^{2-}_3]$

$K_{sp}=s\times s=s^2$

$[Ba^{2+}]=[CO^{2-}_3]=s=4.4\times 10^{-5}mole/L$

Now put all the given values in this expression, we get the value of solubility constant.

$K_{sp}=(4.4\times 10^{-5}mole/L)^2=19.36\times 10^{-10}mole^2/L^2$

Therefore, the value of $K_{sp}$ for $BaCO_3$ is $19.36\times 10^{-10}mole^2/L^2$.