The Ksp for BaCO3 is 5.1x10^-9. how many grams of BaCO3 will dissolve in 1000ml of water

= 0.014 g of BaCO3

Explanation:

Let x = mol/L of BaCO3 that dissolve.

This will give;

x mol/L Ba2 + and x mol/L CO32-

But;

Ksp = 5.1x10^-9.

Therefore;

Ksp = 5.1 x 10^-9 = (x) (x)

Thus;

x = molar solubility

= √ (5.1 x 10^-9)

= 7.1 x 10^-5 M

Therefore;

Mass BaCO3 = 7.1 x 10^-5 M x 1 L x 197.34 g/mol

= 0.014 g