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

it is a good conductor

explanation: