Question

The Ksp for BaCO3 is 5.1 x 10-9. How many grams of BaCO3 will dissolve in 1000.mL of water? THÅR O ? Value | Units Submit Previous Answers Request Answer

Answer 1

At equilibrium:

BaCO3 <----> Ba2+ + CO32-

   s s

Ksp = [Ba2+][CO32-]

5.1*10^-9=(s)*(s)

5.1*10^-9= 1(s)^2

s = 7.141*10^-5 M

Molar mass of BaCO3,

MM = 1*MM(Ba) + 1*MM(C) + 3*MM(O)

= 1*137.3 + 1*12.01 + 3*16.0

= 197.31 g/mol

Molar mass of BaCO3= 197.31 g/mol

s = 7.141*10^-5 mol/L

To covert it to g/L, multiply it by molar mass

s = 7.141*10^-5 mol/L * 197.31 g/mol

s = 1.409*10^-2 g/L

1000 mL of water is 1 L

So,

Amount dissolved in 1 L is 1.409*10^-2 g

Answer: 1.4*10^-2 g