Question

Calculate the molar solubility at 25C of BaSO4 in seawater in which the concentration of sulfate ion is 2.8g/L. Ksp=BaSO4=9.1X10-11

Answer 1

Molar mass of BaSO4 = 233.43 g/mol

solubility in g/L = 2.8 g/L (given)

Solubility in mol/L (c) = 2.8/ 233.43 = 0.0119 M/L

Solubility product for BaSO4:

                                BaSO4    <====>   Ba²⁺ +     SO4^2-

Solubility:                    s                           s                  s

Ksp = [ Ba2][SO4^2- ]

Ksp = [s][s]

Ksp = [s]²       ----------------------(1)

As there is common ion SO4^2- present (with concentration 0.0119 M/L), the molar solubility of BaSO4 will decrease which can be given as sⁱ and at this stage concentration of [Ba²⁺] and [SO4^2-] can be given as:

                                BaSO4    <====>   Ba²⁺ +     SO4^2-
Solubility:                    sⁱ                          sⁱ                 sⁱ

[Ba2+] = sⁱ

[SO4^2-] = sⁱ + c    = c (because common ion of SO4^2-)

Now, Ksp = sⁱ X c     -----------------------(2)

from eq (1) and (2)

sⁱ = Ksp/c

sⁱ = 9.1 X 10^-11/ 0.0119

sⁱ = 7.65 X 10^-9 M/L