Question

What is the concentration of barium ions available in solution if you dissolve 150g of barium sulfate. (Ksp = 1.1 x 10^-10) in 750 mL of water?

Answer 1

Lets calculate the concentration of BaSO4

Molar mass of BaSO4 = 1*MM(Ba) + 1*MM(S) + 4*MM(O)

= 1*137.3 + 1*32.07 + 4*16.0

= 233.37 g/mol

mass of BaSO4 = 150 g

we have below equation to be used:

number of mol of BaSO4,

n = mass of BaSO4/molar mass of BaSO4

=(150.0 g)/(233.37 g/mol)

= 0.6428 mol

volume , V = 750 mL

= 0.75 L

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 0.6428/0.75

= 0.857 M

The salt dissolves as:

BaSO4 <----> Ba2+ + SO42-

   s s

Ksp = [Ba2+][SO42-]

1.1*10^-10=(s)*(s)

1.1*10^-10= 1(s)^2

s = 1.049*10^-5 M

Since solubility is less than concentration,

1.049*10^-5 M will be part of solution

Answer: 1.05*10^-5 M