Question

Solid barium sulfate dissolves into its respective ions at 25°C. Suppose that in a particular solution, [Ba2+ ] = 1.76 x 10-3 M. Find the value of Ksp.

Answer 1

BaSO₄(s) ⇄ Ba²⁺(aq) + SO₄²⁻(aq)

Ksp=[Ba²⁺][SO₄²⁻]

[Ba²⁺]=[SO₄²⁻]

Ksp=[Ba²⁺]²

Ksp=(1.76*10⁻³)² =3.0976×10⁻⁶ ≈3.1×10⁻⁶

Answer 2

The value of solubility product of barium sulfate is
`3.09* 10^(-6)`.

Step-by-step explanation:

`BaSO_4\rightarrow Ba^(2+)+SO_4^(2-)`

S S

1 mole of barium sulfate dissociates into 1 mole of barium ion and one mole sulfate ion.

So,
`[Ba^(2+)]=[SO_4^(2-)]=S`

`[Ba^(2+)]=S=1.76* 10^(-3) M`

The solubility product the barium sulfate will be given by:

`K_(sp)=S* S=S^2`

`K_(sp)= (1.76* 10^(-3))^2=3.09* 10^(-6)`

The value of solubility product of barium sulfate is
`3.09* 10^(-6)`.