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

Answer:

The value of solubility product of barium sulfate is $3.09\times 10^{-6}$.

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, $[Ba^{2+}]=[SO_4^{2-}]=S$

$[Ba^{2+}]=S=1.76\times 10^{-3} M$

The solubility product the barium sulfate will be given by:

$K_{sp}=S\times S=S^2$

$K_{sp}= (1.76\times 10^{-3})^2=3.09\times 10^{-6}$

The value of solubility product of barium sulfate is $3.09\times 10^{-6}$.