Question

The solubility of BaSO₄ in water is 2.42 × 10⁻³ g L⁻¹ at 298 K. The value of its solubility product (K_sp) will be:
(Given molar mass of BaSO₄ = 233 g mol⁻¹)

Options

•  1.08 × 10⁻¹⁴ mol² L⁻²

•  1.08 × 10⁻¹² mol² L⁻²

• 1.08 × 10⁻¹⁰ mol² L⁻²

• 1.08 × 10⁻⁸ mol² L⁻²

Answer 1

1.08 × 10⁻¹⁰ mol² L⁻²

Explanation:

\[\ce{BaSO4 ⇌ Ba^2+ + SO^{2-}_4}\]

K_sp = (s) (s)

K_sp = (s)²

= (2.42 × 10⁻³ g L⁻¹)²

= `((2.42 xx 10^-3  "g L"^-1)/(233  "g mol"^-1))^2`

= (0.01038 × 10⁻³)²

= (1.038 × 10⁻⁵)²

= 1.077 × 10⁻¹⁰

= 1.08 × 10⁻¹⁰ mol² L⁻²