Question

If K_sp of a sparingly soluble salt M₂X at 25°C is 1.0 × 10⁻¹¹, the solubility of the salt in mole litre⁻¹ at this temperature will be ____________.

Options

• `root2((1.0 xx 10^-11)/4)`

• `root3((1.0 xx 10^-11)/4)`

• `root3((1.0 xx 10^-11)/2)`

• `root3(4/(1.0 xx 10^-11))`

Answer 1

If K_sp of a sparingly soluble salt M₂X at 25°C is 1.0 × 10⁻¹¹, the solubility of the salt in mole litre⁻¹ at this temperature will be `underline(root3((1.0 xx 10^-11)/4))`.

Explanation:

For \[\ce{M2X ⇌ 2M^+ + X^{2-}}\]

x = 2, y = 1

K_sp = x^x × y^y × S^x+y

= 2² × 1¹ × S²⁺¹

K_sp = 4S³

1.0 × 10⁻¹¹ = 4S³

S = `root3((1.0 xx 10^-11)/4)` mol/L