Question

The Ksp for bismuth sulphide (Bi₂S₃) is 1.08 × 10^–73. The solubility of Bi₂S₃ in mol L^–1 at 298 K is

• A. 1.0 × 10^–15
• B. 2.7 × 10^–12
• C. 3.2 × 10^–10
• D. 4.2 × 10^–8

Answer 1

The Correct Option is A

To find the solubility of bismuth sulphide (Bi₂S₃), we need to use its solubility product constant (K_sp).

The given dissociation of Bi₂S₃ in solution is:

Bi_2S_3 \rightleftharpoons 2Bi^{3+} + 3S^{2-}

Let the molar solubility of Bi₂S₃ be s mol L^–1. At equilibrium, the concentrations of ions in terms of s will be:

• Concentration of Bi³⁺ = 2s
• Concentration of S^2- = 3s

Therefore, the K_sp expression for Bi₂S₃ is:

K_{sp} = [Bi^{3+}]^2 \times [S^{2-}]^3

Substituting the concentrations in terms of s:

K_{sp} = (2s)^2 \times (3s)^3 = 4s^2 \times 27s^3 = 108s^5

We know that the K_sp for Bi₂S₃ is 1.08 × 10^–73. Setting this equal to the expression above:

108s^5 = 1.08 \times 10^{-73}

Solving for s, we get:

s^5 = \frac{1.08 \times 10^{-73}}{108}

s^5 = 1 \times 10^{-75}

Now, taking the fifth root to solve for s:

s = (1 \times 10^{-75})^{1/5} = 1 \times 10^{-15} \text{ mol L}^{-1}

Thus, the solubility of Bi₂S₃ in mol L^–1 at 298 K is 1.0 × 10^–15.