If the solubility product of PbS is 8 × 10⁻²⁸, then the solubility of PbS in pure water at 298 K is x × 10⁻¹⁶ mol L⁻¹. The value of x is __________. (Nearest Integer)
[Given : √2 = 1.41]

Solution

Chemistry ma’am · 4 steps

PbS gives one Pb2+ and one S2-, so Ksp is just s squared; take the root, then match the power of ten.

1. Set up the equilibrium. PbS(s) ⇌ Pb²⁺(aq) + S²⁻(aq). If s mol of PbS dissolves per litre, it gives s of Pb²⁺ and s of S²⁻.

   Kₛₚ = [Pb²⁺][S²⁻] = s²

2. Take the square root. s = √(8 × 10⁻²⁸) = 2√2 × 10⁻¹⁴ = 2 × 1.41 × 10⁻¹⁴ = 2.82 × 10⁻¹⁴ mol L⁻¹.
3. Match the given form. The answer must be written as x × 10⁻¹⁶. Moving the decimal two places gives 2.82 × 10⁻¹⁴ = 282 × 10⁻¹⁶.
4. So. The solubility is 282 × 10⁻¹⁶ mol L⁻¹, so x = 282.

The common slip: 3. Stopping at 2.82 × 10⁻¹⁴ and rounding to 3 forgets that the answer must be in units of 10⁻¹⁶, which makes it 282.

The idea to keep. For a 1:1 salt like AB, Kₛₚ = s² and s = √Kₛₚ. Always check which power of ten the question wants before reading off x.

Answer from the NTA final answer key. Solution by PrepWiser.