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Chemistry · Ionic equilibrium: Weak and strong electrolytes, ionization of electrolytes

The solubility of in water at is What is its solubility in solution? Assume the

The solubility of \( P b F_{2} \) in water at \( 25^{\circ} C \) is \( \sim 10^{-3} M . \) What is its solubility in \( \mathbf{0 . 0 5} \boldsymbol{M} \boldsymbol{N} \boldsymbol{a} \boldsymbol{F} \) solution? Assume the latter to be fully ionised.

  • A. \( 1.6 \times 10^{-6} M \)
  • B. \( 1.2 \times 10^{-6} M \)
  • C. \( 1.2 \times 10^{-5} M \)
  • D. \( 1.6 \times 10^{-4} M \)

Step-by-step solution

The solubility of PbF₂ in water is ~10⁻³ M, so Ksp = [Pb²⁺][F⁻]² = (s)(2s)² = 4s³ = 4×10⁻⁹. In 0.05 M NaF, let solubility be x. Then [Pb²⁺] = x, [F⁻] ≈ 0.05 M (common ion effect). Ksp = x(0.05)² = 0.0025 x = 4×10⁻⁹ → x = 1.6×10⁻⁶ M.
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