How much lead iodide, Ksp = 7.1 × 10⁻⁹, dissolves in 0.10 mol/L potassium iodide? …

For $\text{A}_x\text{B}_y$, $K_{sp} = [\text{A}]^x[\text{B}]^y = x^x y^y s^{x+y}$, where $s$ is the molar solubility in mol/L. AB: $K_{sp} = s^2$; $\text{AB}_2$: $K_{sp} = 4s^3$. With a common ion of concentration $c$, $s \approx K_{sp}/c$ for type AB. A precipitate forms when the ionic product after mixing is larger than $K_{sp}$.