Lead(II) iodide is Pb I2 (the digit 2 is a subscript to the right of the symbol of iodine, I).
Pb I2 is a solid ionic compound.
Then, it ionizes in solution as per Pb I2 → Pb (2+) + 2I (1-)
That means that there are 2 ions of I(1-) per each ion of Pb(2+).
Then, if the concentration of I (1-) is 1 * 10 ^-4m, the concentration of Pb(2+) is half 1 * 10 ^ - 4 m.
That is 0.5 * 10^-4m = 5.0 * 10^-5 m
Answer: 5.0 * 10^ -5 m
