If a solution of lead iodine has [i-] = 1.1x10-4 m, what is the lead (ii ion concentration at equilibrium?

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

5.0 * 10^ -5 m  

10.063g / 10.00 ml = 1.0063 g/ml which rounds off to 4 sig figs as: 1.006 g/ml