Question

What is the solubility of PbI2 in 0.4 M KI given the solubility constant of PbI2 is 7.1 x 10-9?

Answer 1

The correct answer is 4.4 x 10^-8 M

Step-by-step explanation:

7.1 x 10^-9 = x * (0.4M)^2

(7.1 x 10^-9)/.16 = x

4.4 x 10^-8 = x

Answer 2

WE are given with the solubility constant of lead (II) iodide of 7.1 x10-9 in a 0.5 M kI. Ksp or solubility constant is equal to the product of the concentrations of each ion raised to their respective number of ions. PbI2 when dissociates results to 1 mole Pb2+ and 2 moles I-. The equation is Ksp = (x)(2*0.4)^2 = 7.1 x10-9 where x is the solubility of PbI2. Answer is 1.1 x10^-18 M.