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

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Chemistry

Sammy Simmons

4 February, 07:35

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

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Louis Patel · Answer 1

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.