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Lead (ll) iodide (PbI2) has a solubility of 1.52×10 to the -3 mol/L.
1. write the dissolution reaction to PbI2 including all states.
2. Write the expression for Ksp for Pbl2.
3. What is the concentration of Pb2+ in the equilibrium solution?
4. What is the concentration of I- in the equilibrium solution?
5. Calculate the solubility product of Pbl2.

Lead ll iodide PbI2 has a solubility of 15210 to the 3 molL 1 write the dissolution reaction to PbI2 including all states 2 Write the expression for Ksp for Pbl class=

Respuesta :

Answer:

A. PbI2(s) ===> Pb2+(aq) + 2I-(aq)

B. Ksp = [Pb2+][I-]^2

C. 1.52 x 10^-3 M. It is equal to the moles/L of PbI2 that go into solution.

D. 2 x 1.52 x 10^-3 = 3.04 x 10^-3 M

E. Ksp = (1.52x10^-3)(2.31x10^-6) = 3.51 x 10^-9

Explanation:

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