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For parts of the free-response question that require calculations, clearly show the method used and the steps involved in arriving at your answers. You must show your work to receive credit for your answer. Examples and equations may be included in your answers where appropriate.
PbI2(s)⇄Pb2+(aq)+2I−(aq)Ksp=7×10−9
The dissolution of PbI2(s) is represented above.
Write a mathematical expression that can be used to determine the value of S, the molar solubility of PbI2(s). (Do not do any numerical calculations.)

Answer :

Pstnonsonjokugo

The molar solubility of lead II iodide is 1.21 × 10^-3 M.

What is solubility product?

The term solubility product refers to the equilibrium constant that shows the extent to which a substance is dissolved in water. For the dissolution of lead II iodide we can write;

PbI2(s) ⇄ Pb^2+(aq) + 2I^-(aq)

Hence;

Ksp = [x] [2x]^2 = 4x^3

x = ∛Ksp/4

x  = ∛7 × 10^-9/4

x = 1.21 × 10^-3 M

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