Answered

Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Get accurate and detailed answers to your questions from a dedicated community of experts on our Q&A platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

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.)

Sagot :

pstnonsonjoku

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

Learn more about solubility product: https://brainly.com/question/857770

We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.