Answered

Welcome to Westonci.ca, the place where your questions are answered by a community of knowledgeable contributors. Discover precise answers to your questions from a wide range of experts on our user-friendly Q&A platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

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

Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.