Get the answers you need at Westonci.ca, where our expert community is dedicated to providing you with accurate information. Connect with professionals on our platform to receive accurate answers to your questions quickly and efficiently. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

A chemist measures the solubility of lead(ii) bromide in water to be 2. 96 g/l. calculate the ksp value for the solid. the ksp value is ___________. group of answer choices

Sagot :

The solubility product, ksp of PbBr₂ is 2.102 × 10⁻⁶

What is solubility?

The solubility of a solute is defined as the maximum amount of that solute that can be dissolved in a known quantity of solvent at a given temperature.

What is a solubility product?

Some salts are sparingly soluble in a solvent. For them, we calculate the solubility product.

It is an equilibrium constant that defines the relationship between a solid and its respective ions in an aqueous solution in equilibrium.

The greater the solubility product, the greater the solubility and vice versa.

Here, the solubility of PbBr₂ = 2.96 g/l

Molar solubility of  PbBr₂ = [tex]\frac{2.96}{molar mass of PbBr_2}[/tex] = 2.96/367 = 8.07 × 10⁻³

At equilibrium,

[tex]PbBr_2\rightarrow Pb^{2+} + 2Br^-[/tex]

1 mole of PbBr2 dissociates into 1 mole of Pb²⁺ ions and 2 moles of Br⁻

Let the molar concentration of Pb²⁺ be x, then the molar concentration of Br⁻ is 2x

Ksp = x.(2x)²

       = 4x³

Substitute, x = 8.07 × 10⁻³

Ksp = 4 (8.07 × 10⁻³)³

      = 2.102 × 10⁻⁶

Thus, The ksp of PbBr₂ is 2.102 × 10⁻⁶

Learn more about solubility product:

https://brainly.com/question/1419865

#SPJ4