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 v
1 answer:
The solubility product, ksp of PbBr₂ is 2.102 × 10⁻⁶
<h3>What is solubility?</h3>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.
<h3>What is a solubility product? </h3>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 = = 2.96/367 = 8.07 × 10⁻³
At equilibrium,
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:
#SPJ4
You might be interested in