Respuesta :
when PbI₂ dissolves, it dissociates as follows
PbI₂ --> Pb²⁺ + 2I⁻
Molar solubility is the number of moles of salt that can be dissolved in 1 L of solution
If molar solubility of PbI₂ is x , then molar solubility of Pb²⁺ is x and I⁻ is 2x .
ksp is solubility product constant.
ksp = [Pb²⁺][I⁻]²
ksp = [x][2x]²
ksp = 4x³
4x³ = 1.4 x 10⁻⁸
x³ = 0.35 x 10⁻⁸
x = 1.51 x 10⁻³ M
since molar solubility of I⁻ is 2x , then molar solubility of I⁻ is 3.03 x 10⁻³ M
PbI₂ --> Pb²⁺ + 2I⁻
Molar solubility is the number of moles of salt that can be dissolved in 1 L of solution
If molar solubility of PbI₂ is x , then molar solubility of Pb²⁺ is x and I⁻ is 2x .
ksp is solubility product constant.
ksp = [Pb²⁺][I⁻]²
ksp = [x][2x]²
ksp = 4x³
4x³ = 1.4 x 10⁻⁸
x³ = 0.35 x 10⁻⁸
x = 1.51 x 10⁻³ M
since molar solubility of I⁻ is 2x , then molar solubility of I⁻ is 3.03 x 10⁻³ M
Answer:
3 * 10^-3
Explanation:
The concentration of iodide ions in a saturated solution of lead (ii) iodide is 3 * 10^-3
