Exciting Chemistry Challenge: Lead(II) Fluoride Solubility
Can you calculate the fluoride ion concentration in a saturated lead(II) fluoride solution?
Given data: At a particular temperature, lead(II) fluoride, PbF2, has a Ksp = 5.06×10-9.
Calculation of Fluoride Ion Concentration
Final Answer: The fluoride ion concentration in a saturated lead(II) fluoride solution is 3.2 x 10^-3 M.
Chemistry enthusiasts, get ready for an exciting challenge in calculating the fluoride ion concentration in a saturated lead(II) fluoride solution! The data provided states that at a specific temperature, the solubility product constant (Ksp) of lead(II) fluoride, PbF2, is 5.06×10-9.
To determine the fluoride ion concentration in the saturated solution of PbF2, we need to set up and solve the Ksp expression for the system. Let's break down the solution step by step:
Step 1: Define the dissolution of PbF2 as follows:PbF2(s) ⇌ Pb²+(aq) + 2F¯(aq)
Step 2: Write the Ksp expression for the system:Ksp = [Pb²+][F¯]²
Step 3: Given Ksp = 5.06×10-9, we substitute [Pb²+] = x and [F¯] = 2x (due to stoichiometry) in the equation and solve for x.5.06×10-9 = x * (2x)²
Step 4: Calculate the concentration of fluoride ions:x = 1.6 x 10^-3 M, [Pb²+] = 1.6 x 10^-3 M
[F¯] = 2x = 2 * 1.6 x 10^-3 M = 3.2 x 10^-3 M
Therefore, the fluoride ion concentration in a saturated lead(II) fluoride solution at the specified temperature is 3.2 x 10^-3 M. This calculation is based on the assumption that all dissolved PbF2 completely ionizes into Pb²+ and F¯ ions.
Challenge yourself with more chemistry problems and enhance your understanding of solubility product constants!