Zinc fluoride (ZnF2) dissociates according to the following equation:
ZnF2(s) ⇌ Zn2+(aq) + 2F–(aq)
Ksp of this equilibrium is 3.0 × 10–2.
If the concentration of zinc ions (Zn2+) in a solution is measured to be 4.4 × 10–1 M, what is the concentration of the fluoride ions (F–)?
2.6 x 10^-1 hope dis helps
To find the concentration of fluoride ions (F–) in the solution, we can use the Ksp expression.
Ksp = [Zn2+][F–]²
Given that the concentration of zinc ions (Zn2+) is 4.4 × 10–1 M, we can substitute this value into the Ksp expression.
3.0 × 10–2 = (4.4 × 10–1)([F–])²
Rearranging the equation:
[F–]² = (3.0 × 10–2) / (4.4 × 10–1)
[F–]² = 6.82 × 10–2
Taking the square root of both sides of the equation:
[F–] ≈ √(6.82 × 10–2)
[F–] ≈ 0.261 M
Therefore, the concentration of fluoride ions (F–) in the solution is approximately 0.261 M.
To find the concentration of fluoride ions (F^-), we can use the Ksp value and the concentration of zinc ions (Zn2+) to calculate the concentration of F^- ions in the solution.
First, let's write the equilibrium expression for the dissociation of zinc fluoride:
Ksp = [Zn2+][F^-]^2
We are given the concentration of zinc ions, which is 4.4 × 10^-1 M. Let's substitute this value into the equilibrium expression:
3.0 × 10^-2 = (4.4 × 10^-1)([F^-]^2)
Now, let's rearrange the equation to solve for [F^-]:
[F^-]^2 = (3.0 × 10^-2) / (4.4 × 10^-1)
[F^-]^2 = 6.82 × 10^-2
Taking the square root of both sides of the equation:
[F^-] ≈ √(6.82 × 10^-2)
[F^-] ≈ 2.61 × 10^-1
Therefore, the concentration of fluoride ions (F^-) in the solution is approximately 2.61 × 10^-1 M.