The concentration of fluoride ion [F⁻] at equilibrium = 0.0009 M
Moles of F⁻ in 75 mL of 0.022 M NaF is first determined:
The dissociation equation of NaF is given as follows:
NaF (aq) ----> Na (aq) + F⁻ (aq)
1 mole of NaF produces 1 mole of F⁻ ions
Number of moles = Molarity * volume
Number of moles of NaF = 0.022 M * 0.075 L = 1.65 * 10⁻³ moles
Therefore, moles of F⁻ = 1.65 * 10⁻³ moles
The dissociation equation of Sr(NO₃)₂ is given as follows:
Sr(NO₃)₂ (aq) ---> Sr²⁺ (aq) + 2 NO₃⁻ (aq)
1 mole Sr(NO₃)₂ produces 1 mole Sr²⁺
moles of Sr²⁺ in 25.0 mL solution = 0.025 L * 0.160 M
moles of Sr²⁺ in 25.0 mL solution = 4.0 * 10⁻³ moles
Equation of reaction of Sr²⁺ and F⁻ is given below:
Sr²⁺ (aq) + 2 F⁻ (aq) ⇄ SrF₂ (s)
The solubility product constant, Ksp = 2.0 x 10⁻¹⁰
1 mole of Sr²⁺ reacts with 2 moles of F⁻
4.0 * 10⁻³ moles of Sr²⁺ will react 2 * 4.0 * 10⁻³ moles of F = 8.0 * 10⁻³ moles of F⁻
Therefore, F⁻ is the limiting reactant
Moles of Sr²⁺ left = 4.0 * 10⁻³ - 1.65 * 10⁻³
Moles of Sr²⁺ left = 2.35 * 10⁻³ moles
Concentration of Sr²⁺ = moles / total volume
total volume = 100 mL = 0.1 L
Concentration of Sr²⁺ at equilibrium = 2.35 * 10⁻³/0.1 = 2.35 * 10⁻² M
Assuming change in concentration at equilibrium is very small
[Sr²⁺][F⁻]² = Ksp = 2.0 x 10⁻¹⁰
[F⁻]² = Ksp/[Sr²⁺]
[F⁻]² = 2.0 x 10⁻¹⁰/2.35 * 10⁻²
[F⁻]² = 8.5 * 10⁻⁷ M
[F⁻] = 0.0009 M
Therefore, the concentration of fluoride ion [F⁻] at equilibrium = 0.0009 M
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