The moles of NaF that must be dissolved in 1.00 liter of saturated solution is 0.1946 moles
From the information given; we know that the solubility equilibrium (ksp) value for PbF₂ = 4.0 × 10⁻⁸
Consider s to be the solubility of the product formed from the dissociation of PbF₂.
Then, the dissociation of PbF₂ can be expressed as:
[tex]\mathbf{PbF_2 \to Pb^2 + 2F^-}[/tex]
s 2s
Solubility product Ksp = [Pb²⁺] [F⁻]²
4.00 × 10⁻⁸ = s(2s)²
4.00 × 10⁻⁸ = 2s³
s³ = (4.00 × 10⁻⁸)/2
s³ = 2.00 × 10⁻⁸
[tex]\mathbf{s = \sqrt[3]{2.00 \times 10^{-8} }}[/tex]
s = 0.0027 M
As such, the value of [F⁻] = 2s = (2 × 0.0027) M
= 0.0054 M
If the needed [Pb²⁺] = 1.0 × 10⁻⁶, then, the concentration of F⁻ is determined:
Ksp = [Pb²⁺] [F⁻]²
4.00 × 10⁻⁸ = (1.0 × 10⁻⁶) × [F⁻]²
[F⁻]² = (4.00 × 10⁻⁸)/(1.0 × 10⁻⁶)
[F⁻]² = 0.04
[F⁻] = √0.04
[F⁻] = 0.2 M
Now, in 1.00 Liter of the saturated solution, the number of moles is equivalent to the molarity, as such:
If the initial moles of F⁻ = 0.0054
Then, the needed moles added = (0.2 M - 0.0054) moles
= 0.1946 moles
Therefore, we can conclude that the needed moles added is equal to the moles of NaF that must be dissolved in 1.00 liter of saturated solution is 0.1946 moles
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