To determine the solubility of [tex]\(PbF_2\)[/tex] in 0.15 M [tex]\(NaF\)[/tex] at [tex]\(25^{\circ}C\)[/tex], we need to consider the dissolution and ion interaction of [tex]\(PbF_2\)[/tex] in solution. We'll go through the process step-by-step:
1. Expression for solubility product (Ksp) of [tex]\(PbF_2\)[/tex]:
The compound [tex]\(PbF_2\)[/tex] dissolves in water according to the following equation:
[tex]\[
PbF_2 (s) \rightleftharpoons Pb^{2+} (aq) + 2F^- (aq)
\][/tex]
The solubility product constant, [tex]\(Ksp\)[/tex], for this equilibrium at [tex]\(25^{\circ}C\)[/tex] is [tex]\(3.3 \times 10^{-8}\)[/tex].
2. Effect of [tex]\(NaF\)[/tex] on the equilibrium:
The concentration of [tex]\(NaF\)[/tex] is given as 0.15 M. When [tex]\(NaF\)[/tex] dissociates, it produces [tex]\(Na^+\)[/tex] and [tex]\(F^-\)[/tex] ions:
[tex]\[
NaF (s) \rightleftharpoons Na^+ (aq) + F^- (aq)
\][/tex]
Therefore, 0.15 M [tex]\(NaF\)[/tex] gives a [tex]\(F^-\)[/tex] ion concentration of 0.15 M. Since there are two [tex]\(F^-\)[/tex] ions produced by the solubility of one mole of [tex]\(PbF_2\)[/tex], the total [tex]\(F^-\)[/tex] concentration in the solution initially will be:
[tex]\[
[F^-] = 0.15 \times 2 = 0.30 \, \text{M}
\][/tex]
3. Establishing the solubility equilibrium:
Let [tex]\(s\)[/tex] be the solubility of [tex]\(PbF_2\)[/tex] in moles per liter (M). At equilibrium, the concentrations will be:
[tex]\[
[Pb^{2+}] = s
\][/tex]
[tex]\[
[F^-] \approx 0.30 \, \text{M} \, (\text{considering } s \ll 0.30)
\][/tex]
4. Setting up the [tex]\(Ksp\)[/tex] expression:
Using the [tex]\(Ksp\)[/tex] for [tex]\(PbF_2\)[/tex]:
[tex]\[
Ksp = [Pb^{2+}] [F^-]^2
\][/tex]
Substituting the concentrations at equilibrium:
[tex]\[
3.3 \times 10^{-8} = s \times (0.30)^2
\][/tex]
5. Solving for solubility, [tex]\(s\)[/tex]:
[tex]\[
3.3 \times 10^{-8} = s \times 0.09
\][/tex]
[tex]\[
s = \frac{3.3 \times 10^{-8}}{0.09}
\][/tex]
[tex]\[
s \approx 3.6667 \times 10^{-7} \, \text{M}
\][/tex]
Hence, the solubility of [tex]\(PbF_2\)[/tex] in 0.15 M [tex]\(NaF\)[/tex] at [tex]\(25^{\circ}C\)[/tex] is approximately [tex]\(3.67 \times 10^{-7}\)[/tex] moles per liter.
