What are the equilibrium concentrations of pb2+ and f– in a saturated solution of lead(ii) fluoride if the ksp of pbf2 is 3.20×1

Question

Karolina [17]

3 years ago

11

What are the equilibrium concentrations of pb2+ and f– in a saturated solution of lead(ii) fluoride if the ksp of pbf2 is 3.20×1

0–8?

Chemistry

2 answers:

erica [24] · Answer 1 · 2022-05-14T07:29:35+03:00

Answer is: the equilibrium concentrations fluorine anion are 0.004 M and lead cation are 0.002 M.
Chemical reaction: PbF₂(aq) → Pb²⁺(aq) + 2F⁻(aq).
Ksp = 3,2·10⁻⁸.
[Pb²⁺] = x.
[F⁻] = 2[Pb²⁺] = 2x
Ksp = [Pb²⁺] · [F⁻]².
Ksp = x · 4x².
3,2·10⁻⁸ = 4x³.
x = ∛3,2·10⁻⁸ ÷ 4.
x = [Pb²⁺] = 0,002M = 2·10⁻³ M.
[F⁻] = 2 · 0,002M = 0,004 M = 4·10⁻³ M.

murzikaleks [220] · Answer 2 · 2022-05-14T09:13:45+03:00

The equilibrium concentrations of ${\text{P}}{{\text{b}}^{2+}}$ and ${{\text{F}}^-}$ in a saturated solution of ${\text{Pb}}{{\text{F}}_{\text{2}}}$ are $\boxed{{\text{0}}{\text{.002 M}}}$ and $\boxed{{\text{0}}{\text{.004 M}}}$ respectively.

Further explanation:

The equilibrium constant between the compound and its ion, when dissolved in solution, is known as solubility product constant. It is denoted by ${{\text{K}}_{{\text{sp}}}}$ . The solubility product constant is used to calculate the product of the concentration of ions at equilibrium. Higher the solubility product constant more will be the solubility of the compound.

The general reaction is as follows:

${\text{AB}}\left({aq}\right)\to{{\text{A}}^+}\left({aq}\right)+{{\text{B}}^-}\left( {aq}\right)$

The expression to calculate the solubility product for the general reaction is as follows:

${{\text{K}}_{{\text{sp}}}}=\left[{{{\text{A}}^+}}\right]\left[{{{\text{B}}^-}}\right]$

Here,

${{\text{K}}_{{\text{sp}}}}$ is the solubility product constant.

$\left[{{{\text{A}}^+}}\right]$ is the concentration of ${{\text{A}}^+}$ ions.

$\left[{{{\text{B}}^-}}\right]$ is the concentration of ${{\text{B}}^-}$ ions.

The dissociation of ${\text{Pb}}{{\text{F}}_2}$ occurs as follows:

${\text{Pb}}{{\text{F}}_2}\left({aq}\right)\rightleftharpoons{\text{P}}{{\text{b}}^{2+}}\left({aq}\right)+2{{\text{F}}^-}\left({aq}\right)$

Consider the molar solubility of ${\text{Pb}}{{\text{F}}_2}$ to be S. Therefore, after dissociation, the concentration of ${\text{P}}{{\text{b}}^{2+}}$ and ${{\text{F}}^-}$ are S and 2S respectively.

The formula to calculate the molar solubility of ${\text{Pb}}{{\text{F}}_2}$ is as follows:

${{\text{K}}_{{\text{sp}}}}=\left[{{\text{P}}{{\text{b}}^{{\text{2+}}}}}\right]{\left[{{{\text{F}}^-}}\right]^2}$ …… (1)

Substitute S for $\left[{{\text{P}}{{\text{b}}^{{\text{2+}}}}}\right]$ , 2S for $\left[{{{\text{F}}^-}}\right]$ and ${\text{3}}{\text{.20}}\times{10^{-8}}$ for ${{\text{K}}_{{\text{sp}}}}$ in equation (1).

$\begin{aligned}{\text{3}}{\text{.20}}\times {10^{ - 8}}&=\left[{\text{S}}\right]{\left[ {{\text{2S}}}\right]^2}\hfill\\{\text{3}}{\text{.20}}\times{10^{-8}}&=\left({\text{S}}\right)\left( {4{{\text{S}}^2}}\right)\hfill\\4{{\text{S}}^3}&={\text{3}}{\text{.20}}\times{10^{-8}}\hfill\\\end{aligned}$

Simplifying the above equation for S,

$\begin{aligned}4{{\text{S}}^3}&={\text{3}}{\text{.20}}\times{10^{-8}}\hfill\\{{\text{S}}^3}&=\frac{{{\text{3}}{\text{.20}}\times{{10}^{ - 8}}}}{4}\hfill\\{\text{S}}&=\sqrt[3]{{\frac{{{\text{3}}{\text{.20}}\times{{10}^{-8}}}}{4}}} \hfill\\\end{aligned}$