The equilibrium concentrations of
and
in a saturated solution of
are
and
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
. 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:

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]](https://tex.z-dn.net/?f=%7B%7B%5Ctext%7BK%7D%7D_%7B%7B%5Ctext%7Bsp%7D%7D%7D%7D%3D%5Cleft%5B%7B%7B%7B%5Ctext%7BA%7D%7D%5E%2B%7D%7D%5Cright%5D%5Cleft%5B%7B%7B%7B%5Ctext%7BB%7D%7D%5E-%7D%7D%5Cright%5D)
Here,
is the solubility product constant.
is the concentration of
ions.
is the concentration of
ions.
The dissociation of
occurs as follows:
Consider the molar solubility of
to be S. Therefore, after dissociation, the concentration of
and
are S and 2S respectively.
The formula to calculate the molar solubility of
is as follows:
…… (1)
Substitute S for
, 2S for
and
for
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}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%7B%5Ctext%7B3%7D%7D%7B%5Ctext%7B.20%7D%7D%5Ctimes%20%7B10%5E%7B%20-%208%7D%7D%26%3D%5Cleft%5B%7B%5Ctext%7BS%7D%7D%5Cright%5D%7B%5Cleft%5B%20%7B%7B%5Ctext%7B2S%7D%7D%7D%5Cright%5D%5E2%7D%5Chfill%5C%5C%7B%5Ctext%7B3%7D%7D%7B%5Ctext%7B.20%7D%7D%5Ctimes%7B10%5E%7B-8%7D%7D%26%3D%5Cleft%28%7B%5Ctext%7BS%7D%7D%5Cright%29%5Cleft%28%20%7B4%7B%7B%5Ctext%7BS%7D%7D%5E2%7D%7D%5Cright%29%5Chfill%5C%5C4%7B%7B%5Ctext%7BS%7D%7D%5E3%7D%26%3D%7B%5Ctext%7B3%7D%7D%7B%5Ctext%7B.20%7D%7D%5Ctimes%7B10%5E%7B-8%7D%7D%5Chfill%5C%5C%5Cend%7Baligned%7D)
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}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D4%7B%7B%5Ctext%7BS%7D%7D%5E3%7D%26%3D%7B%5Ctext%7B3%7D%7D%7B%5Ctext%7B.20%7D%7D%5Ctimes%7B10%5E%7B-8%7D%7D%5Chfill%5C%5C%7B%7B%5Ctext%7BS%7D%7D%5E3%7D%26%3D%5Cfrac%7B%7B%7B%5Ctext%7B3%7D%7D%7B%5Ctext%7B.20%7D%7D%5Ctimes%7B%7B10%7D%5E%7B%20-%208%7D%7D%7D%7D%7B4%7D%5Chfill%5C%5C%7B%5Ctext%7BS%7D%7D%26%3D%5Csqrt%5B3%5D%7B%7B%5Cfrac%7B%7B%7B%5Ctext%7B3%7D%7D%7B%5Ctext%7B.20%7D%7D%5Ctimes%7B%7B10%7D%5E%7B-8%7D%7D%7D%7D%7B4%7D%7D%7D%20%5Chfill%5C%5C%5Cend%7Baligned%7D)
Solving for S,

Therefore the equilibrium concentration of
is 0.002 M.
The equilibrium concentration of
is calculated as follows:
![\begin{aligned}\left[ {{{\text{F}}^-}}\right]&=2\left({{\text{0}}{\text{.002 M}}}\right)\\&=0.00{\text{4 M}}\\\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Cleft%5B%20%7B%7B%7B%5Ctext%7BF%7D%7D%5E-%7D%7D%5Cright%5D%26%3D2%5Cleft%28%7B%7B%5Ctext%7B0%7D%7D%7B%5Ctext%7B.002%20M%7D%7D%7D%5Cright%29%5C%5C%26%3D0.00%7B%5Ctext%7B4%20M%7D%7D%5C%5C%5Cend%7Baligned%7D)
Therefore the equilibrium concentration of
is 0.004 M.
Learn more:
1. Sort the solubility of gas will increase or decrease: <u>brainly.com/question/2802008</u>.
2. What is the pressure of the gas?: <u>brainly.com/question/6340739</u>.
Answer details:
Grade: School School
Subject: Chemistry
Chapter: Chemical equilibrium
Keywords: Molar solubility, saturated solution, PbF2, Pb2+, F-, S, 2S, 0.002 M, 0.004 M, equilibrium concentration, Ksp, 3.2*10^-8.