Answer:
Yes!, it is be possible to prepare the solution of PbCl₂ at 25⁰C
Explanation:
For ionic dissociation of PbCl₂; we have
PbCl₂ ⇄ Pb²⁺(aq) + 2Cl⁻(aq)
At Temperature (T) = 25⁰C; Solubility product constant [tex](K_{sp}) = 1.7*10^{-5}[/tex]
[tex]K_{sp}= [Pb^{2+}][Cl^-]^2[/tex]
[tex]K_{sp}= [x}][2x]^2[/tex]
[tex]K_{sp}= 4x^3[/tex]
[tex]1.7*10^{-5}=4x^3[/tex]
[tex]x^3 =\frac{1.7*10^{-5}}{4}[/tex]
[tex]x^3 =4.25*10^{-6}[/tex]
[tex]x =\sqrt[3]{4.25*10^{-6}}[/tex]
x = 0.016 M
Since equilibrium concentration of PbCl₂ (i.e 0.016 M ) is greater than the given concentration of PbCl₂ (i.e 0.002 M); then it is possible to prepare a solution of PbCl₂ at 25⁰C.
