Answer:
[tex]Ksp=1.07x10^{-8}[/tex]
Explanation:
Hello,
In this case, the dissociation reaction is:
[tex]PbI_2(s)\rightleftharpoons Pb^{2+}(aq)+2I^-(aq)[/tex]
For which the equilibrium expression is:
[tex]Ksp=[Pb^{2+}][I^-]^2[/tex]
Thus, since the saturated solution is 0.064g/100 mL at 20 °C we need to compute the molar solubility by using its molar mass (461.2 g/mol)
[tex]Molar solubility=\frac{0.064g}{100mL}*\frac{1000mL}{1L}*\frac{1mol}{461.2g}=1.39x10^{-3}M[/tex]
In such a way, since the mole ratio between lead (II) iodide to lead (II) and iodide ions is 1:1 and 1:2 respectively, the concentration of each ion turns out:
[tex][Pb^{2+}]=1.39x10^{-3}M[/tex]
[tex][I^-]=1.39x10^{-3}M*2=2.78x10^{-3}M[/tex]
Thereby, the solubility product results:
[tex]Ksp=(1.39x10^{-3}M)(2.78x10^{-3}M)^2\\\\Ksp=1.07x10^{-8}[/tex]
Regards.
Solubility product constant for the product of lead(II) iodide is [tex]\bold { 1.07x 10^-^8}[/tex].
The dissociation reaction for lead (II) iodide
[tex]\bold {Pb I^2 (s) \leftrightharpoons Pb^2^+ + 2I^- }[/tex]
Solubility product constant at equilibrium.
[tex]\bold {Ksp = [Pb^2^++[I^-]^2}[/tex]
The molar solubility of the substance can be calculated by using the molar mass,
[tex]\bold {s = \dfrac {0.064}{100 mL} \times 461.2 g/mol = 1.39x10^-^3}[/tex]
Molar ratio between between PbI to lead and iodide ions is 1:1 and 1:2 respectively.
Thus Ksp will be,
[tex]\bold {Ksp =(1.39x10^-^3)(2.78x10^-^3 )^2}\\\\\bold {Ksp = 1.07x 10^-^8}[/tex]
Therefore, solubility product constant for the product of lead(II) iodide is [tex]\bold { 1.07x 10^-^8}[/tex].
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