Respuesta :
Answer: The molar solubility of [tex]PbI_2[/tex] is [tex]1.25\times 10^{-3}mol/L[/tex]
Explanation:
Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium.
Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.
The balanced equilibrium reaction for the ionization of calcium fluoride follows:
[tex]PbI_2\rightleftharpoons Pb^{2+}+2I^-[/tex]
s 2s
The expression for solubility constant for this reaction will be:
[tex]K_{sp}=[Pb^{2+}][I^-]^2[/tex]
We are given:
[tex]K_{sp}=7.9\times 10^{-9}[/tex]
Putting values in above equation, we get:
[tex]7.9\times 10^{-9}=(s)\times (2s)^2\\\\7.9\times 10^{-9}=4s^3\\\\s=1.25\times 10^{-3}mol/L[/tex]
Hence, the molar solubility of [tex]PbI_2[/tex] is [tex]1.25\times 10^{-3}mol/L[/tex]
Answer:
The molar solubility of distilled water is [tex]$1.25 \times 10-3$[/tex] (Option B)
Explanation:
Given:
[tex]Pbl_2 -7.9 \times 10^-^9.[/tex]
Expression for solubility constant is,
[tex]$K_{s p}=\left[P b^{2+}\right]\left[I^{-}\right]^{2}$[/tex]
The given equation is,
[tex]$K_{s p}=7.9 \times 10^{-9}$[/tex]
Equation is to calculated as,
[tex]$7.9 \times 10^{-9}=(s) \times(2 s)^{2}$[/tex]
[tex]$7.9 \times 10^{-9}=4 s^{3}$[/tex]
[tex]$s=1.25 \times 10^{-3} \mathrm{~mol} / L$[/tex]
The solubility product in distilled water is [tex]$s=1.25 \times 10^{-3} \mathrm{~mol} / L$[/tex].
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