Answer:
[tex]\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}[/tex]
Less than the concentration of Pb2+(aq) in the solution in part ( a )
Explanation:
From the question:
A)
We assume that s to be Ā the solubility of PbIā.
The equation of the reaction is given as :
PbIā(s) ā PbĀ²āŗ(aq) + 2Iā»(aq); Ksp = 7 Ć 10ā»ā¹
Ā [PbĀ²āŗ] = Ā s
Then [Iā»] = 2s
[tex]K_{sp} =\text{[Pb$^{2+}$][I$^{-}$]}^{2} = s\times (2s)^{2} = 4s^{3}\\s^{3} = \dfrac{K_{sp}}{4}\\\\s =\mathbf{ \sqrt [3]{\dfrac{K_{sp}}{4}}}\\\\\text{The mathematical expressionthat can be used to determine the value of }\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}[/tex]
B)
The Concentration of PbĀ²āŗ Ā in water is calculated as :
[tex]\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}[/tex]
[tex]\mathbf{s =\sqrt [3]{\dfrac{7*10^{-9}}{4}}}[/tex]
[tex]\mathbf{s} =\sqrt[3]{1.75*10^{-9}}[/tex]
[tex]\mathbf{s} =\mathbf{1.21*10^{-3} \ mol/L }[/tex]
The Concentration of PbĀ²āŗ Ā in 1.0 molĀ·Lā»Ā¹ NaI
[tex]\mathbf{PbCl{_2}} \leftrightharpoons \ \ \ \ \ \ \ \mathbf{Pb^{2+}} \ \ \ \ \ + \ \ \ \ \ \ \ \mathbf{2 I^-}[/tex]
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā [tex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf0} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{1.0}[/tex]
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā [tex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+x} \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+2x}[/tex]
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā [tex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+x} \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{1.0+2x}[/tex]
The equilibrium constant:
[tex]K_{sp} =[Pb^{2+}}][I^-]^2 \\ \\ K_{sp} = s*(1.0*2s)^2 =7*1.0^{-9} \\ \\ s = 7*10^{-9} \ \ m/L[/tex]
It is now clear that maximum possible concentration of PbĀ²āŗ in the solution is less than that in the solution in part (A). This happens due to the Ā common ion effect. The added iodide ion forces the position of equilibrium to shift to the left, reducing the concentration of PbĀ²āŗ.
