Respuesta :
Answer:
The silver iodide will precipitate first.
At concentration more than [tex]4.15\times 10^{-13} M[/tex] of iodide ion will result in precipitation of silver iodide.
At concentration more than [tex]0.0023 M[/tex] of iodide ion will result in precipitation of lead iodide.
Explanation:
Solubility of silver iodide = [tex]K_{sp}=8.3\times 10^{-17}[/tex]
[tex]AgI\rightleftharpoons Ag^++I^-[/tex]
Concentration of silver ions = [tex][Ag^+]=2.0\times 10^{-4} M[/tex]
Suppose concentration of iodide ion due NaI added in solution be = [tex][I^-][/tex]
The expression of solubility product will be given as :
[tex]K_{sp}=[Ag^+][I^-][/tex]
[tex][I^-]=\frac{8.3\times 10^{-17}}{2.0\times 10^{-4} M}=4.15\times 10^{-13} M[/tex]
At concentration more than [tex]4.15\times 10^{-13} M[/tex] of iodide ion will result in precipitation of silver iodide.
Solubility of lead iodide = [tex]K_{sp}'=7.9\times 10^{-9}[/tex]
[tex]PbI_2\rightleftharpoons Pb^{2+}+2I^-[/tex]
Concentration of silver ions = [tex][Pb^{2+}]=1.5\times 10^{-3} M[/tex]
Suppose concentration of iodide ion due NaI added in solution be = [tex][I^-][/tex]
The expression of solubility product will be given as :
[tex]K_{sp}'=[Pb^{2+}][I^-]^2[/tex]
[tex][I^-]^2=\frac{7.9\times 10^{-9}}{1.5\times 10^{-3} M}=4.15\times 10^{-13}[/tex]
[tex][I^-]=0.0023 M[/tex]
At concentration more than [tex]0.0023 M[/tex] of iodide ion will result in precipitation of lead iodide.
For silver iodide , we need concentration of iodide more than [tex]4.15\times 10^{-13} M[/tex] and for lead iodide , we need concentration of iodide more than 0.0023 M. So , this indicates that silver iodide will precipitate out first.
Silver iodide is the first precipitate. silver iodide or lead iodide precipitates at concentrations greater than 4.15× 10⁻¹³ M or 2.3 ×10⁻³ M of iodide ions.
Concentration calculation:
Calculating the value for the first question that is the AgI ppts first:
⇒ AgI
⇒ K = [Ag] [I]
⇒ 8.3 × 10⁻¹⁷= [2e⁻⁴][I]
⇒ [I] = 4.15× 10⁻¹³ Molar
Calculating the value for the second question, which is AgI to ppt, [I] > 4.15× 10⁻¹³ Molar
⇒ PbI₂
⇒ K = [Pb] [I]²
⇒ 7.9× 10⁻⁹ = [1.5 × 10⁻³] [I]²
⇒ I² = 5.26 × 10⁻⁶
⇒ I = 2.3 ×10⁻³ Molar
Find out more about the concentration here:
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