Answers: 2
of iodide ion will result in precipitation of silver iodide.
of iodide ion will result in precipitation of lead iodide.
![[Ag^+]=2.0\times 10^{-4} M](/tpl/images/0549/9170/f3b77.png)
![[I^-]](/tpl/images/0549/9170/5a647.png)
![K_{sp}=[Ag^+][I^-]](/tpl/images/0549/9170/9b82b.png)
![[I^-]=\frac{8.3\times 10^{-17}}{2.0\times 10^{-4} M}=4.15\times 10^{-13} M](/tpl/images/0549/9170/c1577.png)
![[Pb^{2+}]=1.5\times 10^{-3} M](/tpl/images/0549/9170/35d28.png)
![K_{sp}'=[Pb^{2+}][I^-]^2](/tpl/images/0549/9170/d5497.png)
![[I^-]^2=\frac{7.9\times 10^{-9}}{1.5\times 10^{-3} M}=4.15\times 10^{-13}](/tpl/images/0549/9170/819e7.png)
![[I^-]=0.0023 M](/tpl/images/0549/9170/2822d.png)
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A solution contains 2.0 ⨯ 10−4 M Ag+(aq) and 1.5 ⨯ 10−3 M Pb2+(aq). If NaI is added, will AgI (Ksp =...
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