Equilibrium - Result Question 80
####84. In qualitative analysis, the metals of Group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains $Ag^{+}$and $Pb^{2+}$ at a concentration of $0.10 M$. Aqueous $HCl$ is added to this solution until the $Cl^{-}$concentration is $0.10 M$. What will the concentrations of $Ag^{+}$and $Pb^{2+}$ be at equilibrium?
$(K _{s p}.$ for $AgCl=1.8 \times 10^{-10}$,
$K _{s p}$ for $PbCl_2=1.7 \times 10^{-5}$ )
[2011M]
(a) $[Ag^{+}]=1.8 \times 10^{-7} M ;[Pb^{2+}]=1.7 \times 10^{-6} M$
(b) $[Ag^{+}]=1.8 \times 10^{-11} M ;[Pb^{2+}]=8.5 \times 10^{-5} M$
(c) $[Ag^{+}]=1.8 \times 10^{-9} M ;[Pb^{2+}]=1.7 \times 10^{-3} M$
(d) $[Ag^{+}]=1.8 \times 10^{-11} M ;[Pb^{2+}]=8.5 \times 10^{-4} M$
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Solution:
- (c) $K _{s p}=[Ag^{+}][Cl^{-}]$
$1.8 \times 10^{-10}=[Ag^{+}][0.1]$
$[Ag^{+}]=1.8 \times 10^{-9} M$
$K _{s p}=[Pb^{+2}][Cl^{-}]^{2}$
$1.7 \times 10^{-5}=[Pb^{+2}][0.1]^{2}$
$[Pb^{+2}]=1.7 \times 10^{-3} M$
