Question

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 concentration of $ Ag^+$ and $Pb^{2+}$ be at equilibrium?
($K_{sp}$ for $ AgCl = 1.8 \times 10^{-10}, K_{sp}$ for $Pb Cl_2 = 1.7 \times 10^{-5})$

• A. $ [Ag^+] = 1.8 \times 10^{-11} M;$
  $ [pb^{2+} ] = 1.7 \times 10^{-4} M$
• B. $ [Ag^+] = 1.8 \times 10^{-7} M;$
  $ [pb^{2+} ] = 1.7 \times 10^{-6} M$
• C. $ [Ag^+] = 1.8 \times 10^{-11} M;$
  $ [pb^{2+} ] = 1.7 \times 10^{-5} M$
• D. $ [Ag^+] = 1.8 \times 10^{-9} M;$
  $ [pb^{2+} ] = 1.7 \times 10^{-3} M$

Answer 1

Answer: (D)

${\left[ Ag ^{+}\right]\left[ Cl ^{-}\right]=1 \cdot 8 \times 10^{-10} }$
${\left[ Ag ^{+}\right]=\frac{1 \cdot 8 \times 10^{-10}}{0 \cdot 1}=1 \cdot 8 \times 10^{-9} M }$
${\left[ Pb ^{+2}\right]\left[ Cl ^{-}\right]^2=1 \cdot 7 \times 10^{-5} }$
${\left[ pb ^{2+}\right]=\frac{1 \cdot 7 \times 10^{-5}}{0 \cdot 1 \times 0 \cdot 1} 1 \cdot 7 \times 10^{-3} M }$