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 $A{g}^{+}$ and $P{b}^{2+}$ at a concentration of $0.10M$. Aqueous $HCl$ is added to this solution until the $C{l}^{-}$ concentration is $0.10M$. What will the concentration of $A{g}^{+}$ and $P{b}^{2+}$ be at equilibrium?
($K_{sp}$ for $AgCl=1.8\times 10^{-10},K_{sp}$ for $PbC{l}_2=1.7\times 10^{-5})$

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

Answer 1

Answer: (D)

$\left[A{g}^{+}\right]\left[C{l}^{-}\right]=1\cdot 8\times 10^{-10}$
$\left[A{g}^{+}\right]=\frac{1\cdot 8\times 10^{-10}}{0\cdot 1}=1\cdot 8\times 10^{-9}M$
$\left[P{b}^{+2}\right]{\left[C{l}^{-}\right]}^2=1\cdot 7\times 10^{-5}$
$\left[p{b}^{2+}\right]=\frac{1\cdot 7\times 10^{-5}}{0\cdot 1\times 0\cdot 1}1\cdot 7\times 10^{-3}M$