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 0.10 M. Aqueous HCl is added to this solution until the $C{l}^{-}$ concentration is 0.10 M. 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. $\left[A{g}^{+}\right]=1.8\times {10}^{-11}M$;
  $\left[P{b}^{2+}\right]=1.7\times {10}^{-4}M$
• B. $\left[A{g}^{+}\right]=1.8\times {10}^{-7}M$;
  $\left[P{b}^{2+}\right]=1.7\times {10}^{-6}M$
• C. $\left[A{g}^{+}\right]=1.8\times {10}^{-11}M$;
  $\left[P{b}^{2+}\right]=8.5\times {10}^{-5}M$
• D. $\left[A{g}^{+}\right]=1.8\times {10}^{-9}M$;
  $\left[P{b}^{2+}\right]=1.7\times {10}^{-3}M$

Answer 1

The correct option is D $\left[A{g}^{+}\right]=1.8\times {10}^{-9}M$;

$\left[P{b}^{2+}\right]=1.7\times {10}^{-3}M$
$A{g}^{+}+C{l}^{-}⟶AgCl$

$0.1$ $0$ $0$

$x$ $0.1$ $0$

$Ksp=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$ $[A{g}^{+}{]}_{final}=x$

$\Rightarrow 1.8\times {10}^{-10}=x\times 0.1$

$\Rightarrow x=1.8\times {10}^{-9}$

$\Rightarrow \left[A{g}^{+}\right]=1.8\times {10}^{-9}M$

$P{b}^{+2}+2C{l}^{-}⟶PbC{l}_2$

$0.1$ $0$ $0$

$y$ $0.1$ $0$ $[P{b}^{+2}{]}_{final}=y$

$Ksp=\left[P{b}^{+2}\right][C{l}^{-}{]}^2$

$\Rightarrow 1.7\times {10}^{-5}=y\times [0.1{]}^2$

$\Rightarrow \left[P{b}^{+2}\right]=y=1.7\times {10}^{-3}$

$\Rightarrow \left[P{b}^{+2}\right]=1.7\times {10}^{-3}M$