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?
( $K_{s p}$ for AgCl $=1.8\times 10^{- 10},K_{s p}$ for $PbCl_{2}=1.7\times 10^{- 5}$ )

• A. $\left[\right.Ag^{+}\left]\right.=1.8\times 10^{- 7}M,\left[\right.Pb^{2 +}\left]\right.=1.7\times 10^{- 6}M$
• B. $\left[\right.Ag^{+}\left]\right.=1.8\times 10^{- 11}M,\left[\right.Pb^{2 +}\left]\right.=8.5\times 10^{- 5}M$
• C. $\left[\right.Ag^{+}\left]\right.=1.8\times 10^{- 9}M,\left[\right.Pb^{2 +}\left]\right.=1.7\times 10^{- 3}M$
• D. $\left[\right.Ag^{+}\left]\right.=1.8\times 10^{- 11}M,\left[\right.Pb^{2 +}\left]\right.=1.7\times 10^{- 4}M$

Answer 1

Answer: (C)

$K_{s p}\left[\right.AgCl\left]\right.=\left[\right.Ag^{+}\left]\right.\left[\right.Cl^{-}\left]\right.$
$\left[\right.Ag^{+}\left]\right.=\frac{1.8 \times 1 0^{- 10}}{1 0^{- 1}}=1.8\times 10^{- 9}M$
$\text{K}_{\text{sp}} \left[\text{PbCl}_{2} \left]=\right[ \text{Pb}^{2 +}\right] \left[\text{Cl}^{-}\right]^{2}$
$\left[\right.Pb^{2 +}\left]\right.=\frac{1.7 \times 1 0^{- 5}}{1 0^{- 1} \times 1 0^{- 1}}=1.7\times 10^{- 3}M$