Question

Among the solubility rules is the statement that all chlorides are soluble except $H{g}_{2}C{l}_2,AgCl,PbC{l}_2$ and $CuCl$.

• A. 1.$K=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$=1, 2. $K=1/\left[P{b}^{2+}\right][C{l}^{-}{]}^2$ =1
• B. 1.$K=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$=1, 2. $K=1/\left[P{b}^{2+}\right][C{l}^{-}{]}^2$ >1
• C. 1.$K=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$>1, 2. $K=1/\left[P{b}^{2+}\right][C{l}^{-}{]}^2$ <1
• D. 1.$K=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$<1, 2. $K=1/\left[P{b}^{2+}\right][C{l}^{-}{]}^2$ >1

Answer 1

The correct option is D 1.$K=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$<1, 2. $K=1/\left[P{b}^{2+}\right][C{l}^{-}{]}^2$ >1

1. $K=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$ is less than 1 so
AgCl is insoluble thus the concentration of ions are much less than 1 M
2. $K=1/\left[P{b}^{2+}\right][C{l}^{-}{]}^2$ is greater than one because $PbC{l}_2$ is insoluble and formation of the solid will reduce the concentration of ions to a low level.