Question

A solution is 0.0010 M in both Ag${}^{+}$ and Au${}^{+}$. Some solid NaCl is added slowly until the solid AgCl just begins to precipitate. What is the concentration of Au${}^{+}$ ions at this point?

K${}_{sp}$ for AgCl $=1.8\times {10}^{-10}$ and for AuCl $=2.0\times {10}^{-13}$.

• A. $2.0\times {10}^{-10}M$
• B. $4.5\times {10}^{-7}M$
• C. $1.8\times {10}^{-7}M$
• D. $3.0\times {10}^{-4}M$
• E. $1.1\times {10}^{-6}M$

Answer 1

The correct option is E $1.1\times {10}^{-6}M$

The solubility product expression for AgCl is $K_{SP,AgCl}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$

But $\left[A{g}^{+}\right]=0.0010M$ and $K_{SP,AgCl}=1.8\times {10}^{-10}$

Substitute values in the above expression.

$1.8\times {10}^{-10}=0.0010\times \left[C{l}^{-}\right]$

Hence, $\left[C{l}^{-}\right]=1.8\times {10}^{-7}M$

The solubility product expression for AuCl is $K_{SP,AuCl}=\left[A{u}^{+}\right]\left[C{l}^{-}\right]$

But $\left[C{l}^{-}\right]=1.8\times {10}^{-7}M$ and $K_{SP,AuCl}=2.0\times {10}^{-13}$

Substitute values in the above expression.

$2.0\times {10}^{-13}=\left[A{u}^{+}\right]\times 1.8\times {10}^{-7}$

Hence, $\left[A{u}^{+}\right]=1.1\times {10}^{-6}M$