Question

The solubility product of $AgCl$ is $1.8\times {10}^{-10}$. Precipitation of $AgCl$ will occur by mixing which of the following solutions when mixed in equal volumes?

• A. ${10}^{-8}$ M $A{g}^{+}$ and ${10}^{-8}$ M $C{l}^{-}$ions
• B. ${10}^{-3}$ M $A{g}^{+}$ and ${10}^{-3}$ M $C{l}^{-}$ ions
• C. ${10}^{-6}$ M $A{g}^{+}$ and ${10}^{-6}$ M $C{l}^{-}$ ions
• D. ${10}^{-10}$ M $A{g}^{+}$ and ${10}^{-10}$ M $C{l}^{-}$ ions

Answer 1

The correct option is B ${10}^{-3}$ M $A{g}^{+}$ and ${10}^{-3}$ M $C{l}^{-}$ ions

$Agcl\to A{g}^{+}+C{l}^{-}$

$K_{sp}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$

For precipition $\left[A{g}^{+}\right]\left[C{l}^{-}\right]>K_{sp}$

$\left(A\right)\left[A{g}^{+}\right]\left[C{l}^{-}\right]={10}^{-8}\times {10}^{-8}={10}^{-16}$

precipitation not occure

$\left(B\right)\left[A{g}^{+}\right]\left[C{l}^{-}\right]={10}^{-3}\times {10}^{-3}={10}^{-6}$

${10}^{-6}>K_{sp}$

precipitation occur

$\left(C\right){10}^{-6}\times {10}^{-6}={10}^{-12}$

precipitation not occure

$\left(D\right){10}^{-10}\times {10}^{-10}={10}^{-20}$

precipitation not occure