Question

Explain the following phenomenon: You have a test tube with an aqueous solution of silver nitrate as shown in test tube 1 below. A few drops of aqueous sodium chromate solution was added with the end result shown in test tube 2. A few drops of aqueous sodium chloride solution was then added with the end result shown in test tube 3

Use the${\mathit{K}}_{\mathbf{s}\mathbf{p}}$values in the book to support your explanation and include the balanced equations. Also list the ions that are present in solution in each test tube.

Short answer

Solubility of silver chromate is higher than the solubility of silver chloride.

Step-by-step solution

Definition of solubility product constant

The equilibrium constant for the dissolution of a solid substance into an aqueous solution is defined as solubility product constant

Calculations

When sodium chromate is added to silver nitrate solution $A{g}_{2}Cr{O}_4$precipitate is formed. This precipitate partially dissolves in water leaving $A{g}^{+}$and $Cr{O_4}^{2-}$ions.

When sodium chloride is added, $\mathrm{AgCl}$precipitates. This lowers the concentration of $A{g}^{+}$ions in the solution. So $A{g}_{2}Cr{O}_4$formed will dissolve to replenish the $A{g}^{+}$ions disappearing the red solid turning the solution into yellow color due to chromate ions. It can be inferred that $A{g}_{2}Cr{O}_4$is more soluble than n$A{g}_{2}Cl$. This can be confirmed by the $K_{sp}$values.

$\begin{array}{c}{K}_{sp}=\left[A{g}^{+}\right]\left[Cr{O}_4^{2-}\right]\\ 9.0\times {10}^{-12}=\left(2x^2\right)x\\ 4x^3=9.0\times {10}^{-12}\\ x=1.3\times {10}^{-4}M\end{array}$

$\begin{array}{c}{K}_{sp}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]\\ 1.6\times {10}^{-10}=x^2\\ x=1.3\times {10}^{-5}M\end{array}$

Solubility of silver chromate is higher than the solubility of silver chloride