Q19.72P - Chemistry: Molecular Nature Of Matter And Change

Question

The solubility of silver dichromate at $\mathbf{15}\mathbf{ }\mathbf{°}\mathbf{C}$ is $\mathbf{8}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{ }\mathbf{g}\mathbf{/}\mathbf{100}\mathbf{ }\mathbf{mL}$ solution. Calculate its ${\mathit{K}}_{\mathbf{s}\mathbf{p}}$.

Step-by-Step Solution

Answer

The $K_{sp}$ of silver dichromate is $2.84\times {10}^{-11}$.

Concept Introduction

The chemical substance silver dichromate has the formula ${\mathbf{\text{Ag}}}_{\mathbf{2}}{\mathbf{\text{Cr}}}_{\mathbf{2}}{\mathbf{\text{O}}}_{\mathbf{7}}$. 

It is water insoluble and decomposes when exposed to hot water. 

Its anion has a $\mathbf{-}\mathbf{2}$ charge.

Constructing the ICE table

To calculate the ${\mathrm{K}}_{\mathrm{sp}}$of silver dichromate, we first need to write the reaction equation of the dissolution of ${\text{Ag}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$.

${\text{Ag}}_2{\text{Cr}}_2{\text{O}}_7\rightleftharpoons 2{\text{Ag}}^{+}+{\text{Cr}}_2{\text{O}}_7^{2-}$

Write the ${\mathrm{K}}_{\mathrm{sp}}$expression for the reaction.

${\text{K}}_{\text{sp}}\text{ = }{\left[{\text{Ag}}^{\text{ + }}\right]}^{\text{2}}\left[{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2 - }}\right]$

Constructing the ICE table,

Now, we need to substitute it for the $K_{sp}$expression.

$$\begin{gathered} {\text{K}}_{\text{sp}}{\text{ = [2s]}}^{\text{2}}\text{[s]} \\ {\text{K}}_{\text{sp}}{\text{ = 4s}}^{\text{3}} \end{gathered}$$

Solving for $s$ from the solubility given,

$$\begin{gathered} \mathrm{s}=\frac{\mathrm{mol}}{\mathrm{L}} \\ =\frac{8.3\times {10}^{-3} \mathrm{g}{\mathrm{Ag}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\times \frac{1 \mathrm{mol}{\mathrm{Ag}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7}{431.76 \mathrm{g}{\mathrm{Ag}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7}}{100 \mathrm{mL}\times \frac{1 \mathrm{L}}{1000 \mathrm{mL}}} \\ =1.92\times {10}^{-4}. \end{gathered}$$

Calculating the K s p of silver dichromate

Now solve for ${\mathrm{K}}_{\mathrm{sp}}$,

$$\begin{gathered} K_{sp}=4s^3 \\ =4{\left(1.92\times {10}^{-4}\right)}^3 \\ =2.84\times {10}^{-11} \end{gathered}$$

Therefore, the $K_{sp}$ of silver dichromate is $2.84\times {10}^{-11}$.