Q19.64P - Chemistry: Molecular Nature Of Matter And Change

Question

A list of ${\mathbf{\text{K}}}_{\mathbf{\text{sp}}}$ values like that in Appendix C can be used to compare the solubility of silver chloride directly with that of silver bromide but not with that of silver chromate. Explain.

Step-by-Step Solution

Answer

Since both $\text{Ksp}$ values are identical to ${\text{S}}^{\text{2}}$, the solubility of $\text{AgBr}$ can be directly compared to $\text{AgCl}$solubility.

Because we can't compare squared ${\text{S(S}}^{\text{2}}\text{)}$ and $\text{S}\left({\text{S}}^{\text{3}}\right)$ cubed  values, we can't compare the solubilities of $\text{AgBr}$  and $\text{AgCl}$ with the solubility of ${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}.$ 

Step 1: Concept Introduction

The amount of solute that dissolves in a solvent is known as solubility. If there is a rise in the temperature of the solvent, then there is also a rise in the solubility of many solutes.

Step 2: Checking the solubility of AgCl and AgBr in water

Let us write the dissolution equation of $\text{AgCl and AgBr}$ in water.

 ${\text{AgCl > > Ag}}^{\text{ + }}{\text{ + Cl}}^{\text{ - }}$

 S - solubility of $\text{AgCl}$ in water

 $$\begin{gathered} \mathrm{Ksp}\mathrm{ }=\mathrm{ }\left[{\mathrm{Ag}}^{+}\right]\left[{\mathrm{Cl}}^{-}\right]\mathrm{ } \\ =\mathrm{ }1.8\times {10}^{-0}\mathrm{ } \\ =\mathrm{ }{\mathrm{S}}^2 \\ \left[{\mathrm{Ag}}^{+}\right]\mathrm{ }=\mathrm{ }\mathrm{S} \\ \left[{\mathrm{Cl}}^{-}\right]\mathrm{ }=\mathrm{ }\mathrm{S}\mathrm{ } \end{gathered}$$

Now we can write the dissolution equation for $\text{AgBr}$:

 ${\text{AgBr > > Ag}}^{\text{ + }}{\text{ + Br}}^{\text{ - }}$

S- solubility of $\text{AgBr}$ in water

$$\begin{gathered} \text{ Ksp= }\left[{\mathrm{Ag}}^{+}\right]\left[{\mathrm{Br}}^{-}\right]\mathrm{ } \\ =\mathrm{ }5\times {10}^{-13} \\ =\mathrm{ }{\mathrm{S}}^2 \\ \left[{\mathrm{Ag}}^{+}\right]\mathrm{ }=\mathrm{ }\mathrm{S} \\ \left[{\mathrm{Br}}^{-}\right]\mathrm{ }=\mathrm{ }\mathrm{S} \end{gathered}$$ 

The solubility of $\text{AgBr}$ can be directly compared with solubility of $\text{AgCl}$ because both $\text{Ksp}$ values are equal to ${\text{S}}^{\text{2}}$.

Step 3: Checking the solubility of Ag 2 CrO 4 in water

Now we can write the dissolution equation for  

 $\text{b) Ksp = }\left[{\text{Ba}}^{\text{2 + }}\right]{\left[{\text{F}}^{\text{ - }}\right]}^{\text{2}}\text{ c) Ksp = }\left[{\text{Cu}}^{\text{2 + }}\right]\left[{\text{s}}^{\text{2 - }}\right]$

${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}{\text{ > > 2Ag}}^{\text{ + }}{\text{ + CrO}}_{\text{4}}^{\text{2 - }}$

S - solubility of ${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}$ in water

$$\begin{gathered} \text{ Ksp = }{\left[{\mathrm{Ag}}^{+}\right]}^2\left[{\mathrm{CrOH}}_4^{2-}\right] \\ =\mathrm{ }2.6\times {10}^{-12} \\ =\mathrm{ }(2\mathrm{S}{)}^2\mathrm{S} \\ =\mathrm{ }4{\mathrm{S}}^3 \\ \left[{\mathrm{Ag}}^{+}\right]=\mathrm{ }2\mathrm{S} \\ \left[{\mathrm{CrOH}}^{2-}\right]=\mathrm{ }\mathrm{S} \end{gathered}$$ 

The $\left[{\text{Ag}}^{\text{ + }}\right]$ is squared because the equation contains two mols of  ${\text{Ag}}^{\text{ + }}$. 

Since we can't compare squared ${\text{(S}}^{\text{2}}\text{)}$ and cubed $\left({\text{S}}^{\text{3}}\right)$ values, we can't compare solubilities of $\text{AgBr and AgCl}$ with solubilities of ${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}$.