Question

1. The solubility of $\text{Ag(I)}$ in aqueous solutions containing different concentrations of${\mathbf{\text{Cl}}}^{\mathbf{\text{-}}}$is based on the following equilibria:

${\text{Ag}}^{\text{+}}{\text{(aq) + Cl}}^{\text{-}}\text{(aq)}\to \text{AgCl(s)}{\text{K}}_{\text{sp}}{\text{=1.8×10}}^{-10}$

${\text{Ag}}^{\text{+}}{\text{(aq) + 2Cl}}^{\text{-}}\text{(aq)}\rightleftharpoons {\text{AgCl}}_{\text{2}}^{\text{-}}\text{(aq)}{\text{K=1.8×10}}^5$

When solid $\text{AgCl}$is shaken with a solution containing ${\text{Cl}}^{\text{-}}\text{,Ag(I)}$is present as both ${\text{Ag}}^{\text{+}}$and ${\text{AgCl}}_{\text{2}}^{\text{-}}$. The solubility of $\text{AgCl}$ is the sum of the concentrations of ${\text{Ag}}^{\text{+}}$and ${\text{AgCl}}_{\text{2}}^{\text{-}}$.

(a) Show that $\left[{\text{Ag}}^{\text{+}}\right]$in solution is given by

$\left[{\text{Ag}}^{\text{+}}\right]{\text{=1.8×10}}^{-10}\text{/}\left[{\text{Cl}}^{\text{-}}\right]$

and that $\left[{\text{AgCl}}_{\text{2}}^{\text{-}}\right]$in solution is given by

$\left[{\text{AgCl}}_{\text{2}}^{\text{-}}\right]\text{=}\left({\text{3.2×10}}^{-5}\right)\left(\left[{\text{Cl}}^{\text{-}}\right]\right)$

(b) Find the $\left[{\text{Cl}}^{\text{-}}\right]$at which $\left[{\text{Ag}}^{\text{+}}\right]\text{=}\left[\text{AgC}{{\text{l}}_{\text{2}}}^{\text{-}}\right]$

(c) Explain the shape of a plot of $\text{AgCl}$ solubility vs. $\left[{\text{Cl}}^{\text{-}}\right]$.

(d) Find the solubility of$\text{AgCl}$at the$\left[{\text{Cl}}^{\text{-}}\right]$of part (b), which is the minimum solubility of$\text{AgCl}$in the presence of${\text{Cl}}^{\text{-}}$.

Short answer

1. It is proved that $\left[{\text{AgCl}}^{\text{2}}\right]\text{=}\left(3.2\times {10}^{-5}\right)\text{×}\left[{\text{Cl}}^{\text{-}}\right]$and $\left[{\text{Ag}}^{\text{+}}\right]\text{=}\frac{1.8\times {10}^{-5}}{{\displaystyle \left[{\text{Cl}}^{\text{-}}\right]}}$. The value of ${\text{K}}_{\text{sp}}$of MZ is$1.0\times {10}^{-10}\mathrm{M}$.
2. The value of$\text{[Cl]}\text{=}2.37\times {100}^{-3}\mathrm{M}$.

3. The shape is:

   
   

   
   

4. The solubility ofAgCl $1.52\times {100}^{-7}\mathrm{M}$.

Step-by-step solution

Definition of Concept

Solubility: The maximum amount of solute that can be dissolved in the solvent at equilibrium is defined as solubility.

Constant of soluble product: For equilibrium between solids and their respective ions in a solution, the solubility product constant is defined. In general, the term "solubility product" refers to water-equilibrium insoluble or slightly soluble ionic substances.

 Show the given statements

a. Considering the given information:

$$\begin{gathered} \left[{\text{Ag}}^{\text{+}}\right]\text{+}\left[{\text{Cl}}^{\text{-}}\right]\to \text{[AgCl];}\left({\text{K}}_{\text{sp}}{\text{=1.8×10}}^{-10}\right) \\ \left[{\text{Ag}}^{\text{+}}\right]\text{+}\left[{\text{Cl}}^{\text{-}}\right]\to \left[{\text{AgCl}}_{\text{2}}^{\text{-}}\right]\text{;}\left({\text{K}}_{\text{f}}{\text{=1.8×10}}^5\right) \end{gathered}$$

Subtraction of Equations (1) and (2) yields

$$\begin{gathered} \mathrm{Ag}+2{\mathrm{Cl}}^{-}\to \mathrm{AgC}{{\mathrm{l}}_2}^{-} \\ {\text{K}}_{\text{f}}\text{=}\frac{\left|{\displaystyle {{\displaystyle \text{AgCl}}}_{\text{2}}}\right|}{\left[{\displaystyle \left[\text{ACl}\right]\left[\text{Cl}\right]}\right]} \\ \left[{\mathrm{Ag}}^{+}\right]\text{=}\frac{{\displaystyle {\text{K}}_{\text{sp}}}}{{\displaystyle \text{|Cl|}}} \\ \left[{\text{Ag}}^{\text{+}}\right]\text{=}\frac{1.8\times {10}^{-10}}{{\displaystyle \text{[Cl]}}} \end{gathered}$$

Substituting the value of $\left[{\text{Ag}}^{\text{+}}\right]$in the equation (3) we will get,

$\begin{array}{rcl}\left[{\text{AgCl}}^{\text{-}}\right]& =& {\text{K}}_{\text{sp}}\times {\text{K}}_{\text{f}}\times \left[{\text{Cl}}^{\text{-}}\right]\\ & =& \left({\text{1.8×10}}^{\text{- 10}}\right)\times \left({\text{1.8×10}}^{\text{5}}\right)\times \left[{\text{Cl}}^{\text{-}}\right]\\ & =& \left(\text{3.2}\times {\text{10}}^{\text{- 5}}\right)\times \left[{\text{Cl}}^{\text{-}}\right]\end{array}$

Hence, it is proved that$\left[{\text{AgCl}}^{\text{2}}\right]\text{=}\left(\text{3.2}\times {\text{10}}^{\text{- 5}}\right)\text{×}\left[{\text{Cl}}^{\text{-}}\right]$

 Find the  [Cl - ].

(b)

Considering the given information:

Now, the condition stated is that $\left[{\text{Ag}}^{\text{+}}\right]\text{=}\left[{\text{AgCl}}_{\text{2}}\right]\text{.}$

Using the data from Part , we get the following:

$\begin{array}{rcl}\left[{\text{Ag}}^{\text{+}}\right]& =& \left[{\text{AgCl}}_{\text{2}}\right]\\ \frac{1.8\times {10}^{-10}}{{\displaystyle \left[{\text{Cl}}^{\text{-}}\right]}}& =& {\text{(3.2×10}}^{-5}\text{)×}\left[{\mathrm{Cl}}^{-}\right]\\ & & {\text{[Cl]=2.37×100}}^{-3}\text{M}\\ & & \end{array}$

Therefore, the required value$\left[\mathrm{Cl}\right]=2.37\times {100}^{-3}\mathrm{M}$

  Explain the shape of a plot of AgCl solubility vs. [Cl-]

(c)

As the $\left[{\mathrm{Cl}}^{-}\right]$starts to increase the the formation of $\left[\mathrm{AgCl}\right]$also increases, until the $\left[{\mathrm{Ag}}^{+}\right]$is fully consumed. As $\left[\mathrm{AgCl}\right]$increases, then its solubility decreases in order to maintain equilibrium.

 Find the solubility of AgCl

The solubility of AgCI can be calculated using the formula:

$$\begin{gathered} \text{AgCl=}\frac{1.8\times {10}^{-10}}{{\displaystyle \left[{\text{Cl}}^{\text{-}}\right]}}\text{+}\left(3.2\times {10}^{-5}\right)\text{×}\left[{\text{Cl}}^{\text{-}}\right] \\ \text{AgCl=}\frac{{\displaystyle 1.8\times {10}^{-10}}}{{\displaystyle \left[{\text{Cl}}^{\text{-}}\right]}}\text{+}\left(3.2\times {10}^{-5}\right)\text{×}\left(\text{2.37}\times {10}^{-3}\right) \\ {\text{AgCl=1.52×100}}^{-7}\text{M} \end{gathered}$$

Therefore, the required value is$1.52\times {10}^{-7}\mathrm{M}$.