Q19.140 CP - Chemistry: Molecular Nature Of Matter And Change

Question

Quantitative analysis of  ${\mathbf{Cl}}^{\mathbf{-}}$ion is often performed by a titration with silver nitrate, using sodium chromate as an indicator. As standardized ${\mathbf{AgNO}}_{\mathbf{3}}$is added, both white $\mathbf{AgCl}$ and red ${\mathbf{Ag}}_{\mathbf{2}}{\mathbf{CrO}}_{\mathbf{4}}$ precipitate, but so long as some ${\mathbf{Cl}}^{\mathbf{-}}$remains, the ${\mathbf{Ag}}_{\mathbf{2}}{\mathbf{CrO}}_{\mathbf{4}}$redissolves as the mixture is stirred. When the red color is permanent, the equivalence point has been reached.

(a) Calculate the equilibrium constant for the reaction

data-custom-editor="chemistry" $\mathbf{2}\mathbf{AgCl}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{Cr}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{\rightleftharpoons }{\mathbf{Ag}}_{\mathbf{2}}{\mathbf{CrO}}_{\mathbf{4}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{C}{\mathbf{l}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$

(b) Explain why the silver chromate redissolves.

(c) If $\mathbf{25}\mathbf{.}\mathbf{00}\mathbf{ }{\mathbf{cm}}^{\mathbf{3}}$ of $\mathbf{0}\mathbf{.}\mathbf{1000}\mathbf{ }\mathbf{M}\mathbf{ }\mathbf{NaCl}$is mixed with $\mathbf{25}\mathbf{.}\mathbf{00}\mathbf{ }{\mathbf{cm}}^{\mathbf{3}}$of $\mathbf{0}\mathbf{.}\mathbf{1000}\mathbf{ }\mathbf{M}\mathbf{ }{\mathbf{AgNO}}_{\mathbf{3}}$, what is the concentration of ${\mathbf{Ag}}^{\mathbf{+}}$remaining in solution? Is this sufficient to precipitate any silver chromate?

Step-by-Step Solution

Answer

1. The equilibrium constant for the given reaction is $1.2\times {10}^{-8}$.
2. The value of K is very small. So the equilibrium will shift towards left which means silver chromate dissolves.
3. The concentration of ${\mathrm{Ag}}^{+}$ ion remaining in the solution is  $1.3\times {10}^{-3}\mathrm{M} {\mathrm{Ag}}_2{\mathrm{CrO}}_4$ will precipitate at this concentration.

Step 1: Definition of Concept.

Heat Capacity: The amount of heat energy required to raise the temperature of a body by a certain amount is known as heat capacity. The amount of heat in joules required to raise the temperature 1 Kelvin is known as heat capacity (symbol: C) in SI units.

 

Step 2: Calculate the equilibrium constant for the reaction.

(a)

Considering the given information:

$\begin{array}{c} {\mathrm{K}}_{\mathrm{sp}} \mathrm{of} \mathrm{AgCl}\left(\mathrm{s}\right)=1.8\times {10}^{-10}\\ {\mathrm{K}}_{\mathrm{sp}} \mathrm{of} {\mathrm{Ag}}_2{\mathrm{CrO}}_4\left(\mathrm{s}\right)=2.6\times {10}^{-12}\end{array}$

The formula for a balanced chemical reaction is 

$$\begin{gathered} \mathrm{AgCl}\left(\mathrm{s}\right)\rightleftarrows {\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)+\mathrm{C}{\mathrm{l}}^{-}\left(\mathrm{aq}\right) \\ {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ag}}^{+}\right]\left[{\mathrm{Cl}}^{-}\right]=1.8\times {10}^{-10} \\ {\mathrm{Ag}}_2{\mathrm{CrO}}_4\left(\mathrm{s}\right)\rightleftarrows 2{\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)+\mathrm{Cr}{{\mathrm{O}}_4}^{2-}\left(\mathrm{aq}\right) \\ {\mathrm{K}}_{\mathrm{sp}}={\left[{\mathrm{Ag}}^{+}\right]}^2\left[\mathrm{Cr}{{\mathrm{O}}_4}^{2-}\right]=2.6\times {10}^{-12} \end{gathered}$$

The ideal equilibrium is as follows:

$2\mathrm{AgCl}\left(\mathrm{s}\right)+\mathrm{Cr}{{\mathrm{O}}_4}^{2-}\left(\mathrm{aq}\right)\rightleftarrows \mathrm{A}{\mathrm{g}}_2{\mathrm{CrO}}_4\left(\mathrm{s}\right)+2\mathrm{C}{\mathrm{l}}^{-}\left(\mathrm{aq}\right)$

To achieve the desired equilibrium, combine two separate equilibria.

$2\mathrm{AgCl}\left(\mathrm{s}\right)\to 2{\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)+2\mathrm{C}{\mathrm{l}}^{-}\left(\mathrm{aq}\right) \mathrm{K}"=\frac{{\left[{\mathrm{Ag}}^{+}\right]}^2{\left[{\mathrm{Cl}}^{-}\right]}^2}{{\left|{\mathrm{Ag}}_{\mathrm{g}}\mathrm{Cl}\left(\mathrm{s}\right)\right|}^2}$

$\begin{array}{c}={\left(K_{sp}\right)}^2\left[\right[\mathrm{AgCl}\left(s\right)]=1]\\ \left.2{\mathrm{Ag}}_8+\sqrt{aq}\right)+\mathrm{Cr}{{\mathrm{O}}_4}^{2-}\left(\mathrm{aq}\right)\to {\mathrm{Ag}}_2{\mathrm{CrO}}_4\left(s\right) K^{\text{'}\text{'}}=\frac{\left[{\mathrm{Ag}}_2{\mathrm{CrO}}_4\left(s\right)\right]}{\left[{\left.{\mathrm{Ag}}^{+}\right|}^2\left[\mathrm{Cr}{{\mathrm{O}}_4}^{2-}\right]\right.}\\ 2\mathrm{AgCl}\left(s\right)+\mathrm{Cr}{{\mathrm{O}}_4}^{2-}\left(aq\right)\rightleftarrows {\mathrm{Ag}}_2{\mathrm{CrO}}_4\left(s\right)+2{\mathrm{Cl}}^{-}\left(aq\right) \mathrm{K}=K^{\text{'}}{K}^{\text{'}\text{'}}\\ K^{\text{'}}={\left(K_{sp}\right)}^2={\left(1.8\times {10}^{-10}\right)}^2=3.24\times {10}^{-20}\\ K^{\text{'}\text{'}}=\frac{1}{K_{\mathrm{sp}}}=\frac{1}{2.6\times {10}^{-12}}=3.8462\times {10}^{11}\\ K=K^{\text{'}}{K}^{\text{'}\text{'}}=\left(3.24\times {10}^{-20}\right)\left(3.8462\times {10}^{11}\right)\\ =1.2462\times {10}^{-8}\\ =1.2\times {10}^{-8}\end{array}$

Therefore, the equilibrium constant for the given reaction is $1.2\times {10}^{-8}$.

Step 3: Explain why the silver chromate redissolves.

(b)

The equations are:

$\begin{array}{l}2\mathrm{AgCl}\left(\mathrm{s}\right)+\mathrm{Cr}{{\mathrm{O}}_4}^{2-}\left(\mathrm{aq}\right)\rightleftarrows \mathrm{A}{\mathrm{g}}_2{\mathrm{CrO}}_4\left(\mathrm{s}\right)+2 {\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)\\ \mathrm{K} = \frac{{\left[{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)\right]}^2}{\left[\mathrm{Cr}{{\mathrm{O}}_4}^{2-\left(\mathrm{aq}\right)}\right]}\\ = 1.2\times {10}^{-8}\end{array}$

The K is a very small amount of money. As a result, the equilibrium will shift to the left, indicating that silver chromate will redissolve.

Step 4: Find the concentration of Ag + remaining in solution.

(c)

Considering the given information:

$25.00 {\mathrm{cm}}^3$ of $0.1000\mathrm{MNaCl}$

$25.00{\mathrm{cm}}^3$ of  $0.1000{\mathrm{MAgNO}}_3$

Because mixing equal amounts of equal molar solutions would precipitate all of the $\text{AgCL}$, the ${\mathrm{Ag}}^{+}$ ion concentration is entirely derived from $\text{AgCL's}$ ${\text{K}}_{sp}$.

$\begin{array}{c}\mathrm{AgCl}\left(\mathrm{s}\right)\rightleftarrows \mathrm{A}{\mathrm{g}}^{+}\left(\mathrm{aq}\right)+\mathrm{C}{\mathrm{l}}^{-}\left(\mathrm{aq}\right)\\ {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ag}}^{+}\right]\left[{\mathrm{Cl}}^{-}\right]\\ =(\mathrm{S}{)}^2\\ =1.8\times {10}^{-10}\\ \mathrm{S}=\left[{\mathrm{Ag}}^{+}\right]\\ =\sqrt{{\mathrm{K}}_{\mathrm{sp}}}\\ =\sqrt{1.8\times {10}^{-10}}\\ =1.3\times {10}^{-5}.\end{array}$

The concentration of remaining ${\mathrm{Ag}}^{+}$ion in the solution is calculated as $1.3\times {10}^{-3}\mathrm{M}$.

data-custom-editor="chemistry" ${\mathrm{Ag}}_2{\mathrm{CrO}}_4\left(\mathrm{s}\right)\rightleftarrows 2\mathrm{A}{\mathrm{g}}^{+}{\left(\mathrm{aq}\right)+\mathrm{CrO}}_4^{2-}\left(\mathrm{aq}\right)$

$\begin{array}{c}{\mathrm{K}}_{\mathrm{sp}}={\left[{\mathrm{Ag}}^{+}\right]}^2\left[\mathrm{Cr}{{\mathrm{O}}_4}^{2-}\right]\\ =2.6\times {10}^{-12}\\ \left[{\mathrm{CrOO}}^{2-}\right]=\frac{{\mathrm{K}}_{\mathrm{sp}}}{{\left|{\mathrm{Ag}}^{+}\right|}^2}\\ =\frac{2.6\times {10}^{-12}}{{\left(1.3\times {10}^{-5}\right)}^2}\\ =0.01444\\ =0.014\mathrm{M}\end{array}$

If the chromate ion concentration is greater than $0.014\mathrm{M},{\mathrm{Ag}}_2{\mathrm{CrO}}_4$ precipitates.