Chapter 15, Problem 4SP

(a)

Interpretation Introduction

Interpretation:

The minimum mass of $\text{NaCl}$ which is to be added to initiate precipitation has to be determined.

Concept Introduction:

The solubility product for an equilibrium reaction is defined as the product of ion concentration raised to power their stoichiometric coefficient.   The symbol for solubility product is $K_{\text{sp}}$.  If $K_{\text{sp}}$ value is higher, it means that compound is higher is the solubility.

Answer to Problem 4SP

The minimum mass of $\text{NaCl}$ required to initiate precipitation is $\underset{\_}{2.62\times 10^{-6}g}$.

Explanation of Solution

As per the given data the number of moles of silver ion is $0.0010\text{mol}$ in $0.500\text{L}$ of solution.

The concentration of ${\text{Ag}}^{+}$ is calculated by the formula shown below.

  $M=\frac{n}{V}$        (1)

Where,

• $M$ is the molarity.
• $n$ is the number of moles.
• $V$ is the volume.

The value of $n$ for ${\text{Ag}}^{+}$ is $0.0010\text{mol}$.

The value of $V$ for ${\text{Ag}}^{+}$ is $0.500\text{L}$.

Substitute the values of $n$ and $V$ for ${\text{Ag}}^{+}$ in the equation (1).

  $\begin{array}{c}\left[{\text{Ag}}^{+}\right]=\frac{\left(0.0010\text{ mol}\right)}{\left(\text{0}\text{.500}\text{L}\right)}\\ =0.002\text{M}\end{array}$

Thus, the concentration of ${\text{Ag}}^{+}$ is $0.002\text{M}$.

The values of $K_{\text{sp}}$ for $\text{AgCl}$ is $1.8\times {10}^{-10}$.

The dissociation reaction for $\text{AgCl}$ is shown as follows:

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

The $K_{\text{sp}}$ expression for the above reaction is shown below.

  $K_{\text{sp}}\text{=}\left[{\text{Ag}}^{+}\right]\left[{\text{Cl}}^{-}\right]$        (2)

Where,

• $K_{\text{sp}}$ is the solubility product.
• $\left[{\text{Ag}}^{+}\right]$ is the concentration of silver ions.
• $\left[{\text{Cl}}^{-}\right]$ is the concentration of chloride ions.

The value of $K_{\text{sp}}$ is $1.8\times {10}^{-10}$.

The value of $\left[{\text{Ag}}^{+}\right]$ is $0.002\text{M}$.

Substitute the valueof $K_{\text{sp}}$ and $\left[{\text{Ag}}^{+}\right]$ in equation (2).

  $\begin{array}{c}1.8\times {10}^{-10}=\left(0.002\right)\left[{\text{Cl}}^{-}\right]\\ \left[{\text{Cl}}^{-}\right]=\frac{1.8\times {10}^{-10}}{\left(0.002\right)}\\ =9\times {10}^{-8}\text{M}\end{array}$

Since the sodium chloride is the salt of strong acid and strong base, therefore, it dissociates completely into ${\text{Na}}^{+}$ ions and ${\text{Cl}}^{-}$ ions having equal concentrations.

Therefore, the concentration of $\text{NaCl}$ that is added is $9\times {10}^{-8}\text{M}$.

The number of moles of $\text{NaCl}$ is calculated by the formula shown below.

  $n=M\times V$        (3)

Where,

• $M$ is the molarity.
• $n$ is the number of moles.
• $V$ is the volume.

The value of $M$ for $\text{NaCl}$ is $9\times {10}^{-8}\text{M}$.

The value of $V$ for $\text{NaCl}$ is $0.500\text{L}$.

Substitute the values of $M$ and $V$ for $\text{NaCl}$ in the equation (3).

  $\begin{array}{c}{n}_{\text{NaCl}}=\left(9\times {10}^{-8}\text{M}\right)\left(0.500\text{ L}\right)\\ =4.5\times {10}^{-8}\text{mol}\end{array}$

Therefore, the number of moles of $\text{NaCl}$ is $4.5\times {10}^{-8}\text{mol}$.

The mass of $\text{NaCl}$ is calculated by the formula shown below.

  $m=n\times M_{\text{w}}$        (4)

Where,

• $n$ is the number of moles.
• $m$ is the mass.
• $M_{\text{w}}$ is the molar mass.

The value of $n$ for $\text{NaCl}$ is $4.5\times {10}^{-8}\text{mol}$.

The value of $M_{\text{w}}$ for $\text{NaCl}$ is $58.44\text{g/mol}$.

Substitute the values of $n$ and $M_{\text{w}}$ for $\text{NaCl}$ in the equation (4).

  $\begin{array}{c}{m}_{\text{NaCl}}=\left(4.5\times {10}^{-8}\text{mol}\right)\left(58.44\text{g/mol}\right)\\ =2.62\times 10^{-6}g\end{array}$

Thus, the minimum mass of $\text{NaCl}$ which is to be added to initiate precipitation is $\underset{\_}{2.62\times 10^{-6}g}$.

(b)

Interpretation Introduction

Interpretation:

The amount of chloride ion in moles that is added to dissolve the precipitate has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 4SP

Amount of ${\text{Cl}}^{-}$ that is added to dissolve propitiate is $\underset{\_}{0.0033mol}$.

Explanation of Solution

The standard $K_{\text{sp}}$ value for $\text{AgCl}$ is $1.8\times {10}^{-10}$.

The dissociation reaction for $\text{AgCl}$ is shown below:

  $\begin{array}{l}\text{AgCl}\rightleftharpoons {\text{Ag}}^{+}+{\text{Cl}}^{-}\\ \begin{array}{cc}\text{s}& \text{s}\end{array}\end{array}$

Where,

• $\text{s}$ is the solubility in mole per liter of ${\text{Ag}}^{+}$ and ${\text{Cl}}^{-}$ ions.

The $K_{\text{sp}}$ expression for $\text{AgCl}$ is shown below.

  $K_{\text{sp}}=\left[{\text{Ag}}^{+}\right]\left[{\text{Cl}}^{-}\right]$        (5)

Where,

• $K_{\text{sp}}$ is the solubility product.
• $\left[{\text{Ag}}^{+}\right]$ is the concentration of silver ions.
• $\left[{\text{Cl}}^{-}\right]$ is the concentration of nitrate ions.

The value of $\left[{\text{Ag}}^{+}\right]$ is $\text{s}$.

The value of $\left[{\text{Cl}}^{-}\right]$ is $\text{s}$.

The value of $K_{\text{sp}}$ is $2.0\times {10}^{-25}$.

Substitute the value of $K_{\text{sp}}$, $\left[{\text{Ag}}^{+}\right]$  and  $\left[{\text{Cl}}^{-}\right]$ in equation (5).

  $\begin{array}{c}1.8\times {10}^{-10}=\left(\text{s}\right)\left(\text{s}\right)\\ {\text{s}}^{\text{2}}=1.8\times {10}^{-10}\\ s=\sqrt{1.8\times {10}^{-10}}\\ =1.34\times {10}^{-5}\text{mol/L}\end{array}$

The net chemical reaction for dissolving $\text{AgCl}$ by the formation of complex ion that is ${\left[{\text{AgCl}}_2\right]}^{-}$ is the addition of $K_{\text{sp}}$ and $K_{\text{f}}$ equations.

Reaction:1 $\text{AgCl}\left(\text{aq}\right)\rightleftharpoons {\text{ Ag}}^{+}\left(\text{aq}\right)+{\text{Cl}}^{-}\left(\text{aq}\right)K_{\text{sp}}=1.8\times {10}^{-10}$

Reaction:2 ${\text{Ag}}^{+}\left(\text{aq}\right)+2{\text{Cl}}^{-}\rightleftharpoons {\left[{\text{AgCl}}_2\right]}^{-}{K}_{\text{f}}=1.8\times {10}^5$

Reaction:3 $\text{AgCl}\left(\text{aq}\right)+{\text{Cl}}^{-}\rightleftharpoons {\left[{\text{AgCl}}_2\right]}^{-}K=K_{\text{sp}}\times K_{\text{f}}$

The net equilibrium constant used to define the formation of complex is the product of solubility product and the formation constant.  The symbol of net equilibrium constant is $K_{\text{net}}$.

The net equilibrium constant for the formation of ${\left[{\text{AgCl}}_2\right]}^{-}$ is calculated using the relation shown below:

  $K_{\text{net}}=K_{\text{sp}}\times K_{\text{f}}$        (6)

Where,

• $K_{\text{sp}}$ is the solubility product.
• $K_{\text{net}}$ is the net equilibrium constant.
• $K_{\text{f}}$ is the formation constant.

The value of $K_{\text{sp}}$ is $1.8\times {10}^{-10}$.

The value of $K_{\text{f}}$ is $1.8\times {10}^5$.

Substitute the value of $K_{\text{sp}}$ and $K_{\text{f}}$ in equation (6).

  $\begin{array}{c}{K}_{\text{net}}=\left(1.8\times {10}^{-10}\right)\left(1.8\times {10}^5\right)\\ =3.24\times {10}^{-5}\end{array}$

The equilibrium constant expression for second reaction is shown below.

  $K_{\text{f}}=\frac{{\left[{\text{AgCl}}_2\right]}^{-}}{\left[{\text{Ag}}^{+}\right]{\left[{\text{Cl}}^{-}\right]}^2}$        (7)

Where,

• $K_{\text{f}}$ is the formation equilibrium constant.
• ${\left[{\text{AgCl}}_2\right]}^{-}$ is the concentration of complex.
• $\left[{\text{Ag}}^{+}\right]$ is the concentration of silver ion.
• $\left[{\text{Cl}}^{-}\right]$ is the concentration of chloride ion.

The value of $K_{\text{f}}$ is $1.8\times {10}^5$.

The value of $\left[{\text{Cl}}^{-}\right]$ is $9.0\times {10}^{-8}\text{M}$.

The value of $\left[{\text{Ag}}^{+}\right]$ is $0.002\text{M}$.

Substitute the value of $K_{\text{f}}$, $\left[{\text{Cl}}^{-}\right]$, and $\left[{\text{Ag}}^{+}\right]$ in equation (7).

    $\begin{array}{c}1.8\times {10}^5=\frac{{\left[{\text{AgCl}}_2\right]}^{-}}{\left(0.002\right){\left(9\times {10}^{-8}\right)}^2}\\ {\left[{\text{AgCl}}_2\right]}^{-}=\left(1.8\times {10}^5\right)\left(0.002\right){\left(9\times {10}^{-8}\right)}^2\\ =2.9\times {10}^{-12}\text{M}\end{array}$

The equilibrium constant expression for third reaction is shown below.

  $K_{\text{net}}=\frac{{\left[{\text{AgCl}}_2\right]}^{-}}{\left[\text{AgCl}\right]\left[{\text{Cl}}^{-}\right]}$        (8)

Where,

• $K_{\text{net}}$ is the net equilibrium constant.
• ${\left[{\text{AgCl}}_2\right]}^{-}$ is the concentration of complex.
• $\left[{\text{Cl}}^{-}\right]$ is the concentration of chloride ion.
• $\left[\text{AgCl}\right]$ is the concentration of silver chloride.

The value of $K_{\text{net}}$ is $3.4\times {10}^{-18}$.

The value of ${\left[{\text{AgCl}}_2\right]}^{-}$ is $2.9\times {10}^{-12}\text{M}$.

The value of $\left[\text{AgCl}\right]$ is $1.34\times {10}^{-5}\text{M}$.

Substitute the value of $K_{\text{net}}$, ${\left[{\text{AgCl}}_2\right]}^{-}$ and $\left[\text{AgCl}\right]$ in equation (8).

  $\begin{array}{c}3.24\times {10}^{-5}=\frac{\left(2.9\times {10}^{-12}\right)}{\left(1.34\times {10}^{-5}\right)\left[{\text{Cl}}^{-}\right]}\\ \left[{\text{Cl}}^{-}\right]=\frac{\left(2.9\times {10}^{-12}\right)}{\left(1.34\times {10}^{-5}\right)\left(3.24\times {10}^{-5}\right)}\\ =6.67\times {10}^{-3}\text{M}\end{array}$

The value of $M$ for ${\text{Cl}}^{-}$ is $6.67\times {10}^{-3}\text{M}$.

The value of $V$ for ${\text{Cl}}^{-}$ is $0.500\text{L}$.

Substitute the values of $M$ and $V$ for ${\text{Cl}}^{-}$ in the equation (3).

  $\begin{array}{c}{n}_{{\text{Cl}}^{-}}=\left(6.67\times {10}^{-3}\text{M}\right)\left(0.500\text{ L}\right)\\ =0.0033\text{mol}\end{array}$

Thus, the amount of ${\text{Cl}}^{-}$ that is added to dissolve precipitate is $\underset{\_}{0.0033mol}$.