Chapter 17, Problem 109AP

Interpretation Introduction

Interpretation:

The concentration of silver ion, in $\text{mol}/\text{L}$, in a saturated solution of each salt is to be calculated.

Concept Introduction:

The solubility product is represented by ${\text{K}}_{\text{sp}}$. It is the equilibrium constant that exists between solid substances and their dissociated ions.

The solubility product describes the solubility of the solid substance in the solvent at the particular temperature. The solubility product of the substance is directly related to its solubility in the solvent.

Answer to Problem 109AP

The concentration of silver ion in a saturated solution of $\text{AgCl}\left(\text{s}\right)$, $\text{AgBr}\left(\text{s}\right)$ and $\text{AgI}\left(\text{s}\right)$ is $1.3\times {10}^{-5}\text{mol}/\text{L}$, $7.1\times {10}^{-\text{7}}\text{mol}/\text{L}$ and $9.1\times {10}^{-9}\text{mol}/\text{L}$ respectively.

Explanation of Solution

The given salt is $\text{AgCl}$. Its ${\text{K}}_{\text{sp}}$ value is given to be $1.8\times {10}^{-10}$.

The balanced chemical equation for ionization of $\text{AgCl}\left(\text{s}\right)$ is shown below.

$\text{AgCl}\left(\text{s}\right)\rightleftharpoons {\text{Ag}}^{+}\left(\text{a}\text{q}\right)+{\text{Cl}}^{-}\left(\text{a}\text{q}\right)$

The expression of ${\text{K}}_{\text{sp}}$ for the above equation is shown below.

${\text{K}}_{\text{sp}}=\left[{\text{Ag}}^{+}\right]\left[{\text{Cl}}^{-}\right]$

Where,

• ${\text{K}}_{\text{sp}}$ is the solubility product.
• $\left[{\text{Ag}}^{+}\right]$ is the concentration of silver ion.
• $\left[{\text{Cl}}^{-}\right]$ is the concentration of chlorine ion.

Both $\left[{\text{Ag}}^{+}\right]$ and $\left[{\text{Cl}}^{-}\right]$ are assumed to be $\text{s}$. Hence, the expression of ${\text{K}}_{\text{sp}}$ is,

$\begin{array}{c}{\text{K}}_{\text{sp}}=\text{s}\times \text{s}\\ ={\text{s}}^2\end{array}$

Substitute the value of ${\text{K}}_{\text{sp}}$ in the above expression.

$\begin{array}{c}{\text{s}}^2=1.8\times {10}^{-10}\\ \text{s}=\sqrt{1.8\times {10}^{-10}}\\ =1.3\times {10}^{-5}\text{mol}/\text{L}\end{array}$

Therefore, the concentration of silver ion in a saturated solution of $\text{AgCl}\left(\text{s}\right)$ is $1.3\times {10}^{-5}\text{mol}/\text{L}$.

The second salt is $\text{AgBr}$. Its ${\text{K}}_{\text{sp}}$ value is given to be $5.0\times {10}^{-13}$.

The balanced chemical equation for ionization of $\text{AgBr}\left(\text{s}\right)$ is shown below.

$\text{AgBr}\left(\text{s}\right)\rightleftharpoons {\text{Ag}}^{+}\left(\text{a}\text{q}\right)+{\text{Br}}^{-}\left(\text{a}\text{q}\right)$

The expression of ${\text{K}}_{\text{sp}}$ for the above equation is shown below.

${\text{K}}_{\text{sp}}=\left[{\text{Ag}}^{+}\right]\left[{\text{Br}}^{-}\right]$

Where,

• ${\text{K}}_{\text{sp}}$ is the solubility product.
• $\left[{\text{Ag}}^{+}\right]$ is the concentration of silver ion.
• $\left[{\text{Br}}^{-}\right]$ is the concentration of bromide ion.

Both $\left[{\text{Ag}}^{+}\right]$ and $\left[{\text{Br}}^{-}\right]$ are assumed to be $\text{s}$. Hence, the expression for ${\text{K}}_{\text{sp}}$ is,

$\begin{array}{c}{\text{K}}_{\text{sp}}=\text{s}\times \text{s}\\ ={\text{s}}^2\end{array}$

Substitute the value of ${\text{K}}_{\text{sp}}$ in the above expression.

$\begin{array}{c}{\text{s}}^2=5.0\times {10}^{-13}\\ \text{s}=\sqrt{5.0\times {10}^{-13}}\\ =\sqrt{50\times {10}^{-14}}\\ =7.1\times {10}^{-\text{7}}\text{mol}/\text{L}\end{array}$

Therefore, the concentration of silver ion in a saturated solution of $\text{AgBr}\left(\text{s}\right)$ is $7.1\times {10}^{-\text{7}}\text{mol}/\text{L}$.

The given salt is $\text{AgI}$. Its ${\text{K}}_{\text{sp}}$ value is given to be $8.3\times {10}^{-17}$.

The balanced chemical equation for ionization of $\text{AgI}\left(\text{s}\right)$ is shown below.

$\text{AgI}\left(\text{s}\right)\rightleftharpoons {\text{Ag}}^{+}\left(\text{a}\text{q}\right)+{\text{I}}^{-}\left(\text{a}\text{q}\right)$

The expression of ${\text{K}}_{\text{sp}}$ for the above equation is shown below.

${\text{K}}_{\text{sp}}=\left[{\text{Ag}}^{+}\right]\left[{\text{I}}^{-}\right]$

Where,

• ${\text{K}}_{\text{sp}}$ is the solubility product.
• $\left[{\text{Ag}}^{+}\right]$ is the concentration of silver ion.
• $\left[{\text{I}}^{-}\right]$ is the concentration of chlorine ion.

Both $\left[{\text{Ag}}^{+}\right]$ and $\left[{\text{I}}^{-}\right]$ are assumed to be $\text{s}$. Hence, the expression for ${\text{K}}_{\text{sp}}$ is,

$\begin{array}{c}{\text{K}}_{\text{sp}}=\text{s}\times \text{s}\\ ={\text{s}}^2\end{array}$

Substitute the value for ${\text{K}}_{\text{sp}}$ in the above expression.

$\begin{array}{c}{\text{s}}^2=8.3\times {10}^{-17}\\ \text{s}=\sqrt{8.3\times {10}^{-17}}\\ =\sqrt{83\times {10}^{-18}}\\ =9.1\times {10}^{-9}\text{mol}/\text{L}\end{array}$

Therefore, the concentration of silver ion in a saturated solution of $\text{AgI}\left(\text{s}\right)$ is $9.1\times {10}^{-9}\text{mol}/\text{L}$.

Conclusion

The concentration of silver ion in a saturated solution of $\text{AgCl}\left(\text{s}\right)$, $\text{AgBr}\left(\text{s}\right)$ and $\text{AgI}\left(\text{s}\right)$ is $1.3\times {10}^{-5}\text{mol}/\text{L}$, $7.1\times {10}^{-\text{7}}\text{mol}/\text{L}$ and $9.1\times {10}^{-9}\text{mol}/\text{L}$ respectively.