Chapter 6, Problem 6I.2E

(a)

Interpretation Introduction

Interpretation:

The molar solubility of $\text{AgI}$ at ${\text{25}}^{\text{o}}\text{C}$ is $\text{9}{\text{.1×10}}^{\text{-9}}\text{mol}\cdot {\text{L}}^{\text{-1}}$, the value of ${\text{K}}_{\text{sp}}$ has to be determined.

Concept Introduction:

Solubility product:

The equilibrium constant is defined as solubility product and it denoted by the symbol of ${\text{K}}_{\text{sp}}$.  It is expressed by the chemical reaction equilibrium between an ionic solid compound and its ions solubility nature in the solution.

Answer to Problem 6I.2E

The molar solubility of $\text{AgI}$ at ${\text{25}}^{\text{o}}\text{C}$ is $\text{9}{\text{.1×10}}^{\text{-9}}\text{mol}\cdot {\text{L}}^{\text{-1}}$, the value of ${\text{K}}_{\text{sp}}$ is $8.3\times {10}^{-17}$.

Explanation of Solution

The solubility equilibrium of silver iodide, chemical equation is given below.

  ${\text{AgI}}_{\text{(s)}}\rightleftharpoons {\text{Ag}}^{\text{+}}{}_{\text{(aq)}}{\text{+I}}^{\text{-}}{}_{\text{(aq)}}$

The equation for solubility product is

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

$\begin{array}{c}\text{For,}\text{1}{\text{mol Ag}}^{\text{+}}\text{=1}\text{mole}\text{AgI}\\ {\text{Ag}}^{\text{+}}\text{=s}\\ \\ \text{For,}\text{1}\text{mol}{\text{I}}^{\text{-}}\text{=1}\text{mole}\text{AgI}\\ {\text{I}}^{\text{-}}\text{=s}\end{array}$

Substitute the obtained values in below solubility product equation.

  $\begin{array}{c}{\text{K}}_{\text{sp}}{\text{=[Ag}}^{\text{+}}{\text{][I}}^{\text{-}}\text{]}\\ \text{=(s)(s)}\\ {\text{=s}}^{\text{2}}\end{array}$

The molar solubility of silver iodide is $9.1\times {10}^{-9}\text{mol}\cdot {\text{L}}^{\text{-1}}=9.1\times {10}^{-9}$

Molar solubility = s

Substitute s value in above expression,

  $\begin{array}{c}{\text{K}}_{\text{sp}}\text{=}{\text{s}}^2\\ ={(9.1\times {10}^{-9})}^2\\ =8.3\times {10}^{-17}\end{array}$

Therefore, the ${\text{K}}_{\text{sp}}$ value of silver bromide is $8.3\times {10}^{-17}$.

(b)

Interpretation Introduction

Interpretation:

The molar solubility of ${\text{Ca(OH)}}_2$ at ${\text{25}}^{\text{o}}\text{C}$ is $\text{0}\text{.011}\text{mol}\cdot {\text{L}}^{\text{-1}}$, the value of ${\text{K}}_{\text{sp}}$ has to be determined.

Concept introduction:

Refer to part (a).

Answer to Problem 6I.2E

The molar solubility of ${\text{Ca(OH)}}_2$ at ${\text{25}}^{\text{o}}\text{C}$ is $\text{0}\text{.011}\text{mol}\cdot {\text{L}}^{\text{-1}}$, the value of ${\text{K}}_{\text{sp}}$ is $5.3\times {10}^{-6}$.

Explanation of Solution

The solubility equilibrium of ${\text{Ca(OH)}}_2$ chemical equation is given below.

  ${\text{Ca(OH)}}_{\text{2}}{}_{\text{(s)}}\rightleftharpoons {\text{Ca}}^{\text{2+}}{}_{\text{(aq)}}{\text{+2OH}}^{\text{-}}{}_{\text{(aq)}}$

The equation for solubility product is

  ${\text{K}}_{\text{sp}}{\text{=[Ca}}^{\text{2+}}{\text{][OH}}^{\text{-}}{\text{]}}^{\text{2}}$

$\begin{array}{c}\text{For,}\text{1}{\text{mol Ca}}^{\text{2+}}\text{=1}\text{mole}{\text{Ca(OH)}}_2\\ {\text{Ca}}^{\text{2+}}\text{=s}\\ \\ \text{For,}\text{2}\text{mol}{\text{OH}}^{\text{-}}\text{=1}\text{mole}{\text{Ca(OH)}}_2\\ {\text{OH}}^{\text{-}}\text{=2s}\end{array}$

Substitute the obtained values in below solubility product equation.

  $\begin{array}{c}{\text{K}}_{\text{sp}}{\text{=[Ca}}^{\text{2+}}{\text{][OH}}^{\text{-}}{\text{]}}^2\\ {\text{=(s)(2s)}}^2\\ {\text{=4s}}^3\end{array}$

The molar solubility of ${\text{Ca(OH)}}_2$ is $0.011\text{mol}\cdot {\text{L}}^{\text{-1}}=0.011$

Molar solubility = s

Substitute s value in above expression,

  $\begin{array}{c}{\text{K}}_{\text{sp}}\text{=}4{\text{s}}^3\\ =4\times {(0.011)}^3\\ =5.3\times {10}^{-6}\end{array}$

Therefore, the ${\text{K}}_{\text{sp}}$ value of ${\text{Ca(OH)}}_2$ is $5.3\times {10}^{-6}$.

(c)

Interpretation Introduction

Interpretation:

The molar solubility of ${\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4}}$ at ${\text{25}}^{\text{o}}\text{C}$ is $\text{2}\text{.7}\times {\text{10}}^{-6}\text{ mol}\cdot {\text{L}}^{\text{-1}}$, the value of ${\text{K}}_{\text{sp}}$ has to be determined.

Concept introduction:

Refer to part (a).

Answer to Problem 6I.2E

The molar solubility of ${\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4}}$ at ${\text{25}}^{\text{o}}\text{C}$ is $\text{2}\text{.7}\times {\text{10}}^{-6}\text{ mol}\cdot {\text{L}}^{\text{-1}}$, the value of ${\text{K}}_{\text{sp}}$ is $1.4\times {10}^{-21}$.

Explanation of Solution

The solubility equilibrium of ${\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4}}$ chemical equation is given below.

  ${\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4}}{}_{\text{(s)}}\rightleftharpoons {\text{3Ag}}^{\text{+}}{}_{\text{(aq)}}{\text{+PO}}_{\text{4}}{{}^{\text{3-}}}_{\text{(aq)}}$

The equation for solubility product is

  ${\text{K}}_{\text{sp}}{\text{=[Ag}}^{\text{+}}{\text{]}}^3{\text{[PO}}_4{}^{\text{3-}}\text{]}$

$\begin{array}{c}\text{For,}3{\text{mol Ag}}^{\text{+}}\text{=1}\text{mole}{\text{Ag}}_3{\text{PO}}_4\\ {\text{Ag}}^{\text{+}}\text{=3s}\\ \\ \text{For,}1\text{mol}{\text{PO}}_4{}^{\text{3-}}\text{=1}\text{mole}{\text{Ag}}_3{\text{PO}}_4\\ {\text{PO}}_4{}^{\text{3-}}\text{=s}\end{array}$

Substitute the obtained values in below solubility product equation.

  $\begin{array}{c}{\text{K}}_{\text{sp}}{\text{=[Ag}}^{\text{+}}{\text{]}}^3{\text{[PO}}_4{}^{\text{3-}}\text{]}\\ {\text{=(3s)}}^3\text{(s)}\\ {\text{=27s}}^4\end{array}$

The molar solubility of ${\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4}}$ is $2.7\times {10}^{-6}\text{mol}\cdot {\text{L}}^{\text{-1}}=2.7\times {10}^{-6}$

Molar solubility = s

Substitute s value in above expression,

  $\begin{array}{c}{\text{K}}_{\text{sp}}\text{=}27{\text{s}}^4\\ =27\times {(2.7\times {10}^{-6})}^4\\ =1.4\times {10}^{-21}\end{array}$

Therefore, the ${\text{K}}_{\text{sp}}$ value of ${\text{Ba(OH)}}_{\text{2}}$ is $1.4\times {10}^{-21}$.

(d)

Interpretation Introduction

Interpretation:

The molar solubility of ${\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}$ at ${\text{25}}^{\text{o}}\text{C}$ is $5.2\times {\text{10}}^{-7}\text{ mol}\cdot {\text{L}}^{\text{-1}}$, the value of ${\text{K}}_{\text{sp}}$ has to be determined.

Concept introduction:

Refer to part (a).

Answer to Problem 6I.2E

The molar solubility of ${\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}$ at ${\text{25}}^{\text{o}}\text{C}$ is $5.2\times {\text{10}}^{-7}\text{ mol}\cdot {\text{L}}^{\text{-1}}$, the value of ${\text{K}}_{\text{sp}}$ is $5.6\times {10}^{-19}$.

Explanation of Solution

The solubility equilibrium of ${\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}$ chemical equation is given below.

  ${\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}{}_{\text{(s)}}\rightleftharpoons {\text{Hg}}_{\text{2}}{{}^{\text{2+}}}_{\text{(aq)}}{\text{+2Cl}}^{\text{-}}{}_{\text{(aq)}}$

The equation for solubility product is

  ${\text{K}}_{\text{sp}}{\text{=[Hg}}_2{}^{\text{2+}}{\text{][Cl}}^{\text{-}}{\text{]}}^{\text{2}}$

$\begin{array}{c}\text{For,}\text{1}{\text{mol Hg}}_{\text{2}}{}^{\text{2+}}\text{=1}\text{mole}{\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}\\ {\text{Hg}}_{\text{2}}{}^{\text{2+}}\text{=s}\\ \\ \text{For,}\text{2}\text{mol}{\text{Cl}}^{\text{-}}\text{=1}\text{mole}{\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}\\ {\text{Cl}}^{\text{-}}\text{=2s}\end{array}$

Substitute the obtained values in below solubility product equation.

  $\begin{array}{c}{\text{K}}_{\text{sp}}{\text{=[Hg}}_2{}^{\text{2+}}{\text{][Cl}}^{\text{-}}{\text{]}}^{\text{2}}\\ {\text{=(s)(2s)}}^{\text{2}}\\ {\text{=4s}}^{\text{3}}\end{array}$

The molar solubility of ${\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}$ is $5.2\times {10}^{-7}\text{mol}\cdot {\text{L}}^{\text{-1}}=5.2\times {10}^{-7}$

Molar solubility = s

Substitute s value in above expression,

  $\begin{array}{c}{\text{K}}_{\text{sp}}\text{=}4{\text{s}}^3\\ =4\times {(5.2\times {10}^{-7})}^3\\ =5.6\times {10}^{-19}\end{array}$

Therefore, the ${\text{K}}_{\text{sp}}$ value of ${\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}$ is $5.6\times {10}^{-19}$.