Chapter 6, Problem 6I.1E

(a)

Interpretation Introduction

Interpretation:

The molar solubility of $\text{AgBr}$ at ${\text{25}}^{\text{o}}\text{C}$ is $\text{8}{\text{.8×10}}^{\text{-7}}\text{mol}\cdot {\text{L}}^{\text{-1}}$, the value of ${\text{K}}_{\text{sp}}$ has to be determined.

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.1E

The molar solubility of $\text{AgBr}$ at ${\text{25}}^{\text{o}}\text{C}$ is $\text{8}{\text{.8×10}}^{\text{-7}}\text{mol}\cdot {\text{L}}^{\text{-1}}$, the value of ${\text{K}}_{\text{sp}}$ is $7.7\times {10}^{-13}$

Explanation of Solution

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

  ${\text{AgBr}}_{\text{(s)}}\rightleftharpoons {\text{Ag}}^{\text{+}}{}_{\text{(aq)}}{\text{+Cl}}^{\text{-}}{}_{\text{(aq)}}$

The equation for solubility product is

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

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

Substitute the obtained values in below solubility product equation.

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

The molar solubility of silver bromide is $\text{8}\text{.8}\times {10}^{-7}\text{mol}\cdot {\text{L}}^{\text{-1}}=8.8\times {10}^{-7}$

Molar solubility = s

Substitute s value in above expression,

  $\begin{array}{c}{\text{K}}_{\text{sp}}\text{=}{\text{s}}^2\\ ={(8.8\times {10}^{-7})}^2\\ =7.7\times {10}^{-13}\end{array}$

Therefore, the ${\text{K}}_{\text{sp}}$ value of silver bromide is $7.7\times {10}^{-13}$.

(b)

Interpretation Introduction

Interpretation:

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

Concept introduction:

Refer to part (a).

Answer to Problem 6I.1E

The molar solubility of ${\text{PbCrO}}_{\text{4}}$ at ${\text{25}}^{\text{o}}\text{C}$ is $\text{1}{\text{.3×10}}^{\text{-7}}\text{mol}\cdot {\text{L}}^{\text{-1}}$, the value of ${\text{K}}_{\text{sp}}$ is $1.7\times {10}^{-14}$.

Explanation of Solution

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

  ${\text{PbCrO}}_{\text{4}}{}_{\text{(s)}}\rightleftharpoons {\text{Pb}}^{\text{2+}}{}_{\text{(aq)}}{\text{+CrO}}_4{{}^{\text{2-}}}_{\text{(aq)}}$

The equation for solubility product is

  ${\text{K}}_{\text{sp}}{\text{=[Pb}}^{\text{2+}}{\text{][CrO}}_4{}^{\text{2-}}\text{]}$

$\begin{array}{c}\text{For,}\text{1}{\text{mol Pb}}^{\text{2+}}\text{=1}\text{mole}{\text{PbCrO}}_{\text{4}}\\ {\text{Pb}}^{\text{2+}}\text{=s}\\ \\ \text{For,}\text{1}\text{mol}{\text{CrO}}_{\text{4}}{}^{\text{2-}}\text{=1}\text{mole}{\text{CrO}}_{\text{4}}\\ {\text{CrO}}_{\text{4}}{}^{\text{2-}}\text{=s}\end{array}$

Substitute the obtained values in below solubility product equation.

  $\begin{array}{c}{\text{K}}_{\text{sp}}{\text{=[Pb}}^{\text{2+}}{\text{][CrO}}_4{}^{\text{2-}}\text{]}\\ \text{=(s)(s)}\\ {\text{=s}}^{\text{2}}\end{array}$

The molar solubility of silver bromide is $\text{1}\text{.3}\times {10}^{-7}\text{mol}\cdot {\text{L}}^{\text{-1}}=1.3\times {10}^{-7}$

Molar solubility = s

Substitute s value in above expression,

  $\begin{array}{c}{\text{K}}_{\text{sp}}\text{=}{\text{s}}^2\\ ={(1.3\times {10}^{-7})}^2\\ =1.7\times {10}^{-14}\end{array}$

Therefore, the ${\text{K}}_{\text{sp}}$ value of silver bromide is $1.7\times {10}^{-14}$.

(c)

Interpretation Introduction

Interpretation:

The molar solubility of ${\text{Ba(OH)}}_{\text{2}}$ at ${\text{25}}^{\text{o}}\text{C}$ is $0.11\text{mol}\cdot {\text{L}}^{\text{-1}}$, the value of ${\text{K}}_{\text{sp}}$ has to be determined.

Concept introduction:

Refer to part (a).

Answer to Problem 6I.1E

The molar solubility of ${\text{Ba(OH)}}_{\text{2}}$ at ${\text{25}}^{\text{o}}\text{C}$ is $0.11\text{mol}\cdot {\text{L}}^{\text{-1}}$, the value of ${\text{K}}_{\text{sp}}$ is $5.3\times {10}^{-3}$.

Explanation of Solution

The solubility equilibrium of ${\text{Ba(OH)}}_{\text{2}}$ chemical equation is given below.

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

The equation for solubility product is

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

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

Substitute the obtained values in below solubility product equation.

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

The molar solubility of ${\text{Ba(OH)}}_{\text{2}}$ is $0.11\text{mol}\cdot {\text{L}}^{\text{-1}}=0.11$

Molar solubility = s

Substitute s value in above expression,

  $\begin{array}{c}{\text{K}}_{\text{sp}}\text{=}4{\text{s}}^3\\ =4\times {(0.11)}^3\\ =0.0053\\ =5.3\times {10}^{-3}\end{array}$

Therefore, the ${\text{K}}_{\text{sp}}$ value of ${\text{Ba(OH)}}_{\text{2}}$ is $5.3\times {10}^{-3}$.

(d)

Interpretation Introduction

Interpretation:

The molar solubility of ${\text{MgF}}_{\text{2}}$ at ${\text{25}}^{\text{o}}\text{C}$ is $1.2\times {10}^{-3}\text{mol}\cdot {\text{L}}^{\text{-1}}$, the value of ${\text{K}}_{\text{sp}}$ has to be determined.

Concept introduction:

Refer to part (a).

Answer to Problem 6I.1E

The molar solubility of ${\text{MgF}}_{\text{2}}$ at ${\text{25}}^{\text{o}}\text{C}$ is $1.2\times {10}^{-3}\text{mol}\cdot {\text{L}}^{\text{-1}}$, the value of ${\text{K}}_{\text{sp}}$ is $6.9\times {10}^{-9}$.

Explanation of Solution

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

  ${\text{MgF}}_{\text{2}}{}_{\text{(s)}}\rightleftharpoons {\text{Mg}}^{\text{2+}}{}_{\text{(aq)}}{\text{+2F}}^{\text{-}}{}_{\text{(aq)}}$

The equation for solubility product is

  ${\text{K}}_{\text{sp}}{\text{=[Mg}}^{\text{2+}}{\text{][F}}^{\text{-}}{\text{]}}^{\text{2}}$

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

Substitute the obtained values in below solubility product equation.

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

The molar solubility of ${\text{MgF}}_{\text{2}}$ is $1.2\times {10}^{-3}\text{mol}\cdot {\text{L}}^{\text{-1}}=1.2\times {10}^{-3}$

Molar solubility = s

Substitute s value in above expression,

  $\begin{array}{c}{\text{K}}_{\text{sp}}\text{=}4{\text{s}}^3\\ =4\times {(1.2\times {10}^{-3})}^3\\ =6.9\times {10}^{-9}\end{array}$

Therefore, the ${\text{K}}_{\text{sp}}$ value of ${\text{Ba(OH)}}_{\text{2}}$ is $6.9\times {10}^{-9}$.