Chapter 17, Problem 17.50QP

(a)

Interpretation Introduction

Interpretation:

From the given data, the molar solubility of ${\text{PbBr}}_{\text{2}}$ dissolved in various solutions have to be calculated.

Concept introduction:

• Molar solubility is defined as amount of solute that can be dissolved in one litre of solution before it attains saturation.
• Solubility of a compound is expressed as concentration of its ions in saturated solution.
• The solubility product constant (${\text{K}}_{\text{sp}}$) is defined as the equilibrium between compound and its ions in an aqueous solution.
• Solubility product is the multiplication of concentration of dissolved ion,  raised to the power of coefficients.
• Ionic compound ${\text{A}}_{\text{3}}\text{B}$ ${\text{K}}_{\text{sp}}\text{=}{\left[\text{A}\right]}^{\text{3}}\left[\text{B}\right]$.
• When the concentration of one of the ions of a chemical solution got higher, it reacts with counter charged ions and precipitated out as salt till the ion product equals solubility product is called common ion effect.

To calculate: The molar solubility of ${\text{PbBr}}_{\text{2}}$ in pure water.

Answer to Problem 17.50QP

The molar solubility of ${\text{PbBr}}_{\text{2}}$ in pure water is $\text{1}\text{.3}\text{×}{\text{10}}^{\text{-2}}\text{M}$

Explanation of Solution

Given,

The ${\text{K}}_{\text{sp}}\text{of}{\text{PbBr}}_{\text{2}}\text{is}\text{8}\text{.9}\text{×}{\text{10}}^{\text{-6}}$

Strength of $\text{KBr}\text{is}\text{0}\text{.20M}$

Strength of ${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{is}\text{0}\text{.20M}$

  $\begin{array}{l}{\text{PbBr}}_{\text{2}}{}_{\left(\text{s}\right)}\to {\text{Pb}}^{\text{2+}}{}_{\left(\text{aq}\right)}\text{ + }{\text{2Br}}^{\text{-}}{}_{\left(\text{aq}\right)}\\ \text{Initial concentration }\left(\text{M}\right)\text{: }\text{0 }\text{0}\\ \text{Change in concentration }\left(\text{M}\right)\text{: }-s\text{ +s +2s}\\ \text{Equilibrium concentration}\left(\text{M}\right)\text{: }\text{s}\text{+2s}\\ \\ \text{s}\text{is}\text{molar}\text{solubility}\\ {\text{K}}_{\text{sp}}\text{ = }\left[{\text{Pb}}^{\text{2+}}\right]{\left[{\text{Br}}^{\text{-}}\right]}^{\text{2}}\\ \text{8}{\text{.9×10}}^{\text{-}}{}^{\text{6}}\text{= }\left(\text{s}\right){\text{(2s)}}^{\text{2}}\\ \text{8}{\text{.9×10}}^{\text{-}}{}^{\text{6}}\text{=}{\text{4s}}^{\text{3}}\\ \text{s}\text{=molar solubility = 0}\text{.013 M}\text{or}\text{1}\text{.3}\text{×}{\text{10}}^{\text{-2}}\text{M}\end{array}$

The molar solubility can be calculated using solubility product constant.  From the concentrations of lead and bromine ions obtained in equilibrium table and substituting in solubility product expression, the molar solubility is determined.

(b)

Interpretation Introduction

Interpretation:

From the given data, the molar solubility of ${\text{PbBr}}_{\text{2}}$ dissolved in various solutions have to be calculated.

Concept introduction:

• Molar solubility is defined as amount of solute that can be dissolved in one litre of solution before it attains saturation.
• Solubility of a compound is expressed as concentration of its ions in saturated solution.
• The solubility product constant (${\text{K}}_{\text{sp}}$) is defined as the equilibrium between compound and its ions in an aqueous solution.
• Solubility product is the multiplication of concentration of dissolved ion,  raised to the power of coefficients.
• Ionic compound ${\text{A}}_{\text{3}}\text{B}$ ${\text{K}}_{\text{sp}}\text{=}{\left[\text{A}\right]}^{\text{3}}\left[\text{B}\right]$.
• When the concentration of one of the ions of a chemical solution got higher, it reacts with counter charged ions and precipitated out as salt till the ion product equals solubility product is called common ion effect.

To calculate: The molar solubility of ${\text{PbBr}}_{\text{2}}$ in $\text{0}\text{.20 M KBr}$.

Answer to Problem 17.50QP

The molar solubility of ${\text{PbBr}}_{\text{2}}$ in $\text{0}\text{.20 M KBr}$ is $\text{2}\text{.2}\text{×}{\text{10}}^{\text{-4}}\text{M}$

Explanation of Solution

  $\begin{array}{l}{\text{PbBr}}_{\text{2}}{}_{\left(\text{s}\right)}\to {\text{Pb}}^{\text{2+}}{}_{\left(\text{aq}\right)}\text{+ }{\text{2Br}}^{\text{-}}{}_{\left(\text{aq}\right)}\\ \text{Initial concentration }\left(\text{M}\right)\text{: }\text{0 }\text{0}\\ \text{Change in concentration }\left(\text{M}\right)\text{: }-\text{s +s }\text{+2s}\\ \text{Equilibrium concentration}\left(\text{M}\right)\text{: }\text{s}\text{0}\text{.20+2s}\\ \\ \text{s}\text{is}\text{molar}\text{solubility}\\ {\text{K}}_{\text{sp}}\text{ = }\left[{\text{Pb}}^{\text{2+}}\right]{\left[{\text{Br}}^{\text{-}}\right]}^{\text{2}}\\ \text{8}{\text{.9×10}}^{\text{-}}{}^{\text{6}}\text{= }\left(\text{s}\right){\text{(0}\text{.20+2s)}}^{\text{2}}\\ \text{8}{\text{.9×10}}^{\text{-}}{}^{\text{6}}\text{=}\left(\text{s}\right){\text{(0}\text{.20)}}^{\text{2}}\\ \text{s}\text{=molar solubility}\text{=}\text{2}\text{.2}\text{×}{\text{10}}^{\text{-4}}\text{M}\end{array}$

The molar solubility can be calculated using solubility product constant.  The molar solubility of ${\text{PbBr}}_{\text{2}}$ in $\text{0}\text{.20 M KBr}$ is less than in pure water because of common ion (bromide ions)

(c)

Interpretation Introduction

Interpretation:

From the given data, the molar solubility of ${\text{PbBr}}_{\text{2}}$ dissolved in various solutions have to be calculated.

Concept introduction:

• Molar solubility is defined as amount of solute that can be dissolved in one litre of solution before it attains saturation.
• Solubility of a compound is expressed as concentration of its ions in saturated solution.
• The solubility product constant (${\text{K}}_{\text{sp}}$) is defined as the equilibrium between compound and its ions in an aqueous solution.
• Solubility product is the multiplication of concentration of dissolved ion,  raised to the power of coefficients.
• Ionic compound ${\text{A}}_{\text{3}}\text{B}$ ${\text{K}}_{\text{sp}}\text{=}{\left[\text{A}\right]}^{\text{3}}\left[\text{B}\right]$.
• When the concentration of one of the ions of a chemical solution got higher, it reacts with counter charged ions and precipitated out as salt till the ion product equals solubility product is called common ion effect.

The molar solubility of ${\text{PbBr}}_{\text{2}}$ in $\text{0}{\text{.20 M Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{.}$

Answer to Problem 17.50QP

The molar solubility of ${\text{PbBr}}_{\text{2}}$ in $\text{0}{\text{.20 M Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is $\text{3}\text{.3}\text{×}{\text{10}}^{\text{-3}}\text{M}$

Explanation of Solution

  $\begin{array}{l}{\text{PbBr}}_{\text{2}}{}_{\left(\text{s}\right)}\to {\text{Pb}}^{\text{2+}}{}_{\left(\text{aq}\right)}\text{ + }{\text{2Br}}^{\text{-}}{}_{\left(\text{aq}\right)}\\ \text{Initial concentration }\left(\text{M}\right)\text{:}\text{0}\text{.20 }\text{0 }\\ \text{Change in concentration }\left(\text{M}\right)\text{: }\text{-s +s }\text{+2s}\\ \text{Equilibrium concentration}\left(\text{M}\right)\text{: }\text{0}\text{.20+2s}\text{+2s}\\ \\ \text{s}\text{is}\text{molar}\text{solubility}\\ {\text{K}}_{\text{sp}}\text{ = }\left[{\text{Pb}}^{\text{2+}}\right]{\left[{\text{Br}}^{\text{-}}\right]}^{\text{2}}\\ \text{8}{\text{.9×10}}^{\text{-}}{}^{\text{6}}\text{= }\left(\text{s}\right){\text{(0}\text{.20+2s)}}^{\text{2}}\\ \text{8}{\text{.9×10}}^{\text{-}}{}^{\text{6}}\text{=}\left(\text{0}\text{.20}\right){\text{(2s)}}^{\text{2}}\\ \text{s}\text{=molar solubility}\text{=}\text{3}\text{.3}\text{×}{\text{10}}^{\text{-3}}\text{M}\end{array}$

The molar solubility can be calculated using solubility product constant.  The molar solubility of ${\text{PbBr}}_{\text{2}}$ in $\text{0}{\text{.20 M Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is less than in pure water because of common ion (lead ions)