Chapter 17, Problem 15CR

Interpretation Introduction

Interpretation:

The reason as to why dissolving a slightly soluble salt to form a saturated solution is an equilibrium process is to be explained. Three balanced chemical equations for solubility processes and the expressions for ${\text{K}}_{\text{sp}}$ of the chosen reactions are to be stated. The reason as to why the concentration of the sparingly soluble salt itself is not included in the expression is to be explained. The calculation of molar solubility and the solubility in $\text{g}/\text{L}$, for a given value of solubility product, is to be explained.

Concept Introduction:

The equilibrium state of a chemical reaction is the state at which the rate of reaction going into the forward direction becomes equal to the rate of reaction going into the backward direction. A general equilibrium reaction is represented as,

$\text{A}\rightleftharpoons \text{B}+\text{C}$.

Answer to Problem 15CR

When a slightly soluble salt is dissolved to form a saturated solution, the amount of salt dissociating in aqueous medium becomes equal to the amount of salt ions associating to form solid salts. The rate of forward process become equal to rate of backward process therefore, this process is an equilibrium process.

The balanced chemical equation for solubility process of ${\text{PbCl}}_2$ is represented as,

${\text{PbCl}}_2\left(\text{s}\right)\rightleftharpoons {\text{Pb}}^{2+}\left(\text{a}\text{q}\right)+2{\text{Cl}}^{-}\left(\text{a}\text{q}\right)$

The expression for solubility product of the above equation is given as,

${\text{K}}_{\text{sp}}=\left[{\text{Pb}}^{2+}\right]{\left[{\text{Cl}}^{-}\right]}^2$

The balanced chemical equation for solubility process of ${\text{Ag}}_2{\text{CrO}}_4$ is represented as,

${\text{Ag}}_2{\text{CrO}}_4\left(\text{s}\right)\rightleftharpoons 2{\text{Ag}}^{+}\left(\text{a}\text{q}\right)+{\text{CrO}}_4{}^{2-}\left(\text{a}\text{q}\right)$

The expression for solubility product of the above equation is given as,

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

The balanced chemical equation for solubility process of ${\text{CaF}}_2$ is represented as,

${\text{CaF}}_2\left(\text{s}\right)\rightleftharpoons {\text{Ca}}^{2+}\left(\text{a}\text{q}\right)+2{\text{F}}^{-}\left(\text{a}\text{q}\right)$

The expression for solubility product of the above equation is given as,

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

The concentration of the sparingly soluble salt itself is not included in the expression because the salt is present in the solid state and its concentration is taken as unity.

The molar solubility of salt can be can be calculated as,

$\text{x}={\left(\frac{{\text{K}}_{\text{sp}}}{\text{m}+\text{n}}\right)}^{\frac{1}{\text{m}+\text{n}}}$

Where,

• ${\text{K}}_{\text{sp}}$ represents the solubility product of the salt.
• $\text{m}$ represents the stoichiometric coefficient of anion.
• $\text{n}$ represents the stoichiometric coefficient of cation.

The solubility in $\text{g}/\text{L}$ can be calculated by multiplying the molar mass of the ions with molar solubility.

Explanation of Solution

When a slightly soluble salt is dissolved to form a saturated solution, initially the salt starts dissolving into the solution till a particular point. After saturation point of the solution is reached the ions starts colliding to form solid salt. The amount of salt dissociating in aqueous medium becomes equal to the amount of salt ion associating to form solid salts. The rate of forward process become equal to rate of backward process therefore, this process is an equilibrium process. The general equilibrium reaction for dissolution of a salt in aqueous medium is represented as,

$\text{AB}\left(\text{s}\right)\rightleftharpoons {\text{A}}^{+}\left(\text{a}\text{q}\right)+{\text{B}}^{-}\left(\text{a}\text{q}\right)$

The balanced chemical equation for solubility process of ${\text{PbCl}}_2$ is represented as,

${\text{PbCl}}_2\left(\text{s}\right)\rightleftharpoons {\text{Pb}}^{2+}\left(\text{a}\text{q}\right)+2{\text{Cl}}^{-}\left(\text{a}\text{q}\right)$

The expression for solubility product of the above equation is given as,

${\text{K}}_{\text{sp}}=\left[{\text{Pb}}^{2+}\right]{\left[{\text{Cl}}^{-}\right]}^2$

The balanced chemical equation for solubility process of ${\text{Ag}}_2{\text{CrO}}_4$ is represented as,

${\text{Ag}}_2{\text{CrO}}_4\left(\text{s}\right)\rightleftharpoons 2{\text{Ag}}^{+}\left(\text{a}\text{q}\right)+{\text{CrO}}_4{}^{2-}\left(\text{a}\text{q}\right)$

The expression for solubility product of the above equation is given as,

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

The balanced chemical equation for solubility process of ${\text{CaF}}_2$ is represented as,

${\text{CaF}}_2\left(\text{s}\right)\rightleftharpoons {\text{Ca}}^{2+}\left(\text{a}\text{q}\right)+2{\text{F}}^{-}\left(\text{a}\text{q}\right)$

The expression for solubility product of the above equation is given as,

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

The concentration of the sparingly soluble salt is itself not included in the expression because the salt is present in the solid state and its concentration is taken as unity.

The chemical equation for solubility process of a salt is given as,

${\text{A}}_{\text{n}}{\text{B}}_{\text{m}}\left(\text{s}\right)\rightleftharpoons \text{n}{\text{A}}^{\text{m}+}\left(\text{a}\text{q}\right)+\text{m}{\text{B}}^{\text{n}-}\left(\text{a}\text{q}\right)$

The molar solubility of salt can be can be calculated as,

$\text{x}={\left(\frac{{\text{K}}_{\text{sp}}}{\text{m}+\text{n}}\right)}^{\frac{1}{\text{m}+\text{n}}}$

Where,

• ${\text{K}}_{\text{sp}}$ represents the solubility product of the salt.
• $\text{m}$ represents the stoichiometric coefficient of anion.
• $\text{n}$ represents the stoichiometric coefficient of cation.

The solubility in $\text{g}/\text{L}$ can be calculated by multiplying the molar mass of the ion with molar solubility.

Conclusion

When a slightly soluble salt is dissolved to form a saturated solution, the amount of salt dissociating in aqueous medium becomes equal to the amount of salt ion associating to form solid salts. The rate of forward process become equal to rate of backward process therefore, this process is an equilibrium process.

The balanced chemical equation for solubility process of ${\text{PbCl}}_2$ is represented as,

${\text{PbCl}}_2\left(\text{s}\right)\rightleftharpoons {\text{Pb}}^{2+}\left(\text{a}\text{q}\right)+2{\text{Cl}}^{-}\left(\text{a}\text{q}\right)$

The expression for solubility product of the above equation is given as,

${\text{K}}_{\text{sp}}=\left[{\text{Pb}}^{2+}\right]{\left[{\text{Cl}}^{-}\right]}^2$

The balanced chemical equation for solubility process of ${\text{Ag}}_2{\text{CrO}}_4$ is represented as,

${\text{Ag}}_2{\text{CrO}}_4\left(\text{s}\right)\rightleftharpoons 2{\text{Ag}}^{+}\left(\text{a}\text{q}\right)+{\text{CrO}}_4{}^{2-}\left(\text{a}\text{q}\right)$

The expression for solubility product of the above equation is given as,

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

The balanced chemical equation for solubility process of ${\text{CaF}}_2$ is represented as,

${\text{CaF}}_2\left(\text{s}\right)\rightleftharpoons {\text{Ca}}^{2+}\left(\text{a}\text{q}\right)+2{\text{F}}^{-}\left(\text{a}\text{q}\right)$

The expression for solubility product of the above equation is given as,

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

The concentration of the sparingly soluble salt is itself not included in the expression because the salt is present in the solid state and its concentration is taken as unity.

The molar solubility of salt can be can be calculated as,

$\text{x}={\left(\frac{{\text{K}}_{\text{sp}}}{\text{m}+\text{n}}\right)}^{\frac{1}{\text{m}+\text{n}}}$

Where,

• ${\text{K}}_{\text{sp}}$ represents the solubility product of the salt.
• $\text{m}$ represents the stoichiometric coefficient of anion.
• $\text{n}$ represents the stoichiometric coefficient of cation.

The solubility in $\text{g}/\text{L}$ can be calculated by multiplying the molar mass of the ion with molar solubility.