Chapter 17, Problem 90AP

Interpretation Introduction

(a)

Interpretation:

The ionization equation of the given sparingly soluble salt is to be stated and the solubility of the salt in $\text{mol/L}$ is to be calculated.

Concept Introduction:

The solubility product is represented by ${\text{K}}_{\text{sp}}$. The solubility product describes the solubility of the solid substance in the solvent at the particular temperature. The high value of solubility product indicates that the substance will be more soluble in water, whereas the low value of solubility product is the indication of the low solubility of the solid in the solvent.

Answer to Problem 90AP

The ionization reaction of ${\text{BaCO}}_{\text{3}}$ is,

${\text{BaCO}}_{\text{3}}\left(\text{s}\right)\rightleftharpoons {\text{Ba}}^{2+}\left(\text{a}\text{q}\right)+{\text{CO}}_3^{2-}\left(\text{a}\text{q}\right)$

The solubility of ${\text{BaCO}}_{\text{3}}$ is $7.14\times {10}^{-5}\text{ mol/L}$.

Explanation of Solution

The given sparingly soluble substance is ${\text{BaCO}}_{\text{3}}$. On dissolution in water, salts dissociate to form metallic cations and hydroxide anions. The ionization reaction of ${\text{BaCO}}_{\text{3}}$ is,

${\text{BaCO}}_{\text{3}}\left(\text{s}\right)\rightleftharpoons {\text{Ba}}^{2+}\left(\text{a}\text{q}\right)+{\text{CO}}_3^{2-}\left(\text{a}\text{q}\right)$

The solubility of the salt is assumed to be $\text{s}\text{ mol/L}$.

According to the law of chemical equilibrium, the solubility product for ${\text{BaCO}}_{\text{3}}$ is,

$\begin{array}{c}{\text{K}}_{\text{sp}}=\left[{\text{Ba}}^{2+}\right]\left[{\text{CO}}_3^{2-}\right]\\ =\text{s}\times \text{s}\\ ={\text{s}}^2\end{array}$

The salt, ${\text{BaCO}}_{\text{3}}$ is not included in the equilibrium expression because it is a pure solid.

The solubility product, ${\text{K}}_{\text{sp}}$ for ${\text{BaCO}}_{\text{3}}$ is $5.1\times {10}^{-9}$.

Substitute the value of solubility product in the above expression.

$\begin{array}{c}{\text{K}}_{\text{sp}}={\text{s}}^2\\ 5.1\times {10}^{-9}={\text{s}}^2\\ \text{s}=\sqrt{5.1\times {10}^{-9}}\\ \text{s}=7.14\times {10}^{-5}\text{ mol/L}\end{array}$

The solubility of ${\text{BaCO}}_{\text{3}}$ is $7.14\times {10}^{-5}\text{ mol/L}$.

Interpretation Introduction

(b)

Interpretation:

The ionization equation of the given sparingly soluble salt is to be stated and the solubility of the salt in $\text{mol/L}$ is to be calculated.

Concept Introduction:

The solubility product is represented by ${\text{K}}_{\text{sp}}$. The solubility product describes the solubility of the solid substance in the solvent at the particular temperature. The high value of solubility product indicates that the substance will be more soluble in water, whereas the low value of solubility product is the indication of the low solubility of the solid in the solvent.

Answer to Problem 90AP

The ionization reaction of ${\text{CdCO}}_{\text{3}}$ is,

${\text{CdCO}}_{\text{3}}\left(\text{s}\right)\rightleftharpoons {\text{Cd}}^{2+}\left(\text{a}\text{q}\right)+{\text{CO}}_3^{2-}\left(\text{a}\text{q}\right)$

The solubility of ${\text{CdCO}}_{\text{3}}$ is $2.28\times {10}^{-6}\text{ mol/L}$.

Explanation of Solution

The given sparingly soluble substance is ${\text{CdCO}}_{\text{3}}$. On dissolution in water, salts dissociate to form metallic cations and hydroxide anions. The ionization reaction of ${\text{CdCO}}_{\text{3}}$ is,

${\text{CdCO}}_{\text{3}}\left(\text{s}\right)\rightleftharpoons {\text{Cd}}^{2+}\left(\text{a}\text{q}\right)+{\text{CO}}_3^{2-}\left(\text{a}\text{q}\right)$

The solubility of the salt is assumed to be $\text{s}\text{ mol/L}$.

According to the law of chemical equilibrium, the solubility product for ${\text{CdCO}}_{\text{3}}$ is,

$\begin{array}{c}{\text{K}}_{\text{sp}}=\left[{\text{Cd}}^{2+}\right]\left[{\text{CO}}_3^{2-}\right]\\ =\text{s}\times \text{s}\\ ={\text{s}}^2\end{array}$

The salt, ${\text{CdCO}}_{\text{3}}$ is not included in the equilibrium expression because it is a pure solid.

The solubility product, ${\text{K}}_{\text{sp}}$ for ${\text{CdCO}}_{\text{3}}$ is $5.2\times {10}^{-12}$.

Substitute the value of solubility product in the above expression.

$\begin{array}{c}{\text{K}}_{\text{sp}}={\text{s}}^2\\ 5.2\times {10}^{-12}={\text{s}}^2\\ \text{s}=\sqrt{5.2\times {10}^{-12}}\\ \text{s}=2.28\times {10}^{-6}\text{ mol/L}\end{array}$

The solubility of ${\text{CdCO}}_{\text{3}}$ is $2.28\times {10}^{-6}\text{ mol/L}$.

Interpretation Introduction

(c)

Interpretation:

The ionization equation of the given sparingly soluble salt is to be stated and the solubility of the salt in $\text{mol/L}$ is to be calculated.

Concept Introduction:

The solubility product is represented by ${\text{K}}_{\text{sp}}$. The solubility product describes the solubility of the solid substance in the solvent at the particular temperature. The high value of solubility product indicates that the substance will be more soluble in water, whereas the low value of solubility product is the indication of the low solubility of the solid in the solvent.

Answer to Problem 90AP

The ionization reaction of ${\text{CaCO}}_{\text{3}}$ is,

${\text{CaCO}}_{\text{3}}\left(\text{s}\right)\rightleftharpoons {\text{Ca}}^{2+}\left(\text{a}\text{q}\right)+{\text{CO}}_3^{2-}\left(\text{a}\text{q}\right)$

The solubility of ${\text{CaCO}}_{\text{3}}$ is $5.3\times {10}^{-5}\text{ mol/L}$.

Explanation of Solution

The given sparingly soluble substance is ${\text{CaCO}}_{\text{3}}$. On dissolution in water, salts dissociate to form metallic cations and hydroxide anions. The ionization reaction of ${\text{CaCO}}_{\text{3}}$ is,

${\text{CaCO}}_{\text{3}}\left(\text{s}\right)\rightleftharpoons {\text{Ca}}^{2+}\left(\text{a}\text{q}\right)+{\text{CO}}_3^{2-}\left(\text{a}\text{q}\right)$

The solubility of the salt is assumed to be $\text{s}\text{ mol/L}$.

According to the law of chemical equilibrium, the solubility product for ${\text{CaCO}}_{\text{3}}$ is,

$\begin{array}{c}{\text{K}}_{\text{sp}}=\left[{\text{Ca}}^{2+}\right]\left[{\text{CO}}_3^{2-}\right]\\ =\text{s}\times \text{s}\\ ={\text{s}}^2\end{array}$

The salt, ${\text{CaCO}}_{\text{3}}$ is not included in the equilibrium expression because it is a pure solid.

The solubility product, ${\text{K}}_{\text{sp}}$ for ${\text{CaCO}}_{\text{3}}$ is $2.8\times {10}^{-9}$.

Substitute the value of solubility product in the above expression.

$\begin{array}{c}{\text{K}}_{\text{sp}}={\text{s}}^2\\ 2.8\times {10}^{-9}={\text{s}}^2\\ \text{s}=\sqrt{2.8\times {10}^{-9}}\\ \text{s}=5.3\times {10}^{-5}\text{ mol/L}\end{array}$

The solubility of ${\text{CaCO}}_{\text{3}}$ is $5.3\times {10}^{-5}\text{ mol/L}$.

Interpretation Introduction

(d)

Interpretation:

The ionization equation of the given sparingly soluble salt is to be stated and the solubility of the salt in $\text{mol/L}$ is to be calculated.

Concept Introduction:

The solubility product is represented by ${\text{K}}_{\text{sp}}$. The solubility product describes the solubility of the solid substance in the solvent at the particular temperature. The high value of solubility product indicates that the substance will be more soluble in water, whereas the low value of solubility product is the indication of the low solubility of the solid in the solvent.

Answer to Problem 90AP

The ionization reaction of ${\text{CoCO}}_{\text{3}}$ is,

${\text{CoCO}}_{\text{3}}\left(\text{s}\right)\rightleftharpoons {\text{Co}}^{2+}\left(\text{a}\text{q}\right)+{\text{CO}}_3^{2-}\left(\text{a}\text{q}\right)$

The solubility of ${\text{CoCO}}_{\text{3}}$ is $3.87\times {10}^{-7}\text{ mol/L}$.

Explanation of Solution

The given sparingly soluble substance is ${\text{CoCO}}_{\text{3}}$. On dissolution in water, salts dissociate to form metallic cations and hydroxide anions. The ionization reaction of ${\text{CoCO}}_{\text{3}}$ is,

${\text{CoCO}}_{\text{3}}\left(\text{s}\right)\rightleftharpoons {\text{Co}}^{2+}\left(\text{a}\text{q}\right)+{\text{CO}}_3^{2-}\left(\text{a}\text{q}\right)$

The solubility of the salt is assumed to be $\text{s}\text{ mol/L}$.

According to the law of chemical equilibrium, the solubility product for ${\text{CoCO}}_{\text{3}}$ is,

$\begin{array}{c}{\text{K}}_{\text{sp}}=\left[{\text{Co}}^{2+}\right]\left[{\text{CO}}_3^{2-}\right]\\ =\text{s}\times \text{s}\\ ={\text{s}}^2\end{array}$

The salt, ${\text{CoCO}}_{\text{3}}$ is not included in the equilibrium expression because it is a pure solid.

The solubility product, ${\text{K}}_{\text{sp}}$ for ${\text{CoCO}}_{\text{3}}$ is $1.5\times {10}^{-13}$.

Substitute the value of solubility product in the above expression.

$\begin{array}{c}{\text{K}}_{\text{sp}}={\text{s}}^2\\ 1.5\times {10}^{-13}={\text{s}}^2\\ \text{s}=\sqrt{1.5\times {10}^{-13}}\\ \text{s}=3.87\times {10}^{-7}\text{ mol/L}\end{array}$

The solubility of ${\text{CoCO}}_{\text{3}}$ is $3.87\times {10}^{-7}\text{ mol/L}$.