Chapter 16, Problem 19PE

(a)

Interpretation Introduction

Interpretation:

Concentration ${\text{H}}^{+}$ for solution of $1.2M$${\text{H}}_2{\text{CO}}_3$ has to be calculated.

Concept Introduction:

Consider ionization of a weak acid $\text{HA}$ that is in equilibrium with its ions and is as follows:

  $\text{HA}\left(aq\right)\rightleftharpoons {\text{H}}^{\text{+}}\left(aq\right)+{\text{A}}^{-}\left(aq\right)$

The equilibrium constant for ionization of weak acid is called ionization constant $K_a$ and is expressed as follows:

  $K_a=\frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$

$K_a$ is defined as the ratio of concentration of products to concentration of reactants.

Explanation of Solution

The ionization of ${\text{H}}_2{\text{CO}}_3$ is as follows:

  ${\text{H}}_2{\text{CO}}_3\left(aq\right)\rightleftharpoons 2{\text{H}}^{\text{+}}\left(aq\right)+{\text{CO}}_3^{2-}\left(aq\right)$

The expression for $K_a$ for given reaction is as follows:

  $K_a=\frac{{\left[{\text{H}}^{\text{+}}\right]}^2\left[{\text{CO}}_3^{2-}\right]}{\left[{\text{H}}_2{\text{CO}}_3\right]}$        (1)

Through chemical equation it is evident ionization of ${\text{H}}_2{\text{CO}}_3$ produces two moles of ${\text{H}}^{\text{+}}$ and one mole of ${\text{CO}}_3^{2-}$. Therefore, concentration of ${\text{H}}^{\text{+}}$ doubles and hence, consider $\text{2x}$ for ${\text{H}}^{\text{+}}$ and $\text{x}$ for ${\text{CO}}_3^{2-}$. Value of $\text{x}$ is small so concentration of ${\text{H}}_2{\text{CO}}_3$ can be considered as $1.2M$.

Rearrange equation (1) for $\left[{\text{H}}^{\text{+}}\right]$.

  ${\left[{\text{H}}^{\text{+}}\right]}^2=\frac{K_a\left[{\text{H}}_2{\text{CO}}_3\right]}{\left[{\text{CO}}_3^{2-}\right]}$        (2)

Substitute $\text{x}$ for $\left[{\text{CO}}_3^{2-}\right]$, $\text{2x}$ for $\left[{\text{H}}^{\text{+}}\right]$, $4.4\times {10}^{-7}$ for $K_a$ and $1.2$ for $\left[{\text{H}}_2{\text{CO}}_3\right]$ in equation (2).

  $\begin{array}{c}{\left(\text{2x}\right)}^2=\frac{\left(4.4\times {10}^{-7}\right)\left(1.2\right)}{\text{x}}\\ {\text{4x}}^3=\left(4.4\times {10}^{-7}\right)\left(1.2\right)\end{array}$

Solve this equation for value of $\text{x}$.

  $\text{x}=0.005$

Hence, $\left[{\text{H}}^{\text{+}}\right]$ in $1.2M$${\text{H}}_2{\text{CO}}_3$ is. $0.005M$.

(b)

Interpretation Introduction

Interpretation:

$\text{pH}$ for solution of $1.2M{\text{ H}}_2{\text{CO}}_3$ has to be calculated.

Concept Introduction:

$\text{pH}$ is defined as the concentration of hydrogen ion. It also explains about the acidity of solution.

Explanation of Solution

The ionization of ${\text{H}}_2{\text{CO}}_3$ is as follows:

  ${\text{H}}_2{\text{CO}}_3\left(aq\right)\rightleftharpoons 2{\text{H}}^{\text{+}}\left(aq\right)+{\text{CO}}_3^{2-}\left(aq\right)$

The expression for $K_a$ for given reaction is as follows:

  $K_a=\frac{{\left[{\text{H}}^{\text{+}}\right]}^2\left[{\text{CO}}_3^{2-}\right]}{\left[{\text{H}}_2{\text{CO}}_3\right]}$        (1)

Through chemical equation it is evident ionization of ${\text{H}}_2{\text{CO}}_3$ produces two moles of ${\text{H}}^{\text{+}}$ and one mole of ${\text{CO}}_3^{2-}$. Therefore, concentration of ${\text{H}}^{\text{+}}$ doubles and hence, consider $\text{2x}$ for ${\text{H}}^{\text{+}}$ and $\text{x}$ for ${\text{CO}}_3^{2-}$. Value of $\text{x}$ is small so concentration of ${\text{H}}_2{\text{CO}}_3$ can be considered as $1.2M$.

Rearrange equation (1) for $\left[{\text{H}}^{\text{+}}\right]$.

  ${\left[{\text{H}}^{\text{+}}\right]}^2=\frac{K_a\left[{\text{H}}_2{\text{CO}}_3\right]}{\left[{\text{CO}}_3^{2-}\right]}$        (2)

Substitute $\text{x}$ for $\left[{\text{CO}}_3^{2-}\right]$, $\text{2x}$ for $\left[{\text{H}}^{\text{+}}\right]$, $4.4\times {10}^{-7}$ for $K_a$ and $1.2$ for $\left[{\text{H}}_2{\text{CO}}_3\right]$ in equation (2).

  $\begin{array}{c}{\left(\text{2x}\right)}^2=\frac{\left(4.4\times {10}^{-7}\right)\left(1.2\right)}{\text{x}}\\ {\text{4x}}^3=\left(4.4\times {10}^{-7}\right)\left(1.2\right)\end{array}$

Solve this equation for value of $\text{x}$.

  $\text{x}=0.005$

Hence, $\left[{\text{H}}^{\text{+}}\right]$ in $1.2M$${\text{H}}_2{\text{CO}}_3$ is. $0.005M$.

$\text{pH}$ is defined as negative logarithm of concentration of ${\text{H}}^{+}$. It is calculated as follows:

  $\text{pH}=-\log \left[{\text{H}}^{\text{+}}\right]$        (3)

Substitute $0.005M$ for ${\text{H}}^{+}$ in equation (3).

  $\begin{array}{c}\text{pH}=-\log \left[0.005\right]\\ =2.30\end{array}$

Hence, $\text{pH}$ in $1.2M$${\text{H}}_2{\text{CO}}_3$ is $2.30$.

(c)

Interpretation Introduction

Interpretation:

Percent ionization of solution of $1.2M{\text{ H}}_2{\text{CO}}_3$ has to be calculated.

Concept Introduction:

Consider a weak acid $\text{HA}$ that is in equilibrium with its ions and is as follows:

  $\text{HA}\left(aq\right)\rightleftharpoons {\text{H}}^{\text{+}}\left(aq\right)+{\text{A}}^{-}\left(aq\right)$

Percent ionization for a weak acid is defined as the ratio of concentration of ${\text{H}}^{\text{+}}$ or ${\text{A}}^{-}$ to initial concentration of $\text{HA}$ that is multiplied with 100 and is expressed as follows:

  $\text{Percent ionization}=\left(\frac{\left[{\text{H}}^{\text{+}}\right]\text{or}\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}\right)\cdot 100$

Explanation of Solution

The ionization of ${\text{H}}_2{\text{CO}}_3$ is as follows:

  ${\text{H}}_2{\text{CO}}_3\left(aq\right)\rightleftharpoons 2{\text{H}}^{\text{+}}\left(aq\right)+{\text{CO}}_3^{2-}\left(aq\right)$

The expression for $K_a$ for given reaction is as follows:

  $K_a=\frac{{\left[{\text{H}}^{\text{+}}\right]}^2\left[{\text{CO}}_3^{2-}\right]}{\left[{\text{H}}_2{\text{CO}}_3\right]}$        (1)

Through chemical equation it is evident ionization of ${\text{H}}_2{\text{CO}}_3$ produces two moles of ${\text{H}}^{\text{+}}$ and one mole of ${\text{CO}}_3^{2-}$. Therefore, concentration of ${\text{H}}^{\text{+}}$ doubles and hence, consider $\text{2x}$ for ${\text{H}}^{\text{+}}$ and $\text{x}$ for ${\text{CO}}_3^{2-}$. Value of $\text{x}$ is small so concentration of ${\text{H}}_2{\text{CO}}_3$ can be considered as $1.2M$.

Rearrange equation (1) for $\left[{\text{H}}^{\text{+}}\right]$.

  ${\left[{\text{H}}^{\text{+}}\right]}^2=\frac{K_a\left[{\text{H}}_2{\text{CO}}_3\right]}{\left[{\text{CO}}_3^{2-}\right]}$        (2)

Substitute $\text{x}$ for $\left[{\text{CO}}_3^{2-}\right]$, $\text{2x}$ for $\left[{\text{H}}^{\text{+}}\right]$, $4.4\times {10}^{-7}$ for $K_a$ and $1.2$ for $\left[{\text{H}}_2{\text{CO}}_3\right]$ in equation (2).

  $\begin{array}{l}{\left(\text{x}\right)}^2=\frac{\left(4.4\times {10}^{-7}\right)\cdot \left(1.2\right)}{\text{x}}\\ {\text{4x}}^3=\left(4.4\times {10}^{-7}\right)\cdot \left(1.2\right)\end{array}$

Solve this equation for value of $\text{x}$.

  $\text{x}=0.005$

Hence, $\left[{\text{H}}^{\text{+}}\right]$ in $1.2M$${\text{H}}_2{\text{CO}}_3$ is. $0.005M$.

The percent ionization of ${\text{H}}_2{\text{CO}}_3$ is calculated as follows:

  $\begin{array}{c}\text{Percent ionization}=\left(\frac{\left[{\text{H}}^{\text{+}}\right]}{\left[{\text{HC}}_3{\text{H}}_5{\text{O}}_2\right]}\right)\cdot 100\\ =\left(\frac{0.005M}{1.2M}\right)\cdot 100\\ =0.42\text{ \%}\end{array}$

Hence, percent ionization of ${\text{H}}_2{\text{CO}}_3$ is $0.42\text{ \%}$.