Chapter 15, Problem 15.67QE

(a)

Interpretation Introduction

Interpretation:

The fraction of acid ionized in $0.33M$${\text{HNO}}_2$ solution has to be determined.

Concept Introduction:

A weak acid in water produces a hydrogen ion and conjugate base. When weak acid dissolves in water, some acid molecules transfer proton to water.

In solution of weak acid, the actual concentration of the acid molecules becomes less because partial dissociation of acid has occurred and lost protons to form hydrogen ions.

The reaction is as follows:

  $\text{HA}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(\text{aq}\right)+{\text{A}}^{-}\left(\text{aq}\right)$

The reaction is as follows:

  $\text{HA}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(\text{aq}\right)+{\text{A}}^{-}\left(\text{aq}\right)$

The expression for $K_{\text{a}}$ is as follows:

  $K_{\text{a}}=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$

For value of $K_{\text{a}}$ of different acids. (Refer to 15.6 in the book).

The fraction ionized is equal to ratio of concentration of ionized acid to analytical concentration multiplied by 100.

    $\text{Fraction ionized}=\left(\frac{\left[{\text{A}}^{-}\right]}{C_{\text{HA}}}\right)\left(100\%\right)$

Answer to Problem 15.67QE

The fraction of acid ionizedin $0.33M$${\text{HNO}}_2$ solution is $\text{4}\text{.03545 }\%$.

Explanation of Solution

The chemical equation for ionization of ${\text{HNO}}_2$ is as follows:

    ${\text{HNO}}_2+{\text{H}}_{\text{2}}\text{O}\to {\text{H}}_{\text{3}}{\text{O}}^{+}+{\text{NO}}_2{}^{-}$

The concentration of ${\text{HNO}}_2$ is $0.33M$.

Also, ${\text{HNO}}_2$ is ionized into ${\text{H}}_{\text{3}}{\text{O}}^{+}$ and ${\text{NO}}_2{}^{-}$. Therefore concentration of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is equal to ${\text{NO}}_2{}^{-}$.

Consider the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ and ${\text{NO}}_2{}^{-}$ be $x$.

The ICE table for the above reaction is as follows:

  $\begin{array}{cccccccc}\text{Equation}& {\text{HNO}}_2& +& {\text{H}}_{\text{2}}\text{O}& \to & {\text{H}}_{\text{3}}{\text{O}}^{+}& +& {\text{NO}}_2{}^{-}\\ \text{Initial}\left(M\right)& 0.33& & 0& & 0& & 0\\ \text{Change}\left(M\right)& -x& & & & +x& & +x\\ \text{Equilibrium}\left(M\right)& 0.33-x& & & & x& & x\end{array}$

The expression for $K_{\text{a}}$ for ionization of ${\text{HNO}}_2$ is as follows:

  $K_{\text{a}}=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{NO}}_2{}^{-}\right]}{\left[{\text{HNO}}_2\right]}$        (1)

Substitute $0.33M$ for $\left[{\text{HNO}}_2\right]$, $x$ for $\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]$, $x$ for $\left[{\text{NO}}_2{}^{-}\right]$ and $5.6\times {10}^{-4}$ for $K_{\text{a}}$ in equation (1), the equation needed to solve for the concentration of the hydrogen ion is as follows:

  $5.6\times {10}^{-4}=\frac{\left(x\right)\left(x\right)}{0.33-x}$

Rearrange above equation to obtain the required quadratic equation to compute the concentration of hydrogen ion as follows:

  $x^2+5.6\times {10}^{-4}x-1.848\times {10}^{-4}=0$

Solve for x, therefore the concentration of hydrogen ion is as calculated follows:

  $x=0.013317$

Or,

  $x=-0.013317$

Neglect, the negative value of x as concentration cannot be negative.

Therefore concentration of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is equal to ${\text{NO}}_2{}^{-}$ and is $0.013317$.

The equation for fraction of acid ionized of $0.075M$${\text{HNO}}_2$ solution is a follows:

  $\text{Fraction ionized}=\left(\frac{\left[{\text{NO}}_2{}^{-}\right]}{C_{\text{HA}}}\right)\left(100\%\right)$        (2)

Substitute $0.013317$ for $\left[{\text{NO}}_2{}^{-}\right]$ and $0.075M$ for $C_{\text{HA}}$ in equation (2).

  $\begin{array}{c}\text{Fraction ionized}=\left(\frac{0.013317}{0.33}\right)\left(100\%\right)\\ =\text{4}\text{.03545 }\%\end{array}$

Hence, the fraction of acid ionized in $0.33M$${\text{HNO}}_2$ solution is $\text{4}\text{.03545 }\%$.

(b)

Interpretation Introduction

Interpretation:

The fraction of acid ionized in $0.016M$${\text{C}}_6{\text{H}}_5\text{OH}$ solution has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 15.67QE

The fraction of acid ionized in $0.016M$${\text{C}}_6{\text{H}}_5\text{OH}$ solution is $\text{0}\text{.007906 }\%$.

Explanation of Solution

The chemical equation for ionization of ${\text{C}}_6{\text{H}}_5\text{OH}$ is as follows:

    ${\text{C}}_6{\text{H}}_5\text{OH}+{\text{H}}_{\text{2}}\text{O}\to {\text{H}}_{\text{3}}{\text{O}}^{+}+{\text{C}}_6{\text{H}}_5{\text{O}}^{-}$

The concentration of ${\text{C}}_6{\text{H}}_5\text{COOH}$ is $0.016M$.

Also, ${\text{C}}_6{\text{H}}_5\text{OH}$ is ionized into ${\text{H}}_{\text{3}}{\text{O}}^{+}$ and ${\text{C}}_6{\text{H}}_5{\text{O}}^{-}$. Therefore, concentration of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is equal to ${\text{C}}_6{\text{H}}_5{\text{O}}^{-}$.

Consider the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ and ${\text{C}}_6{\text{H}}_5{\text{O}}^{-}$ be x.

The ICE table for the above reaction is as follows:

  $\begin{array}{cccccccc}\text{Equation}& {\text{C}}_6{\text{H}}_5\text{OH}& +& {\text{H}}_{\text{2}}\text{O}& \to & {\text{H}}_{\text{3}}{\text{O}}^{+}& +& {\text{C}}_6{\text{H}}_5{\text{O}}^{-}\\ \text{Initial}\left(M\right)& 0.016& & 0& & 0& & 0\\ \text{Change}\left(M\right)& -x& & & & +x& & +x\\ \text{Equilibrium}\left(M\right)& 0.016-x& & & & x& & x\end{array}$

The expression for $K_{\text{a}}$ for ionization of ${\text{C}}_6{\text{H}}_5\text{OH}$  is as follows:

  $K_{\text{a}}=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{C}}_6{\text{H}}_5{\text{O}}^{-}\right]}{\left[{\text{C}}_6{\text{H}}_5\text{OH}\right]}$        (3)

Substitute $0.016M$ for $\left[{\text{C}}_6{\text{H}}_5\text{OH}\right]$, $x$ for $\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]$, $x$ for $\left[{\text{C}}_6{\text{H}}_5{\text{O}}^{-}\right]$ and $1.0\times {10}^{-10}$ for $K_{\text{a}}$ in equation (3), the equation needed to solve for the concentration of the hydrogen ion is as follows:

  $1.0\times {10}^{-10}=\frac{\left(x\right)\left(x\right)}{0.016-x}$

Rearrange above equation to obtain the required quadratic equation to compute the concentration of hydrogen ion as follows:

  $x^2+1.0\times {10}^{-10}x-0.016\times {10}^{-10}=0$

Solve for x, therefore the concentration of hydrogen ion is as calculated follows:

  $x=1.26496\times {10}^{-6}$

Or,

  $x=-1.26496\times {10}^{-6}$

Neglect, the negative value of x as concentration cannot be negative.

Therefore, concentration of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is equal to ${\text{C}}_6{\text{H}}_5{\text{O}}^{-}$ and is $1.26496\times {10}^{-6}$.

The equation for fraction of acid ionized in $0.016M$${\text{C}}_6{\text{H}}_5\text{OH}$ solution is a follows:

  $\text{Fraction ionized}=\left(\frac{\left[{\text{C}}_6{\text{H}}_5{\text{O}}^{-}\right]}{C_{\text{HA}}}\right)\left(100\%\right)$        (4)

Substitute $1.26496\times {10}^{-6}$ for $\left[{\text{C}}_6{\text{H}}_5{\text{O}}^{-}\right]$ and $0.016M$ for $C_{\text{HA}}$ in equation (4), fraction ionized is calculated as follows:

  $\begin{array}{c}\text{Fraction ionized}=\left(\frac{1.26496\times {10}^{-6}}{0.016}\right)\left(100\%\right)\\ =\text{0}\text{.007906 }\%\end{array}$

Hence, the fraction of acid ionized in $0.016M$${\text{C}}_6{\text{H}}_5\text{OH}$ solution is $\text{0}\text{.007906 }\%$.

(c)

Interpretation Introduction

Interpretation:

The fraction of acid ionized in $0.25M$$\text{HF}$ solution has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 15.67QE

The fraction of acid ionized in $0.25M$$\text{HF}$ solution is $\text{4}\text{.94548 }\%$.

Explanation of Solution

The chemical equation for ionization of $\text{HF}$ is as follows:

    $\text{HF}+{\text{H}}_{\text{2}}\text{O}\to {\text{H}}_{\text{3}}{\text{O}}^{+}+{\text{F}}^{-}$

The concentration of $\text{HF}$ is $0.25M$.

Also, $\text{HF}$ is ionized into ${\text{H}}_{\text{3}}{\text{O}}^{+}$ and ${\text{F}}^{-}$. Therefore concentration of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is equal to ${\text{F}}^{-}$.

Let us assume the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ and ${\text{F}}^{-}$ be $x$.

The ICE table for the above reaction is as follows:

  $\begin{array}{cccccccc}\text{Equation}& \text{HF}& +& {\text{H}}_{\text{2}}\text{O}& \to & {\text{H}}_{\text{3}}{\text{O}}^{+}& +& {\text{F}}^{-}\\ \text{Initial}\left(M\right)& 0.25& & 0& & 0& & 0\\ \text{Change}\left(M\right)& -x& & & & +x& & +x\\ \text{Equilibrium}\left(M\right)& 0.25-x& & & & x& & x\end{array}$

The expression for $K_{\text{a}}$ for ionization of $\text{HF}$ is as follows:

  $K_{\text{a}}=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{F}}^{-}\right]}{\left[\text{HF}\right]}$        (5)

Substitute $0.25M$ for $\text{HF}$, $x$ for $\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]$, $x$ for $\left[{\text{F}}^{-}\right]$ and $6.3\times {10}^{-4}$ for $K_{\text{a}}$ in equation (5), the equation needed to solve for the concentration of the hydrogen ion is as follows:

  $6.3\times {10}^{-4}=\frac{\left(x\right)\left(x\right)}{0.255-x}$

Rearrange above equation to obtain the required quadratic equation to compute the concentration of hydrogen ion as follows:

  $x^2+6.3\times {10}^{-4}x-1.6065\times {10}^{-4}=0$

Solve for x, therefore the concentration of hydrogen ion is as calculated follows:

  $x=\text{0}\text{.0123637}$

Or,

  $x=-\text{0}\text{.0123637}$

Neglect, the negative value of x as concentration cannot be negative.

Therefore, concentration of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ and ${\text{F}}^{-}$ is $\text{0}\text{.0123637}$.

The equation for fraction of acid ionized in $0.25M$$\text{HF}$ solution is a follows:

  $\text{Fraction ionized}=\left(\frac{\left[{\text{F}}^{-}\right]}{C_{\text{HA}}}\right)\left(100\%\right)$        (6)

Substitute $\text{0}\text{.0123637}$ for $\left[{\text{F}}^{-}\right]$ and $0.25M$ for $C_{\text{HA}}$ in equation (

  $\begin{array}{c}\text{Fraction ionized}=\left(\frac{\text{0}\text{.0123637}}{0.25}\right)\left(100\%\right)\\ =\text{4}\text{.94548 }\%\end{array}$

Hence, the fraction of acid ionized in $0.25M$$\text{HF}$ solution is $\text{4}\text{.94548 }\%$.

(d)

Interpretation Introduction

Interpretation:

The fraction of acid ionized in $0.010M$$\text{HCOOH}$ solution has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 15.67QE

The fraction of acid ionized in $0.010M$$\text{HCOOH}$ solution $\text{12}\text{.3637 }\%$

Explanation of Solution

The chemical equation for ionization of $\text{HCOOH}$ is as follows:

    $\text{HCOOH}+{\text{H}}_{\text{2}}\text{O}\to {\text{H}}_{\text{3}}{\text{O}}^{+}+{\text{HCOO}}^{-}$

The concentration of $\text{HCOOH}$ is $0.010M$.

Also, $\text{HCOOH}$ is ionized into ${\text{H}}_{\text{3}}{\text{O}}^{+}$ and ${\text{HCOO}}^{-}$. Therefore, concentration of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is equal to ${\text{HCOO}}^{-}$.

Let us assume the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ and ${\text{HCOO}}^{-}$ be $x$.

The ICE table for the above reaction is as follows:

  $\begin{array}{cccccccc}\text{Equation}& \text{HCOOH}& +& {\text{H}}_{\text{2}}\text{O}& \to & {\text{H}}_{\text{3}}{\text{O}}^{+}& +& {\text{HCOO}}^{-}\\ \text{Initial}\left(M\right)& 0.010& & 0& & 0& & 0\\ \text{Change}\left(M\right)& -x& & & & +x& & +x\\ \text{Equilibrium}\left(M\right)& 0.010-x& & & & x& & x\end{array}$

The expression for $K_{\text{a}}$ for ionization of $\text{HCOOH}$  is as follows:

  $K_{\text{a}}=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{HCOO}}^{-}\right]}{\left[\text{HCOOH}\right]}$        (7)

Substitute $0.010M$ for $\text{HCOOH}$, $x$ for $\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]$, $x$ for $\left[{\text{HCOO}}^{-}\right]$ and $1.8\times {10}^{-4}$ for $K_{\text{a}}$ in equation (7), the equation needed to solve for the concentration of the hydrogen ion is as follows:

  $1.8\times {10}^{-4}=\frac{\left(x\right)\left(x\right)}{0.010-x}$

Rearrange above equation to obtain the required quadratic equation to compute the concentration of hydrogen ion as follows:

  $x^2+1.8\times {10}^{-4}x-0.018\times {10}^{-4}=0$

Solve for $x$, therefore the concentration of hydrogen ion is as calculated follows:

  $x=\text{0}\text{.00125466}$

Or,

  $x=-\text{0}\text{.00125466}$

Neglect, the negative value of x as concentration cannot be negative.

Therefore, concentration of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is $\text{0}\text{.00125466}$.

The equation for fraction of acid ionized in $0.010M$$\text{HCOOH}$ solution is a follows:

  $\text{Fraction ionized}=\left(\frac{\left[{\text{HCOO}}^{-}\right]}{C_{\text{HA}}}\right)\left(100\%\right)$        (8)

Substitute $\text{0}\text{.00125466}$ for $\left[{\text{HCOO}}^{-}\right]$ and $0.010M$ for $C_{\text{HA}}$ in equation (8).

  $\begin{array}{c}\text{Fraction ionized}=\left(\frac{\text{0}\text{.00125466}}{0.010}\right)\left(100\%\right)\\ =\text{12}\text{.3637 }\%\end{array}$

Hence, the fraction of acid ionized in $0.010M$$\text{HCOOH}$ solution is $\text{12}\text{.3637 }\%$.