Chapter 15, Problem 15.63QE

(a)

Interpretation Introduction

Interpretation:

$\text{pH}$ for $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 negative logarithm of molar concentration of hydronium ion is called $\text{pH}$. The expression for pH is as follows:

    $\text{pH}=-{\text{log}}_{\text{10}}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$        (1)

Answer to Problem 15.63QE

$\text{pH}$ for $0.33M$${\text{HNO}}_2$ solution is $1.87559$.

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:

  $K_{\text{a}}=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{NO}}_2{}^{-}\right]}{\left[{\text{HNO}}_2\right]}$        (2)

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 (2), 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:

  $x=0.013317$

Therefore, concentration of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is $0.013317$.

Substitute $0.013317$ for $\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]$ in equation (1), $\text{pH}$ is calculated as follows:

  $\begin{array}{c}\text{pH}=-{\text{log}}_{\text{10}}\left[0.013317\right]\\ =1.87559\end{array}$

Hence, $\text{pH}$ for $0.33M$${\text{HNO}}_2$ solution is $1.87559$.

(b)

Interpretation Introduction

Interpretation:

$\text{pH}$ for $0.016M$${\text{C}}_6{\text{H}}_5\text{OH}$ solution has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 15.63QE

$\text{pH}$ for $0.016M$${\text{C}}_6{\text{H}}_5\text{OH}$ solution is 5.90.

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:

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

Rearrangeabove 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 $1.26496\times {10}^{-6}$.

Substitute $1.26496\times {10}^{-6}$ for $\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]$ in equation (1), $\text{pH}$ is calculated as follows:

  $\begin{array}{c}\text{pH}=-{\text{log}}_{\text{10}}\left[1.26496\times {10}^{-6}\right]\\ =5.89792\end{array}$

Hence, $\text{pH}$ for $0.016M$${\text{C}}_6{\text{H}}_5\text{OH}$ solution is $5.89792$.

(c)

Interpretation Introduction

Interpretation:

$\text{pH}$ for $0.25M$$\text{HF}$ solution has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 15.63QE

$\text{pH}$ for $0.25M$$\text{HF}$ solution is $1.90785$.

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:

  $K_{\text{a}}=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{F}}^{-}\right]}{\left[\text{HF}\right]}$        (4)

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 (4), 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 $\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]$ is $\text{0}\text{.0123637}$.

Substitute $\text{0}\text{.0123637}$ for $\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]$ in equation (1), $\text{pH}$ is calculated as follows:

  $\begin{array}{c}\text{pH}=-{\text{log}}_{\text{10}}\left[\text{0}\text{.0123637}\right]\\ =1.90785\end{array}$

Hence, $\text{pH}$ for $0.25M$$\text{HF}$ solution is $1.90785$.

(d)

Interpretation Introduction

Interpretation:

$\text{pH}$ for $0.010M$$\text{HCOOH}$ solution has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 15.63QE

$\text{pH}$ for $0.010M$$\text{HCOOH}$ solution is $2.90147$

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:

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

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 (5), 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 $\text{0}\text{.00125466}$.

Substitute $\text{0}\text{.00125466}$ for $\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]$ in equation (1), $\text{pH}$ is calculated as follows:

  $\begin{array}{c}\text{pH}=-{\text{log}}_{\text{10}}\left[\text{0}\text{.00125466}\right]\\ =2.90147\end{array}$

Hence, $\text{pH}$ for $0.010M$$\text{HCOOH}$ solution is $2.90147$.