Chapter 15, Problem 124QRT

(a)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of initial solution and $\text{pH}$ of the solution after evaporation has to be calcuated.

Concept Introduction:

Molarity is defined as the ratio of number of moles of solute to the volume of solution is liters.  The S.I. unit of molarity is molar and it is represented by $\text{M}$.

Mathematical formulation of molarity is shown as follows:

  $\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution }\left(\text{L}\right)}$

Explanation of Solution

As per the given data, the volume of the solution is $100\text{mL}$.

The unit conversion of volume from $\text{mL}$ to $\text{L}$ is shown below.

  $100.0\text{mL}\times \frac{1\text{L}}{1000\text{mL}}=0.1\text{L}$

The standard value of $K_{\text{a}}$ foranisic acid is  $3.38\times {10}^{-5}$.

The ICE table for anisic acid is shown below:

  $\begin{array}{l}{\text{C}}_{\text{8}}{\text{H}}_{\text{8}}{\text{O}}_3\rightleftharpoons {\text{H}}^{+}+{\text{C}}_{\text{8}}{\text{H}}_{\text{7}}{\text{O}}_3{}^{-}\\ \begin{array}{cccc}\text{I:}& 0.0010& 0& 0\\ \text{C:}& -x& +x& +x\\ \text{E:}& \left(0.0010-x\right)& x& x\end{array}\end{array}$

Where,

• $\text{I}$ is the Initial concentration.
• $\text{C}$ is the Change concentration.
• $\text{E}$ is the Equilibrium concentration.
• $x$ is the change is concentration.

The expression for $K_{\text{a}}$ for the above reaction is shown below.

  $K_{\text{a}}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{C}}_8{\text{H}}_7{\text{O}}_3{}^{-}\right]}{\left[{\text{C}}_8{\text{H}}_8{\text{O}}_3\right]}$        (1)

Where,

• $K_{\text{a}}$ is the dissociation constant.
• $\left[{\text{C}}_8{\text{H}}_7{\text{O}}_3{}^{-}\right]$ is the concentration of anisicate ion.
• $\left[{\text{C}}_8{\text{H}}_8{\text{O}}_3\right]$ is the concentration of anisic acid.
• $\left[{\text{H}}^{+}\right]$ is the concentration of hydrogen ion.

The value of $K_{\text{a}}$ is $3.38\times {10}^{-5}$.

The value of $\left[{\text{C}}_8{\text{H}}_7{\text{O}}_3{}^{-}\right]$ is $x$.

The value of $\left[{\text{C}}_8{\text{H}}_8{\text{O}}_3\right]$ is $\left(0.0010-x\right)$.

The value of $\left[{\text{H}}^{+}\right]$ is $x$.

Substitute the value of $K_{\text{a}}$, $\left[{\text{C}}_8{\text{H}}_7{\text{O}}_3{}^{-}\right]$, $\left[{\text{C}}_8{\text{H}}_8{\text{O}}_3\right]$ and $\left[{\text{H}}^{+}\right]$ in equation (1).

  $3.38\times {10}^{-5}=\frac{\left(x\right)\left(x\right)}{\left(0.0010-x\right)}$

As $x$ is too small with respect to $0.0010$.  Therefore, $\left(0.0010-x\right)$ is approximately equal to $0.0010$.

  $\begin{array}{c}3.38\times {10}^{-5}=\frac{\left(x\right)\left(x\right)}{0.0010}\\ x^2=3.38\times {10}^{-5}\times 0.0010\\ x=\sqrt{3.38\times {10}^{-8}}\\ =1.83\times {10}^{-4}\text{M}\end{array}$

Therefore, the concentration of hydrogen ions in the solution is $1.83\times {10}^{-4}\text{M}$.

The initial $\text{pH}$ of the solution is calculated by using the relation shown below.

  $\text{pH}=-\log \left[{\text{H}}^{+}\right]$        (2)

The value of $\left[{\text{H}}^{+}\right]$ is $1.83\times {10}^{-4}\text{M}$.

Substitute the value of $\left[{\text{H}}^{+}\right]$ in equation (2).

  $\begin{array}{c}\text{pH}=-\log \left(1.83\times {10}^{-4}\right)\\ =3.737\end{array}$

As per the given data, the $50.0\text{mL}$ of water is evaporated.  Therefore, due to this the molarity of the solution also changes but the number of moles remains same.

The number of moles of hydrogen ions acid in solution is calculated by the formula shown below.

  $n=M\times V$        (3)

Where,

• $n$ is the number of moles.
• $M$ is the molarity.
• $V$ is the volume.

The value of $M$ for hydrogen ions is $1.83\times {10}^{-4}\text{M}$.

The value of $V$ is $0.1\text{L}$.

Substitute the values of $M$ and $V$ for hydrogen ions in the equation (3).

  $\begin{array}{c}n=1.83\times {10}^{-4}\text{M}\times 0.1\text{L}\\ =1.83\times {10}^{{}^{-5}}\text{mol}\end{array}$

Thus, the number of moles of hydrogen ions in the solution is $1.83\times {10}^{{}^{-5}}\text{mol}$.

The volume of the solution after evaporation is  $50.0\text{mL}$.

The unit conversion of volume from $\text{mL}$ to $\text{L}$ is shown below.

  $50.0\text{mL}\times \frac{1\text{L}}{1000\text{mL}}=0.05\text{L}$

The concentration of solution after evaporation is calculated by the formula shown below.

  $M=\frac{n}{V}$        (4)

Where,

• $n$ is the number of moles.
• $M$ is the molarity.
• $V$ is the volume.

The value of $n$ is $1.83\times {10}^{-5}\text{mol}$.

The value of $V$ is $0.05\text{L}$.

Substitute the values of $n$ and $V$ in the equation (4).

  $\begin{array}{c}M=\frac{1.83\times {10}^{-5}\text{mol}}{0.05\text{L}}\\ =3.66\times {10}^{-4}\text{M}\end{array}$

Thus, the concentration of solution after evaporation is $3.66\times {10}^{-4}\text{M}$.

The new ICE table for anisic acidis shown below:

  $\begin{array}{l}{\text{C}}_{\text{8}}{\text{H}}_{\text{8}}{\text{O}}_3\rightleftharpoons {\text{H}}^{+}+{\text{C}}_{\text{8}}{\text{H}}_{\text{7}}{\text{O}}_3{}^{-}\\ \begin{array}{cccc}\text{I:}& 3.66\times {10}^{-4}& 0& 0\\ \text{C:}& -x& +x& +x\\ \text{E:}& \left(3.66\times {10}^{-4}-x\right)& x& x\end{array}\end{array}$

Where,

• $\text{I}$ is the Initial concentration.
• $\text{C}$ is the Change concentration.
• $\text{E}$ is the Equilibrium concentration.
• $x$ is the change is concentration

The value of $K_{\text{a}}$ is $3.38\times {10}^{-5}$.

The value of $\left[{\text{C}}_8{\text{H}}_7{\text{O}}_3{}^{-}\right]$ is $x$.

The value of $\left[{\text{C}}_8{\text{H}}_8{\text{O}}_3\right]$ is $\left(3.66\times {10}^{-4}-x\right)$.

The value of $\left[{\text{H}}^{+}\right]$ is $x$.

Substitute the value of $K_{\text{a}}$, $\left[{\text{C}}_8{\text{H}}_7{\text{O}}_3{}^{-}\right]$,  $\left[{\text{C}}_8{\text{H}}_8{\text{O}}_3\right]$  and $\left[{\text{H}}^{+}\right]$  in equation (1).

  $3.38\times {10}^{-5}=\frac{\left(x\right)\left(x\right)}{\left(3.66\times {10}^{-4}-x\right)}$

As $x$ is too small with respect to $3.66\times {10}^{-4}$.  Therefore, $\left(3.66\times {10}^{-4}-x\right)$  is approximately equal to $3.66\times {10}^{-4}$.

  $\begin{array}{c}3.38\times {10}^{-5}=\frac{\left(x\right)\left(x\right)}{3.66\times {10}^{-4}}\\ x^2=3.38\times {10}^{-5}\times 3.66\times {10}^{-4}\\ x=\sqrt{1.237\times {10}^{-8}}\\ =1.11\times {10}^{-4}\text{M}\end{array}$

Therefore the new concentration of hydrogen ions in the solution is $1.11\times {10}^{-4}\text{M}$.

The value of $\left[{\text{H}}^{+}\right]$ is $1.11\times {10}^{-4}\text{M}$.

Substitute the value of $\left[{\text{H}}^{+}\right]$ in equation (2).

  $\begin{array}{c}\text{pH}=-\log \left(1.11\times {10}^{-4}\text{M}\right)\\ =3.95\end{array}$

Thus, after rounding to three significant figures, the initial $\text{pH}$ and the $\text{pH}$ of the solution after evaporation is $\underset{\_}{3.74}$ and $\underset{\_}{3.95}$ respectively.

(b)

Interpretation Introduction

Interpretation:

The degree of dissociation of anisic acid before and after evaporation has to be calculated.

Concept Introduction:

The degree of dissociation is defined as the ratio of the concentration of ion dissociated to the concentration of the initial solution.  Mathematically it can be is represented as shown below.

  $\alpha =\frac{{\text{C}}_{\text{ion}}}{{\text{C}}_{\text{solution}}}$

Answer to Problem 124QRT

The degree of dissociation of anisic acid before and after evaporation is $\underset{\_}{0.183}$ and $\underset{\_}{0.303}$ respectively.

Explanation of Solution

The degree of dissociation when volume of the solution is $0.1\text{L}$ is calculated using the relation shown below.

  $\alpha =\frac{{\text{C}}_{\text{ion}}}{{\text{C}}_{\text{solution}}}$        (5)

Where,

• $\alpha$ is the degree of dissociation.
• ${\text{C}}_{\text{solution}}$ is the concentration of solution.
• ${\text{C}}_{\text{ion}}$ is the concentration of ion dissociated.

The value of ${\text{C}}_{\text{solution}}$ is $0.0010\text{M}$.

The value of ${\text{C}}_{\text{ion}}$ is $1.83\times {10}^{-4}\text{M}$.

Substitute the value of ${\text{C}}_{\text{solution}}$ and ${\text{C}}_{\text{ion}}$ in equation (5).

    $\begin{array}{c}\alpha =\frac{1.83\times {10}^{-4}\text{M}}{0.0010\text{M}}\\ =0.183\end{array}$

The degree of dissociation when volume of the solution is $0.05\text{L}$ is determined as shown below.

The value of ${\text{C}}_{\text{solution}}$ is $3.66\times {10}^{-4}\text{M}$.

The value of ${\text{C}}_{\text{ion}}$ is $1.11\times {10}^{-4}\text{M}$.

Substitute the value of ${\text{C}}_{\text{solution}}$ and ${\text{C}}_{\text{ion}}$ in equation (5).

    $\begin{array}{c}\alpha =\frac{1.11\times {10}^{-4}\text{M}}{3.66\times {10}^{-4}\text{M}}\\ =0.303\end{array}$

Thus, the degree of dissociation of anisic acid before and after evaporation is $\underset{\_}{0.183}$ and $\underset{\_}{0.303}$ respectively.