Chapter 15, Problem 43QRT

(a)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of $\text{30}\text{mL}$ $\text{0}\text{.100}\text{M}$ benzoic acid solution when titrated with $\text{10}\text{mL}$ $\text{0}\text{.100}\text{M}$ $\text{NaOH}$ has to be calculated.

Concept Introduction:

The $\text{pH}$ is defined as the negative logarithm to the hydrogen ion concentration and it tells the nature of the solution. The physical constant which is used to determine the strength of an acid is known as acid dissociation constant. It is denoted by $K_{\text{a}}$. The greater $K_{\text{a}}$ value, stronger is the acid.

Answer to Problem 43QRT

The $\text{pH}$ of $\text{30}\text{mL}$ $\text{0}\text{.100}\text{M}$ benzoic acid solution when titrated with $\text{10}\text{mL}$ $\text{0}\text{.100}\text{M}$ $\text{NaOH}$ is $\underset{\_}{3.62}$.

Explanation of Solution

The given molarity of benzoic acid is $\text{0}\text{.100}\text{M}$.

The given molarity of titrant, $\text{NaOH}$ is $\text{0}\text{.100}\text{M}$.

The given volume of benzoic acid is $\text{30}\text{mL}$.

The given volume of $\text{NaOH}$is $\text{10}\text{mL}$.

The given value of $K_{\text{a}}$ of benzoic acid is $1.2\times {10}^{-4}$.

The total volume of the solution after titration is $\text{30}\text{mL}+\text{10}\text{mL}=\text{40}\text{mL}$.

The titration reaction of benzoic acid and $\text{NaOH}$ is given below.

    ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}\left(\text{aq}\right)+\text{ NaOH}\left(\text{aq}\right)\to {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COONa}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

The conversion of $\text{mL}$ into $\text{L}$ is shown below.

    $1\text{ml}={10}^{-3}\text{L}$

So, the volume of benzoic acid is calculated below.

    $30\text{mL}=0.030\text{L}$

The volume of $\text{NaOH}$is calculated below.

    $10\text{mL}=0.010\text{L}$

The total volume of the solution is calculated below.

    $40\text{mL}=0.040\text{L}$

The mole, $n$, of the given substance is calculated as below.

$n=\text{Volume}\times \text{molarity}$ (1)

Substitute the values of volume and molarity of benzoic acid in equation (1).

    $\begin{array}{c}n=\text{Volume}\times \text{molarity}\\ =0.030\text{L}\times 0.1\text{mol}/\text{L}\\ =3\times {10}^{-3}\text{mol}\end{array}$

Substitute the values of volume and molarity of $\text{NaOH}$ in equation (1).

    $\begin{array}{c}n=\text{Volume}\times \text{molarity}\\ =0.010\text{L}\times 0.1\text{mol}/\text{L}\\ =1\times {10}^{-3}\text{mol}\end{array}$

So, the moles of benzoic acid is $3\times {10}^{-3}\text{mol}$ and the moles $\text{10}\text{mL}$ $\text{NaOH}$ solution is $1\times {10}^{-3}\text{mol}$. So, $1\times {10}^{-3}\text{mol}$ of $\text{NaOH}$ neutralizes $1\times {10}^{-3}\text{mol}$ of benzoic acid solution after titration. Thus, this reaction leads to the formation of $1\times {10}^{-3}\text{mol}$ of sodium benzoate ions and $2\times {10}^{-3}\text{mol}$ benzoic acid exists un-neutralized inside the solution after the complete titration.

The molarity of un-neutralized $2\times {10}^{-3}\text{mol}$ benzoic acid and $1\times {10}^{-3}\text{mol}$ of sodium benzoate ions is calculated by the formula given below.

    $\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}}{\text{Total}\text{volume}}$                                                                          ……(2)

Substitute the values of number of moles of benzoic acid and total volume of the solution in equation (2).

    $\begin{array}{c}\text{Molarity}=\frac{2\times {10}^{-3}\text{mol}}{0.040\text{L}}\\ =0.05\text{M}\end{array}$

Substitute the values of number of moles of sodium benzoate ions and total volume of the solution in equation (2).

    $\begin{array}{c}\text{Molarity}=\frac{1\times {10}^{-3}\text{mol}}{0.040\text{L}}\\ =0.025\text{M}\end{array}$

The Henderson-Hasselbalch equation can be represented as follows.

    $\begin{array}{l}\text{pH }=\text{p}{K}_{\text{a}}+\log \frac{\left[\text{conjugate}\text{base}\right]}{\left[\text{conjugate}\text{acid}\right]}\\ \text{pH }=-\log K_{\text{a}}+\log \frac{\left[\text{conjugate}\text{base}\right]}{\left[\text{conjugate}\text{acid}\right]}\end{array}$                                                          ……(3)

Where,

• $\left[\text{conjugate}\text{base}\right]$ is the concentration of conjugate base that is of sodium benzoate in the solution.
• $\left[\text{conjugate}\text{acid}\right]$ is the concentration of conjugate acid that is of un-neutralized benzoic acid in the solution.

Substitute the value of $K_{\text{a}}$ of benzoic acid, concentration of conjugate base and conjugate acid in equation (3).

$\begin{array}{c}\text{pH }=-\log \left(1.2\times {10}^{-4}\right)+\log \frac{\left[\text{0}\text{.025}\right]}{\left[\text{0}\text{.0500}\right]}\\ \text{pH }=3.92+\left(-0.3010\right)\\ =\underset{\_}{3.62}\end{array}$

Therefore the $\text{pH}$ of $\text{30}\text{mL}$ $\text{0}\text{.100}\text{M}$ benzoic acid solution when titrated with $\text{10}\text{mL}$ $\text{0}\text{.100}\text{M}$ $\text{NaOH}$  is $\underset{\_}{3.62}$.

(b)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of $\text{30}\text{mL}$ $\text{0}\text{.100}\text{M}$ benzoic acid solution when titrated with $\text{30}\text{mL}$ $\text{0}\text{.100}\text{M}$ $\text{NaOH}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 43QRT

The $\text{pH}$ of $\text{30}\text{mL}$ $\text{0}\text{.100}\text{M}$ benzoic acid solution when titrated with $\text{30}\text{mL}$ $\text{0}\text{.100}\text{M}$ $\text{NaOH}$ is $\underset{\_}{8.31}$.

Explanation of Solution

The given molarity of benzoic acid is $\text{0}\text{.100}\text{M}$.

The given molarity of titrant, $\text{NaOH}$ is $\text{0}\text{.100}\text{M}$.

The given volume of benzoic acid is $\text{30}\text{mL}$.

The given volume of $\text{NaOH}$is $\text{30}\text{mL}$.

The given value of $K_{\text{a}}$ of benzoic acid is $1.2\times {10}^{-4}$.

The total volume of the solution after titration is $\text{30}\text{mL}+3\text{0}\text{mL}=\text{60}\text{mL}$.

The titration reaction of benzoic acid and $\text{NaOH}$ is given below.

    ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}\left(\text{aq}\right)+\text{ NaOH}\left(\text{aq}\right)\to {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COONa}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

The conversion of $\text{mL}$ into $\text{L}$ is shown below.

    $1\text{ml}={10}^{-3}\text{L}$

So, the volume of benzoic acid and $\text{NaOH}$ is calculated below.

    $30\text{mL}=0.030\text{L}$

The total volume of the solution is calculated below.

    $60\text{mL}=0.060\text{L}$

The mole, $n$, of the given substance is calculated as below.

$n=\text{Volume}\times \text{molarity}$ (1)

Substitute the values of volume and molarity of benzoic acid in equation (1).

    $\begin{array}{c}n=\text{Volume}\times \text{molarity}\\ =0.030\text{L}\times 0.1\text{mol}/\text{L}\\ =3\times {10}^{-3}\text{mol}\end{array}$

Substitute the values of volume and molarity of $\text{NaOH}$ in equation (1).

    $\begin{array}{c}n=\text{Volume}\times \text{molarity}\\ =0.030\text{L}\times 0.1\text{mol}/\text{L}\\ =3\times {10}^{-3}\text{mol}\end{array}$

So, the moles of benzoic acid is $3\times {10}^{-3}\text{mol}$ and the moles $\text{30}\text{mL}$ $\text{NaOH}$ solution is $3\times {10}^{-3}\text{mol}$. So, $3\times {10}^{-3}\text{mol}$ of $\text{NaOH}$ neutralizes $3\times {10}^{-3}\text{mol}$ of benzoic acid solution and gives the equivalence point. Thus, this reaction leads to the formation of $3\times {10}^{-3}\text{mol}$ of sodium benzoate ions after the complete titration.

The molarity of $3\times {10}^{-3}\text{mol}$ of sodium benzoate ions is calculated by the formula given below.

    $\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}}{\text{Total}\text{volume}}$                                                                          ……(2)

Substitute the values of number of moles of sodium benzoate ions and total volume of the solution in equation (2).

    $\begin{array}{c}\text{Molarity}=\frac{3\times {10}^{-3}\text{mol}}{0.060\text{L}}\\ =0.050\text{M}\end{array}$

Thus, the equivalence point of the reaction of benzoic acid and $\text{NaOH}$ is shown below.

    ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}\left(\text{aq}\right)+\text{ NaOH}\left(\text{aq}\right)\rightleftharpoons {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COONa}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

The value of $K_{\text{w}}$ is $1.0\times {10}^{-14}$.

The expression for calculating the value of $K_{\text{b}}$ of acetate ion in the solution is given below.

    $K_{\text{b}}=\frac{K_{\text{w}}}{K_{\text{a}}}$                                                                                  ……(4)

Where,

• $K_{\text{b}}$ is the equilibrium constant of base.
• $K_{\text{w}}$ is the equilibrium constant of water.
• $K_{\text{a}}$ is the equilibrium constant of acid.

Substitute the value of $K_{\text{a}}$ of benzoic acid and $K_{\text{w}}$ is the equilibrium constant of water in equation (4).

    $\begin{array}{c}{K}_{\text{b}}=\frac{1.0\times {10}^{-14}}{1.2\times {10}^{-4}}\\ =8.3\times {10}^{-11}\end{array}$

The concentration of ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}$ and ${\text{OH}}^{-}$ is supposed to be $\text{x}$and the concentration of sodium benzoate, ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}^{-}$ is supposed to $0.0500-\text{x}$.

The expression for calculating the equilibrium constant, $K_{\text{b}}$ is given below.

    $K_{\text{b}}=\frac{\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}^{-}\right]}$                                                                      ……(5)

Substitute the respective values of concentration and value of $K_{\text{b}}$ equation (5).

    $\begin{array}{c}8.3\times {10}^{-11}=\frac{\left[\text{x}\right]\left[\text{x}\right]}{\left[0.0500-\text{x}\right]}\\ 8.3\times {10}^{-11}=\frac{{\text{x}}^2}{\left[0.0500-\text{x}\right]}\end{array}$

The value of $K_{\text{b}}$ is small, so, $0.0500-\text{x}=\text{x}$. Simplify the above equation as follows.

    $\begin{array}{c}8.3\times {10}^{-11}=\frac{{\text{x}}^2}{\left[0.0500\right]}\\ 8.3\times {10}^{-11}\times 0.0500={\text{x}}^2\\ \text{x}=2.0\times {10}^{-6}\end{array}$

The equation to calculate the value of$\text{pOH}$ is given below.

    $\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$                                                                                             ……(6)

Substitute $2.0\times {10}^{-6}$ for $\left[{\text{OH}}^{-}\right]$ in equation (6).

    $\begin{array}{c}\text{pOH}=-\log \left[{\text{OH}}^{-}\right]\\ =-\log \left(2.0\times {10}^{-6}\right)\\ =5.69\end{array}$

The relation between $\text{pH}$and $\text{pOH}$ is given below.

    $\text{pH}+\text{pOH}=14$                                                                                            ……(7)

Substitute $5.69$ for in equation (7).

    $\begin{array}{c}\text{pH}+\text{5}\text{.69}=14\\ \text{pH}=14-5.69\\ =\underset{\_}{8.31}\end{array}$

Therefore, the $\text{pH}$ of $\text{30}\text{mL}$ $\text{0}\text{.100}\text{M}$ benzoic acid solution when titrated with $\text{30}\text{mL}$ $\text{0}\text{.100}\text{M}$ $\text{NaOH}$ is $\underset{\_}{8.31}$

(c)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of $\text{30}\text{mL}$ $\text{0}\text{.100}\text{M}$ benzoic acid solution when titrated with $\text{40}\text{mL}$ $\text{0}\text{.100}\text{M}$ $\text{NaOH}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 43QRT

The $\text{pH}$ of $\text{30}\text{mL}$ $\text{0}\text{.100}\text{M}$ benzoic acid solution when titrated with $\text{40}\text{mL}$ $\text{0}\text{.100}\text{M}$ $\text{NaOH}$ is $\underset{\_}{12.15}$.

Explanation of Solution

The given molarity of benzoic acid is $\text{0}\text{.100}\text{M}$.

The given molarity of titrant, $\text{NaOH}$ is $\text{0}\text{.100}\text{M}$.

The given volume of benzoic acid is $\text{30}\text{mL}$.

The given volume of $\text{NaOH}$is $\text{40}\text{mL}$.

The given value of $K_{\text{a}}$ of benzoic acid is $1.2\times {10}^{-4}$.

The total volume of the solution after titration is $\text{30}\text{mL}+\text{40}\text{mL}=\text{70}\text{mL}$.

The titration reaction of benzoic acid and $\text{NaOH}$ is given below.

    ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}\left(\text{aq}\right)+\text{ NaOH}\left(\text{aq}\right)\to {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COONa}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

The conversion of $\text{mL}$ into $\text{L}$ is shown below.

    $1\text{ml}={10}^{-3}\text{L}$

So, the volume of benzoic acid is calculated below.

    $9781305498129$

The total volume of the solution is calculated below.

    $70\text{mL}=0.070\text{L}$

So, the concentration of hydroxyl ion is more than the required concentration for equivalence point. The excess volume of ${\text{OH}}^{-}$ ions is $\text{40}\text{mL}-\text{30}\text{mL}=\text{10}\text{mL}$.

So, the volume of excess ${\text{OH}}^{-}$ ions in liters is calculated below.

    $10\text{mL}=0.010\text{L}$

The mole, $n$, of the given substance is calculated as below.

$n=\text{Volume}\times \text{molarity}$ (1)

Substitute the values of volume and molarity of excess ${\text{OH}}^{-}$n equation (1).

    $\begin{array}{c}n=\text{Volume}\times \text{molarity}\\ =0.010\text{L}\times 0.1\text{mol}/\text{L}\\ =1\times {10}^{-3}\text{mol}\end{array}$

The molarity of $1\times {10}^{-3}\text{mol}$ of excess ${\text{OH}}^{-}$is calculated by the formula given below.

    $\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}}{\text{Total}\text{volume}}$                                                                          ……(2)

Substitute the values of number of moles of excess ${\text{OH}}^{-}$and total volume of the solution in equation (2).

    $\begin{array}{c}\text{Molarity}=\frac{1\times {10}^{-3}\text{mol}}{0.070\text{L}}\\ =0.0140\text{M}\end{array}$

The equation to calculate the value of$\text{pOH}$ is given below.

    $\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$                                                                                             ……(6)

Substitute $0.0140\text{M}$ for $\left[{\text{OH}}^{-}\right]$ in equation (6).

    $\begin{array}{c}\text{pOH}=-\log \left[{\text{OH}}^{-}\right]\\ =-\log \left(0.0140\text{M}\right)\\ =1.85\end{array}$

The relation between $\text{pH}$and $\text{pOH}$ is given below.

    $\text{pH}+\text{pOH}=14$                                                                                            ……(7)

Substitute $5.69$ for in equation (7).

    $\begin{array}{c}\text{pH}+1.85=14\\ \text{pH}=14-1.85\\ =\underset{\_}{12.15}\end{array}$

Therefore, the $\text{pH}$ of $\text{30}\text{mL}$ $\text{0}\text{.100}\text{M}$ benzoic acid solution when titrated with $\text{30}\text{mL}$ $\text{0}\text{.100}\text{M}$ $\text{NaOH}$ is $\underset{\_}{12.15}$.