Chapter 16, Problem 16.114QP

Interpretation Introduction

Interpretation: The $\text{pH}$ values at the equivalence point of two solutions are to be compared.

Concept introduction: The value of $\text{pH}$ is calculated by the use of concentration of ${\text{H}}^{+}$ and ${\text{OH}}^{-}$ ions in the solution. $\text{pH}$ is the numeric value which defines acidity and basicity of the solution.

To determine: The $\text{pH}$ values at the equivalence point of two solutions.

Answer to Problem 16.114QP

The value of $\text{pH}$ at equivalence point of strong base and weak acid is $\underset{\_}{8.8}$ and the $\text{pH}$ value in case of strong acid and strong base at equivalence point is $\underset{\_}{7}$.

Explanation of Solution

Given

The concentration of $\text{NaOH}$, $M_1$ is $0.125\text{M}$.

The concentration of weak acid, $M_2$ is $0.50\text{M}$.

The addition of weak acid in a solution, $V_2$ is $25\text{ml}$.

The volume of $\text{NaOH}$, $V_1$ is $x$.

The conversion of units of volume into $\text{L}$.

$\begin{array}{c}1\text{ml}=1000\text{L}\\ 25\text{ml}=0.025\text{L}\end{array}$

The weak acid is assumed to be acetic acid.

The equation of dissociation of acetic acid is given as,

${\text{CH}}_3\text{COOH}\to {\text{CH}}_3{\text{COOH}}^{-}+{\text{H}}^{+}$

The equation of reaction of sodium hydroxide and acetic acid is given as,

$\text{NaOH}+{\text{CH}}_3\text{COOH}\to {\text{NaCH}}_3\text{COO}+{\text{H}}_2\text{O}$

The volume required to neutralize the weak acid is calculated by the formula,

$M_1{V}_1=M_2{V}_2$ (1)

Where,

• $M_1$ is the molarity of the base.
• $V_1$ is the volume of the base.
• $M_2$ is the molarity of the acid.
• $V_2$ is the volume of the acid.

Substitute the values of $M_1,V_1,M_2\text{and}{V}_2$ in equation (1).

$\begin{array}{c}0.125\text{M}\times x=0.50\text{M}\times 0.025\text{L}\\ x=\frac{0.0125}{0.125}\\ =0.1\text{L}\end{array}$

The value of $\text{mol}$ of acid is calculated by the formula,

$\text{mol}\text{of}\text{acid}=M_2{V}_2$ (2)

Where,

• $M_2$ is the molarity of the acid.
• $V_2$ is the volume of the acid.

Substitute the value of $M_2\text{and}{V}_2$ in equation (2).

$\begin{array}{c}\text{mol}\text{of}\text{acid}=0.50\text{M}\times 0.025\text{L}\\ =0.0125\text{mol}\end{array}$

The value of $\text{mol}$ of base is calculated by the formula,

$\text{mol}\text{of}\text{base}=M_1{V}_1$ (3)

Where,

• $M_1$ is the molarity of the base.
• $V_1$ is the volume of the base.

Substitute the value of $M_1\text{and}{V}_1$ in equation (3).

$\begin{array}{c}\text{mol}\text{of}\text{base}=0.125\text{M}\times 0.1\text{L}\\ =0.0125\text{mol}\end{array}$

The concentration of acid and base at the different stages of reaction is evaluated by the $\text{ICE}$ table,

$\begin{array}{cccccccc}\text{Reaction}& \text{NaOH}& +& {\text{CH}}_3\text{COOH}& \to & {\text{NaCH}}_3\text{COO}& +& {\text{H}}_2\text{O}\\ \text{Initial}& 0.0125& & 0.0125& & 0& & \\ \text{Change}& -0.0125& & -0.0125& & +0.0125& & \\ \text{Equilibrium}& 0& & 0& & 0.0125& & \end{array}$

The number of moles of solute and solvent are equal so, the molarity of solute and solvent is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}\text{or}\text{solvent}}{\text{Total}\text{volume}\text{of}\text{solution}}$ (4)

Substitute the values of number of moles and total volume of solution in equation (4).

$\begin{array}{c}\text{Molarity}=\frac{0.0125}{0.125}\text{M}\\ =\text{0}\text{.1}\text{M}\end{array}$

The $\text{ICE}$ table of second step of reaction is,

$\begin{array}{cccccccc}\text{Reaction}& {\text{NaCH}}_3\text{COO}& +& {\text{H}}_2\text{O}& \to & {\text{NaCH}}_3\text{COOH}& +& {\text{OH}}^{-}\\ \text{Initial}& 0.1& & & & 0& & 0\\ \text{Change}& -x& & & & +x& & +x\\ \text{Equilibrium}& \left(0.1-x\right)& & & & x\text{M}& & x\text{}\text{M}\end{array}$

The $K_{\text{a}}$ value of acetic acid is $1.8\times {10}^{-5}$.

The value of $K_{\text{b}}$ is calculated by the formula,

$K_{\text{b}}=\frac{K_{\text{w}}}{K_{\text{a}}}$ (5)

Where,

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

Substitute the values of $K_{\text{b}},K_{\text{a}}\text{and}{K}_{\text{w}}$ in equation (5).

$\begin{array}{c}{K}_{\text{b}}=\frac{{10}^{-10}}{1.8\times {10}^{-5}}\\ =5.6\times {10}^{-10}\end{array}$

The expression of $K_{\text{b}}$ is written as,

$K_b=\frac{\left[\text{BH}\right]\left[{\text{OH}}^{-}\right]}{\left[\text{B}\right]}$ (6)

Where,

• $\left[\text{BH}\right]$ is concentration of ${\text{NaCH}}_3\text{COOH}$ in a solution.
• $\left[\text{B}\right]$ is the concentration of ${\text{NaCH}}_3\text{COO}$ in a solution.
• $\left[{\text{OH}}^{-}\right]$ is the concentration of hydroxide ions.
• $K_{\text{b}}$ is the base dissociation constant.

Substitute the values of $K_{\text{b}},\left[\text{BH}\right],\left[{\text{OH}}^{-}\right]\text{and}\left[\text{B}\right]$ in equation (6).

$\begin{array}{c}5.6\times {10}^{-10}=\frac{x\times x}{0.1}\\ x^2=0.56\times {10}^{-10}\\ x=7.48\times {10}^{-6}\end{array}$

The value of $x=\left[{\text{OH}}^{-}\right]$.

The value of $\text{pOH}$ is calculated by the formula,

$\text{pOH}=-\text{log}\left[x\right]$ (7)

Substitute the value of $x$ in equation (7).

$\begin{array}{c}\text{pOH}=-\text{log}\left(7.48\times {10}^{-6}\right)\\ =5.12\end{array}$

The value of $\text{pH}$ is calculated by the formula,

$\text{pH}=14-\text{pOH}$ (8)

Substitute the value of $\text{pOH}$ in equation (9).

$\begin{array}{c}\text{pH}=14-5.12\\ =\underset{\_}{8.8}\end{array}$

Therefore, the value of $\text{pH}$ at equivalence point of strong base and weak acid is $\underset{\_}{8.8}$.

Similarly,

The concentration of $\text{NaOH}$, $M_1$ is $0.125\text{M}$.

The concentration of strong acid, $M_2$ is $0.50\text{M}$.

The addition of strong acid in a solution, $V_2$ is $25\text{ml}$.

The volume of $\text{NaOH}$, $V_1$ is $x$.

The conversion of units of volume into $\text{L}$.

$\begin{array}{c}1\text{ml}=1000\text{L}\\ 25\text{ml}=0.025\text{L}\end{array}$

The volume required to neutralize the weak acid is calculated by the formula,

$M_1{V}_1=M_2{V}_2$ (1)

Where,

• $M_1$ is the molarity of the base.
• $V_1$ is the volume of the base.
• $M_2$ is the molarity of the acid.
• $V_2$ is the volume of the acid.

Substitute the values of $M_1,V_1,M_2\text{and}{V}_2$ in equation (1).

$\begin{array}{c}0.125\text{M}\times x=0.50\text{M}\times 0.025\text{L}\\ x=\frac{0.0125}{0.125}\\ =0.1\text{L}\end{array}$

Therefore, the volume of sodium hydroxide that is required to titrate the weak acid and strong acid is the same but their $\text{pH}$ value at equivalence point is different.

The strong acids and strong bases combine to form neutralize solution, if their amount is same so, the $\text{pH}$ value in case of strong acid and strong base at equivalence point is $\underset{\_}{7}$.

Conclusion

The value of $\text{pH}$ at equivalence point of strong base and weak acid is $\underset{\_}{8.8}$ and the $\text{pH}$ value in case of strong acid and strong base at equivalence point is $\underset{\_}{7}$.