Chapter 14, Problem 61E

Lactic acid is a common by-product of cellular respiration and is often said to cause the “burn” associated with strenuous activity. A 25.0-mL sample of 0.100 M lactic acid (HC₃H₅O₃, pK_a = 3.86) is titrated with 0.100 M NaOH solution. Calculate the pH after the addition of 0.0 mL, 4.0 mL, 8.0 mL, 12.5 mL, 20.0 mL, 24.0 mL, 24.5 mL, 24.9 mL, 25.0 mL, 25.1 mL, 26.0 mL, 28.0 mL, and 30.0 mL of the NaOH. Plot the results of your calculations as pH versus milliliters of NaOH added.

Interpretation Introduction

Interpretation: The $\text{pH}$ of the solution when $25\text{mL}$ of $0.100\text{M}$ lactic acid is titrated with various given volumes of $0.100\text{M}\text{NaOH}$ and graph between calculated $\text{pH}$ and milliliters of $\text{NaOH}$ added is to be stated.

Concept introduction: Lactic acid $\left({\text{HC}}_3{\text{H}}_5{\text{O}}_3\right)$ is a weak acid and $\text{NaOH}$ is a strong base. When mixed together in a solution, neutralization reaction takes place with the formation of a salt and water molecule.

Answer to Problem 61E

• The $\text{pH}$ when $0.0\text{mL}\text{of}0.100\text{M}\text{NaOH}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}{\text{HC}}_3{\text{H}}_5{\text{O}}_3$ is $\underset{\_}{2.43}$.
• The $\text{pH}$ when $4.0\text{mL}\text{of}0.100\text{M}\text{NaOH}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}{\text{HC}}_3{\text{H}}_5{\text{O}}_3$ is $\underset{\_}{3.06}$.
• The $\text{pH}$ when $8.0\text{mL}\text{of}0.100\text{M}\text{NaOH}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}{\text{HC}}_3{\text{H}}_5{\text{O}}_3$ is $\underset{\_}{3.53}$.
• The $\text{pH}$ when $12.5\text{mL}\text{of}0.100\text{M}\text{NaOH}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}{\text{HC}}_3{\text{H}}_5{\text{O}}_3$ is $\underset{\_}{3.86}$.
• The $\text{pH}$ when $20.0\text{mL}\text{of}0.100\text{M}\text{NaOH}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}{\text{HC}}_3{\text{H}}_5{\text{O}}_3$ is $\underset{\_}{4.46}$.
• The $\text{pH}$ when $24.5\text{mL}\text{of}0.100\text{M}\text{NaOH}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}{\text{HC}}_3{\text{H}}_5{\text{O}}_3$ is $\underset{\_}{4.24}$.
• The $\text{pH}$ when $25.0\text{mL}\text{of}0.100\text{M}\text{NaOH}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}{\text{HC}}_3{\text{H}}_5{\text{O}}_3$ is $\underset{\_}{8.27}$.
• The $\text{pH}$ when $25.1\text{mL}\text{of}0.100\text{M}\text{NaOH}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}{\text{HC}}_3{\text{H}}_5{\text{O}}_3$ is $\underset{\_}{10.29}$.
• The $\text{pH}$ when $26.0\text{mL}\text{of}0.100\text{M}\text{NaOH}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}{\text{HC}}_3{\text{H}}_5{\text{O}}_3$ is $\underset{\_}{11.29}$.
• The $\text{pH}$ when $28.0\text{mL}\text{of}0.100\text{M}\text{NaOH}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}{\text{HC}}_3{\text{H}}_5{\text{O}}_3$ is $\underset{\_}{11.74}$.
• The $\text{pH}$ when $30.0\text{mL}\text{of}0.100\text{M}\text{NaOH}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}{\text{HC}}_3{\text{H}}_5{\text{O}}_3$ is $\underset{\_}{11.95}$.

Explanation of Solution

Explanation

On $\text{0}\text{.0}\text{mL}$ addition of $\text{NaOH}$ to the $25\text{mL}\text{of}\text{0}\text{.100}\text{M}{\text{HC}}_3{\text{H}}_5{\text{O}}_3$, the concentration of lactate ion at equilibrium is calculated by using ICE (Initial Change Equilibrium) table.

$\begin{array}{cccccccc}& {\text{HC}}_5{\text{H}}_5{\text{O}}_3& +& {\text{H}}_2\text{O}& \rightleftharpoons & {\text{C}}_5{\text{H}}_5{\text{O}}_3^{-}& +& {\text{H}}_3{\text{O}}^{+}\\ \text{Initial}\left(\text{M}\right):& 0.100& & & & 0& & 0\\ \text{Change}\left(\text{M}\right):& -x& & & & x& & x\\ \text{Equilibrium}\left(\text{M}\right):& 0.100-x& & & & x& & x\end{array}$

The value of acid dissociation constant at equilibrium is calculated by the formula.

$K_{\text{a}}=\frac{\left[\text{Concentration}\text{of}\text{products}\right]}{\left[\text{Concentration}\text{of}\text{reactants}\right]}$

Therefore, for the above reaction the value of base dissociation constant at equilibrium is,

$K_{\text{b}}=\frac{\left[{\text{C}}_5{\text{H}}_5{\text{NH}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{C}}_5{\text{H}}_5\text{N}\right]}$

The standard value of $K_{\text{a}}$ for lactic acid is $1.38\times {10}^{-4}$.

Substitute the values of concentration of reactants, products and $K_{\text{b}}$ in the above expression.

$\begin{array}{c}1.38\times {10}^{-4}=\frac{x\times x}{0.100-x}\\ x^2+1.38\times {10}^{-4}-1.38\times {10}^{-5}=0\\ x=3.64\times {10}^{-3}\end{array}$

Hence, the value of $\left[{\text{H}}^{+}\right]$ is $3.64\times {10}^{-3}\text{M}$.

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

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

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the above expression.

$\begin{array}{l}\text{pH}=-\log \left[3.64\times {10}^{-3}\right]\\ \text{pH}=\underset{\_}{2.43}\end{array}$

on $\text{4}\text{.0}\text{mL}$ addition of $\text{NaOH}$ to the $25\text{mL}\text{of}\text{0}\text{.100}\text{M}{\text{HC}}_3{\text{H}}_5{\text{O}}_3$,the concentration of lactate ion at equilibrium is calculated by using reaction stoichiometric coefficients.

$\begin{array}{cccccc}& {\text{HC}}_5{\text{H}}_5{\text{O}}_3& +& \text{NaOH}& \rightleftharpoons & {\text{C}}_5{\text{H}}_5{\text{O}}_3\text{Na}\\ \text{Number}\text{of}\text{Moles}\text{before}\text{the}\text{reaction}& 0.0025& & 0.0004& & 0\\ \text{Number}\text{of}\text{Moles}\text{after}\text{the}\text{reaction}& 0.0021& & 0& & 0.0004\end{array}$

The value of $\text{pH}$ is calculated by the Henderson-Hasselbalch equation.

$\text{pH}=\text{p}{K}_{\text{a}}+\text{log}\frac{\left[\text{Salt}\right]}{\left[\text{Acid}\right]}$

Where,

• $\text{p}{K}_{\text{a}}$ is negative logarithm of $K_{\text{a}}$.
• $\text{pH}$ is power of $\left[{\text{H}}^{+}\right]$.
• $\left[\text{Salt}\right]$ is concentration of base after the reaction.
• $\left[\text{Acid}\right]$ is concentration of acid after the reaction.

The standard value of $\text{p}{K}_{\text{a}}$ for pyridine is $3.86$.

Substitute the values of concentration of salt, acid and $p{K}_{\text{a}}$ in the above expression.

$\begin{array}{l}\text{pH}=3.86+\text{log}\left[\frac{0.0004}{0.0025}\right]\\ \text{pH}=\underset{\_}{3.06}\end{array}$

On $\text{8}\text{.0}\text{mL}$ addition of $\text{NaOH}$ to the $25\text{mL}\text{of}\text{0}\text{.100}\text{M}{\text{HC}}_3{\text{H}}_5{\text{O}}_3$, the concentration of lactate ion at equilibrium is calculated by using reaction stoichiometric coefficients.

$\begin{array}{cccccc}& {\text{HC}}_5{\text{H}}_5{\text{O}}_3& +& \text{NaOH}& \rightleftharpoons & {\text{C}}_5{\text{H}}_5{\text{O}}_3\text{Na}\\ \text{Number}\text{of}\text{Moles}\text{before}\text{the}\text{reaction}& 0.0025& & 0.0008& & 0\\ \text{Number}\text{of}\text{Moles}\text{after}\text{the}\text{reaction}& 0.0017& & 0& & 0.0008\end{array}$

The value of $\text{pH}$ is calculated by the Henderson-Hasselbalch equation.

$\text{pH}=\text{p}{K}_{\text{a}}+\text{log}\frac{\left[\text{Salt}\right]}{\left[\text{Acid}\right]}$

Where,

• $\text{p}{K}_{\text{a}}$ is negative logarithm of $K_{\text{a}}$.
• $\text{pH}$ is power of $\left[{\text{H}}^{+}\right]$.
• $\left[\text{Salt}\right]$ is concentration of base after the reaction.
• $\left[\text{Acid}\right]$ is concentration of acid after the reaction.

The standard value of $\text{p}{K}_{\text{a}}$ for pyridine is $3.86$.

Substitute the values of concentration of salt, acid and $\text{p}{K}_{\text{a}}$ in the above expression.

$\begin{array}{l}\text{pH}=3.86+\text{log}\left[\frac{0.0008}{0.0017}\right]\\ \text{pH}=\underset{\_}{3.53}\end{array}$

On $\text{12}\text{.5}\text{mL}$ addition of $\text{NaOH}$ to the $25\text{mL}\text{of}\text{0}\text{.100}\text{M}{\text{HC}}_3{\text{H}}_5{\text{O}}_3$, the concentration of lactate ion at equilibrium is calculated by using reaction stoichiometric coefficients.

$\begin{array}{cccccc}& {\text{HC}}_5{\text{H}}_5{\text{O}}_3& +& \text{NaOH}& \rightleftharpoons & {\text{C}}_5{\text{H}}_5{\text{O}}_3\text{Na}\\ \text{Number}\text{of}\text{Moles}\text{before}\text{the}\text{reaction}& 0.0025& & 0.00125& & 0\\ \text{Number}\text{of}\text{Moles}\text{after}\text{the}\text{reaction}& 0.000125& & 0& & 0.000125\end{array}$

The value of $\text{pH}$ is calculated by the Henderson-Hasselbalch equation.

$\text{pH}=\text{p}{K}_{\text{a}}+\text{log}\frac{\left[\text{Salt}\right]}{\left[\text{Acid}\right]}$

Where,

• $\text{p}{K}_{\text{a}}$ is negative logarithm of $K_{\text{a}}$.
• $\text{pH}$ is power of $\left[{\text{H}}^{+}\right]$.
• $\left[\text{Salt}\right]$ is concentration of base after the reaction.
• $\left[\text{Acid}\right]$ is concentration of acid after the reaction.

The standard value of $\text{p}{K}_{\text{a}}$ for pyridine is $3.86$.

Substitute the values of concentration of salt, acid and $p{K}_{\text{a}}$ in the above expression.

$\begin{array}{l}\text{pH}=3.86+\text{log}\left[\frac{0.00125}{0.00125}\right]\\ \text{pH}=\underset{\_}{3.86}\end{array}$

On $\text{20}\text{.0}\text{mL}$ addition of $\text{NaOH}$ to the $25\text{mL}\text{of}\text{0}\text{.100}\text{M}{\text{HC}}_3{\text{H}}_5{\text{O}}_3$, the concentration of lactate ion at equilibrium is calculated by using reaction stoichiometric coefficients.

$\begin{array}{cccccc}& {\text{HC}}_5{\text{H}}_5{\text{O}}_3& +& \text{NaOH}& \rightleftharpoons & {\text{C}}_5{\text{H}}_5{\text{O}}_3\text{Na}\\ \text{Number}\text{of}\text{Moles}\text{before}\text{the}\text{reaction}& 0.0025& & 0.002& & 0\\ \text{Number}\text{of}\text{Moles}\text{after}\text{the}\text{reaction}& 0.0005& & 0& & 0.002\end{array}$

The value of $\text{pH}$ is calculated by the Henderson-Hasselbalch equation.

$\text{pH}=\text{p}{K}_{\text{a}}+\text{log}\frac{\left[\text{Salt}\right]}{\left[\text{Acid}\right]}$

Where,

• $\text{p}{K}_{\text{a}}$ is negative logarithm of $K_{\text{a}}$.
• $\text{pH}$ is power of $\left[{\text{H}}^{+}\right]$.
• $\left[\text{Salt}\right]$ is concentration of base after the reaction.
• $\left[\text{Acid}\right]$ is concentration of acid after the reaction.

The standard value of $\text{p}{K}_{\text{a}}$ for pyridine is $3.86$.

Substitute the values of concentration of salt, acid and $p{K}_{\text{a}}$ in the above expression.

$\begin{array}{l}\text{pH}=3.86+\text{log}\left[\frac{0.002}{0.0025}\right]\\ \text{pH}=\underset{\_}{4.46}\end{array}$

On $\text{24}\text{.5}\text{mL}$ addition of $\text{NaOH}$ to the $25\text{mL}\text{of}\text{0}\text{.100}\text{M}{\text{HC}}_3{\text{H}}_5{\text{O}}_3$, the concentration of lactate ion at equilibrium is calculated by using reaction stoichiometric coefficients.

$\begin{array}{cccccc}& {\text{HC}}_5{\text{H}}_5{\text{O}}_3& +& \text{NaOH}& \rightleftharpoons & {\text{C}}_5{\text{H}}_5{\text{O}}_3\text{Na}\\ \text{Number}\text{of}\text{Moles}\text{before}\text{the}\text{reaction}& 0.0025& & 0.0024& & 0\\ \text{Number}\text{of}\text{Moles}\text{after}\text{the}\text{reaction}& 0.0001& & 0& & 0.0024\end{array}$

The value of $\text{pH}$ is calculated by the Henderson-Hasselbalch equation.

$\text{pH}=\text{p}{K}_{\text{a}}+\text{log}\frac{\left[\text{Salt}\right]}{\left[\text{Acid}\right]}$

Where,

• $\text{p}{K}_{\text{a}}$ is negative logarithm of $K_{\text{a}}$.
• $\text{pH}$ is power of $\left[{\text{H}}^{+}\right]$.
• $\left[\text{Salt}\right]$ is concentration of base after the reaction.
• $\left[\text{Acid}\right]$ is concentration of acid after the reaction.

The standard value of $\text{p}{K}_{\text{a}}$ for pyridine is $3.86$.

Substitute the values of concentration of salt, acid and $\text{p}{K}_{\text{a}}$ in the above expression.

$\begin{array}{l}\text{pH}=3.86+\text{log}\left[\frac{0.0024}{0.0001}\right]\\ \text{pH}=\underset{\_}{4.24}\end{array}$

On $\text{25}\text{.0}\text{mL}$ addition of $\text{NaOH}$ to the $25\text{mL}\text{of}\text{0}\text{.100}\text{M}{\text{HC}}_3{\text{H}}_5{\text{O}}_3$, the concentration of lactate ion at equilibrium is calculated by using reaction stoichiometric coefficients.

$\begin{array}{cccccc}& {\text{HC}}_5{\text{H}}_5{\text{O}}_3& +& \text{NaOH}& \rightleftharpoons & {\text{C}}_5{\text{H}}_5{\text{O}}_3\text{Na}\\ \text{Number}\text{of}\text{Moles}\text{before}\text{the}\text{reaction}& 0.0025& & 0.0025& & 0\\ \text{Number}\text{of}\text{Moles}\text{after}\text{the}\text{reaction}& 0& & 0& & 0.0025\end{array}$

At this stage complete consumption of lactic acid and $\text{NaOH}$ takes place with the formation of salt. The strength of the salt is calculated by considering the dissociation reaction of salt formed as.

${\text{C}}_5{\text{H}}_5{\text{O}}_3\text{Na}\rightleftharpoons {\text{C}}_5{\text{H}}_5{\text{O}}_3^{-}+{\text{Na}}^{+}$

The concentration of lactate ion $\left({\text{C}}_5{\text{H}}_5{\text{O}}_3^{-}\right)$$=\frac{\text{Moles}\text{of}\text{lacate}\text{ion}}{\text{Volume}}$

Substitute the value of moles and volume in the above expression.

$\begin{array}{l}\text{Concentration}=\frac{0.0025}{0.05}\\ \text{Concentration}=0.05\text{M}\end{array}$

The value of $\left[{\text{H}}^{+}\right]$ is calculated by using ICE table at dissociation reaction of lactate ion.

$\begin{array}{cccccccc}& {\text{C}}_5{\text{H}}_5{\text{O}}_3^{-}& +& {\text{H}}_2\text{O}& \rightleftharpoons & {\text{C}}_5{\text{H}}_5{\text{O}}_3\text{H}& +& {\text{OH}}^{-}\\ \text{Initial}\left(\text{M}\right):& 0.05& & & & 0& & 0\\ \text{Change}\left(\text{M}\right):& -y& & & & y& & y\\ \text{Equilibrium}\left(\text{M}\right):& 0.05-y& & & & y& & y\end{array}$

The value of base dissociation constant at equilibrium is calculated by the formula.

$K_{\text{b}}=\frac{\left[\text{Concentration}\text{of}\text{products}\right]}{\left[\text{Concentration}\text{of}\text{reactants}\right]}$

Therefore, for the above reaction the value of acid dissociation constant at equilibrium is,

$K_b=\frac{\left[{\text{C}}_5{\text{H}}_5\text{OH}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{C}}_5{\text{H}}_5{\text{O}}_3^{-}\right]}$

The relation,

$K_{\text{b}}=\frac{K_{\text{w}}}{K_{\text{a}}}$

The value of $K_{\text{a}}$ is $1.38\times {10}^{-4}$.

Substitute the value of $K_{\text{w}}$ and $K_{\text{a}}$ in the above expression.

$\begin{array}{l}{K}_{\text{b}}=\frac{1.0\times {10}^{-14}}{1.38\times {10}^{-4}}\\ K_{\text{b}}=7.24\times {10}^{-11}\end{array}$

Substitute the values of concentration of reactants, products and $K_{\text{b}}$ in the $K_{\text{b}}$ expression.

$\begin{array}{c}7.24\times {10}^{-11}=\frac{x\times x}{0.05-x}\\ x^2+7.24\times {10}^{-11}-3.62\times {10}^{-12}=0\\ x=1.90\times {10}^{-6}\end{array}$

Hence, the value of $\left[{\text{OH}}^{-}\right]$ is $0.000013\text{M}$

The value of $\left[{\text{H}}^{+}\right]$ is calculated by the formula.

$\left[{\text{H}}^{+}\right]=\frac{\left[1.0\times {10}^{-14}\right]}{\left[{\text{OH}}^{-}\right]}$

Substitute the value of $\left[{\text{OH}}^{-}\right]$ in the above expression.

$\begin{array}{l}\left[{\text{H}}^{+}\right]=\frac{\left[1.0\times {10}^{-14}\right]}{\left[1.90\times {10}^{-6}\right]}\\ \left[{\text{H}}^{+}\right]=5.26\times {10}^{-9}\end{array}$

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

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

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the above expression.

$\begin{array}{l}\text{pH}=-\log \left[5.26\times {10}^{-9}\right]\\ \text{pH}=\underset{\_}{8.27}\end{array}$

On $\text{25}\text{.1}\text{mL}$ addition of $\text{NaOH}$ to the $25\text{mL}\text{of}\text{0}\text{.100}\text{M}\text{latic}\text{acid}$, the concentration of lactate ion at equilibrium is calculated by using reaction stoichiometric coefficients.

$\begin{array}{cccccc}& {\text{HC}}_3{\text{H}}_5{\text{O}}_3& +& \text{NaOH}& \rightleftharpoons & {\text{C}}_3{\text{H}}_5{\text{O}}_3\text{Na}\\ \text{Number}\text{of}\text{Moles}\text{before}\text{the}\text{reaction}& 0.0025& & 0.00251& & 0\\ \text{Number}\text{of}\text{Moles}\text{after}\text{the}\text{reaction}& 0& & 0.00001& & 0.0025\end{array}$

The $\left[{\text{OH}}^{-}\right]$ value $\begin{array}{l}=\frac{0.00001}{0.025+0.026}\\ =0.000196\text{M}\end{array}$

The value of $\left[{\text{H}}^{+}\right]$ is calculated by the formula.

$\left[{\text{H}}^{+}\right]=\frac{\left[1.0\times {10}^{-14}\right]}{\left[{\text{OH}}^{-}\right]}$

Substitute the value of $\left[{\text{OH}}^{-}\right]$ in the above expression.

$\begin{array}{l}\left[{\text{H}}^{+}\right]=\frac{\left[1.0\times {10}^{-14}\right]}{\left[1.96\times {10}^{-4}\right]}\\ \left[{\text{H}}^{+}\right]=5.10\times {10}^{-11}\end{array}$

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

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

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the above expression.

$\begin{array}{l}\text{pH}=-\log \left[5.10\times {10}^{-11}\right]\\ \text{pH}=\underset{\_}{10.29}\end{array}$

On $\text{26}\text{mL}$ addition of $\text{NaOH}$ to the $25\text{mL}\text{of}\text{0}\text{.100}\text{M}\text{lactic}\text{acid}$, the concentration of lactate ion at equilibrium is calculated by using reaction stoichiometric coefficients.

$\begin{array}{cccccc}& {\text{HC}}_3{\text{H}}_5{\text{O}}_3& +& \text{NaOH}& \rightleftharpoons & {\text{C}}_3{\text{H}}_5{\text{O}}_3\text{Na}\\ \text{Number}\text{of}\text{Moles}\text{before}\text{the}\text{reaction}& 0.0025& & 0.0026& & 0\\ \text{Number}\text{of}\text{Moles}\text{after}\text{the}\text{reaction}& 0& & 0.0001& & 0.0025\end{array}$

The $\left[{\text{OH}}^{-}\right]$ value $\begin{array}{l}=\frac{0.0001}{0.025+0.026}\\ =0.00196\text{M}\end{array}$

The value of $\left[{\text{H}}^{+}\right]$ is calculated by the formula,

$\left[{\text{H}}^{+}\right]=\frac{\left[1.0\times {10}^{-14}\right]}{\left[{\text{OH}}^{-}\right]}$

Substitute the value of $\left[{\text{OH}}^{-}\right]$ in the above expression.

$\begin{array}{l}\left[{\text{H}}^{+}\right]=\frac{\left[1.0\times {10}^{-14}\right]}{\left[1.96\times {10}^{-3}\right]}\\ \left[{\text{H}}^{+}\right]=5.10\times {10}^{-12}\end{array}$

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

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

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the above expression.

$\begin{array}{l}\text{pH}=-\log \left[1.96\times {10}^{-4}\right]\\ \text{pH}=\underset{\_}{11.29}\end{array}$

On $\text{28}\text{mL}$ addition of $\text{NaOH}$ to the $25\text{mL}\text{of}\text{0}\text{.100}\text{M}\text{lactic}\text{acid}$,the concentration of lactate ion at equilibrium is calculated by using reaction stoichiometric coefficients.

$\begin{array}{cccccc}& {\text{HC}}_3{\text{H}}_5{\text{O}}_3& +& \text{NaOH}& \rightleftharpoons & {\text{C}}_3{\text{H}}_5{\text{O}}_3\text{Na}\\ \text{Number}\text{of}\text{Moles}\text{before}\text{the}\text{reaction}& 0.0025& & 0.0028& & 0\\ \text{Number}\text{of}\text{Moles}\text{after}\text{the}\text{reaction}& 0& & 0.0003& & 0.0025\end{array}$

The $\left[{\text{OH}}^{-}\right]$ value $\begin{array}{l}=\frac{0.0003}{0.025+0.028}\\ =5.6\times {10}^{-3}\text{M}\end{array}$

The value of $\left[{\text{H}}^{+}\right]$ is calculated by the formula.

$\left[{\text{H}}^{+}\right]=\frac{\left[1.0\times {10}^{-14}\right]}{\left[{\text{OH}}^{-}\right]}$

Substitute the value of $\left[{\text{OH}}^{-}\right]$ in the above expression.

$\begin{array}{l}\left[{\text{H}}^{+}\right]=\frac{\left[1.0\times {10}^{-14}\right]}{\left[5.6\times {10}^{-3}\right]}\\ \left[{\text{H}}^{+}\right]=1.78\times {10}^{-12}\end{array}$

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

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

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the above expression.

$\begin{array}{l}\text{pH}=-\log \left[1.78\times {10}^{-12}\right]\\ \text{pH}=\underset{\_}{11.74}\end{array}$

At $30\text{mL}$ addition of $\text{NaOH}$ to the $25\text{mL}\text{of}\text{0}\text{.100}\text{M}\text{lactic}\text{acid}$, the concentration of lactate ion at equilibrium is calculated by using reaction stoichiometric coefficients.

$\begin{array}{cccccc}& {\text{HC}}_3{\text{H}}_5{\text{O}}_3& +& \text{NaOH}& \rightleftharpoons & {\text{C}}_3{\text{H}}_5{\text{O}}_3\text{Na}\\ \text{Number}\text{of}\text{Moles}\text{before}\text{the}\text{reaction}& 0.0025& & 0.0030& & 0\\ \text{Number}\text{of}\text{Moles}\text{after}\text{the}\text{reaction}& 0& & 0.0005& & 0.0025\end{array}$

The $\left[{\text{OH}}^{-}\right]$ value $\begin{array}{l}=\frac{0.0005}{0.025+0.028}\\ =9.09\times {10}^{-3}\text{M}\end{array}$

The value of $\left[{\text{H}}^{+}\right]$ is calculated by the formula.

$\left[{\text{H}}^{+}\right]=\frac{\left[1.0\times {10}^{-14}\right]}{\left[{\text{OH}}^{-}\right]}$

Substitute the value of $\left[{\text{OH}}^{-}\right]$ in the above expression.

$\begin{array}{l}\left[{\text{H}}^{+}\right]=\frac{\left[1.0\times {10}^{-14}\right]}{\left[9.09\times {10}^{-3}\right]}\\ \left[{\text{H}}^{+}\right]=1.1\times {10}^{-12}\end{array}$

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

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

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the above expression.

$\begin{array}{l}\text{pH}=-\log \left[1.1\times {10}^{-12}\right]\\ \text{pH}=\underset{\_}{11.95}\end{array}$

The values of $\text{pH}$ corresponding to the different volumes of $\text{NaOH}$ added are shown in the following table.

$\text{Volume}\text{of}\text{NaOH}\text{added}$	$\text{pH}$
$0$	$2.43$
$4$	$3.06$
$8$	$3.53$
$12.5$	$3.86$
$20$	$4.46$
$24.5$	$4.24$
$25$	$8.27$
$25.1$	$10.29$
$26$	$11.29$
$28$	$11.74$
$30$	$11.95$

Table 1

The graph between the volume of $\text{NaOH}$ added and the corresponding $\text{pH}$ values is stated as follows.

Conclusion

Conclusion

• The $\text{pH}$ when $0.0\text{mL}\text{of}0.100\text{M}\text{NaOH}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}{\text{HC}}_3{\text{H}}_5{\text{O}}_3$ is $\underset{\_}{2.43}$.
• The $\text{pH}$ when $4.0\text{mL}\text{of}0.100\text{M}\text{NaOH}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}{\text{HC}}_3{\text{H}}_5{\text{O}}_3$ is $\underset{\_}{3.06}$.
• The $\text{pH}$ when $8.0\text{mL}\text{of}0.100\text{M}\text{NaOH}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}{\text{HC}}_3{\text{H}}_5{\text{O}}_3$ is $\underset{\_}{3.53}$.
• The $\text{pH}$ when $12.5\text{mL}\text{of}0.100\text{M}\text{NaOH}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}{\text{HC}}_3{\text{H}}_5{\text{O}}_3$ is $\underset{\_}{3.86}$.
• The $\text{pH}$ when $20.0\text{mL}\text{of}0.100\text{M}\text{NaOH}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}{\text{HC}}_3{\text{H}}_5{\text{O}}_3$ is $\underset{\_}{4.46}$.
• The $\text{pH}$ when $24.5\text{mL}\text{of}0.100\text{M}\text{NaOH}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}{\text{HC}}_3{\text{H}}_5{\text{O}}_3$ is $\underset{\_}{4.24}$.
• The $\text{pH}$ when $25.0\text{mL}\text{of}0.100\text{M}\text{NaOH}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}{\text{HC}}_3{\text{H}}_5{\text{O}}_3$ is $\underset{\_}{8.27}$.
• The $\text{pH}$ when $25.1\text{mL}\text{of}0.100\text{M}\text{NaOH}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}{\text{HC}}_3{\text{H}}_5{\text{O}}_3$ is $\underset{\_}{10.29}$.
• The $\text{pH}$ when $26.0\text{mL}\text{of}0.100\text{M}\text{NaOH}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}{\text{HC}}_3{\text{H}}_5{\text{O}}_3$ is $\underset{\_}{11.29}$.
• The $\text{pH}$ when $28.0\text{mL}\text{of}0.100\text{M}\text{NaOH}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}{\text{HC}}_3{\text{H}}_5{\text{O}}_3$ is $\underset{\_}{11.74}$.
• The $\text{pH}$ when $30.0\text{mL}\text{of}0.100\text{M}\text{NaOH}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}{\text{HC}}_3{\text{H}}_5{\text{O}}_3$ is $\underset{\_}{11.95}$