Chapter 15, Problem 64E

Repeat the procedure in Exercise 61, but for the titration of 25.0 mL of 0.100 M pyridine with 0.100 M hydrochloric acid (K_b for pyridine is 1.7 × 10⁻⁹). Do not calculate the points at 24.9 and 25.1 mL.

Interpretation Introduction

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

Concept introduction: Pyridine is a weak base and $\text{HCl}$ is a strong acid. When mixed together in a solution, neutralization reaction takes place with the formation of a salt and water molecule.

Concept introduction: Pyridine is a weak base and $\text{HCl}$ is a strong acid. When mixed together in a solution, neutralization reaction takes place with the formation of a salt and water molecule.

Answer to Problem 64E

Answer

• The $\text{pH}$ when $0.0\text{mL}\text{of}0.100\text{M}\text{HCl}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}\text{pyridine}$ is $\underset{\_}{9.11}$.
• The $\text{pH}$ when $4.0\text{mL}\text{of}0.100\text{M}\text{HCl}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}\text{pyridine}$ is $\underset{\_}{5.92}$.
• The $\text{pH}$ when $8.0\text{mL}\text{of}0.100\text{M}\text{HCl}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}\text{pyridine}$ is $\underset{\_}{5.53}$.
• The $\text{pH}$ when $12.5\text{mL}\text{of}0.100\text{M}\text{HCl}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}\text{pyridine}$ is $\underset{\_}{5.2}$.
• The $\text{pH}$ when $20.0\text{mL}\text{of}0.100\text{M}\text{HCl}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}\text{pyridine}$ is $\underset{\_}{4.6}$.
• The $\text{pH}$ when $24.5\text{mL}\text{of}0.100\text{M}\text{HCl}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}\text{pyridine}$ is $\underset{\_}{3.82}$.
• The $\text{pH}$ when $25.0\text{mL}\text{of}0.100\text{M}\text{HCl}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}\text{pyridine}$ is $\underset{\_}{0.05}$.
• The $\text{pH}$ when $26.0\text{mL}\text{of}0.100\text{M}\text{HCl}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}\text{pyridine}$ is $\underset{\_}{2.71}$.
• The $\text{pH}$ when $28.0\text{mL}\text{of}0.100\text{M}\text{HCl}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}\text{pyridine}$ is $\underset{\_}{2.24}$.
• The $\text{pH}$ when $30.0\text{mL}\text{of}0.100\text{M}\text{HCl}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}\text{pyridine}$ is $\underset{\_}{2.04}$.

Explanation of Solution

Explanation

Given

The value of $K_{\text{b}}$ for pyridine is $1.7\times {10}^{-9}$.

At $\text{0}\text{.0}\text{mL}$, addition of $\text{HCl}$ to the $25\text{mL}\text{of}\text{0}\text{.100}\text{M}\text{pyridine}$, the concentration of pyridinium ion at equilibrium is calculated by using ICE (Initial Change Equilibrium) table.

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

The value of 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 base dissociation constant at equilibrium is,

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

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

$\begin{array}{c}1.7\times {10}^{-9}\times {10}^{-9}=\frac{x\times x}{0.100-x}\\ x^2+1.7\times {10}^{-9}-1.7\times {10}^{-10}=0\\ x=0.000013\end{array}$

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

The ionic-product of water is,

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

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

$\begin{array}{l}\left[{\text{H}}^{+}\right]\times 0.000013=1.0\times {10}^{-14}\\ \left[{\text{H}}^{+}\right]=\frac{1.0\times {10}^{-14}}{0.000013}\\ \left[{\text{H}}^{+}\right]=7.69\times {10}^{-10}\text{M}\end{array}$

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.

$\begin{array}{l}\text{pH}=-\log \left[7.69\times {10}^{-10}\right]\\ \text{pH}=\underset{\_}{9.11}\end{array}$

At $\text{4}\text{.0}\text{mL}$, addition of $\text{HCl}$ to the $25\text{mL}\text{of}\text{0}\text{.100}\text{M}\text{pyridine}$, the concentration of pyridinium ion at equilibrium is calculated by using the stoichiometry of the reaction.

$\begin{array}{cccccc}& {\text{C}}_5{\text{H}}_5\text{N}& +& \text{HCl}& \rightleftharpoons & {\text{C}}_5{\text{H}}_5\text{NHCl}\\ \text{Number}\text{of}\text{Moles}\text{before}\text{the}\text{reaction}& 0.0025& & 0.0004& & 0\\ \text{Number}\text{of}\text{Moles}\text{after}\text{the}\text{reaction}& 0.0021& & 0& & 0.0004\end{array}$

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

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

Where,

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

The standard value of $p{K}_{\text{a}}$ for pyridine is $5.2$.

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

$\begin{array}{l}\text{pH}=5.2+\text{log}\left[\frac{0.0021}{0.0004}\right]\\ \text{pH}=5.92\end{array}$

At $\text{8}\text{.0}\text{mL}$, addition of $\text{HCl}$ to the $25\text{mL}\text{of}\text{0}\text{.100}\text{M}\text{pyridine}$,the concentration of pyridinium ion at equilibrium is calculated by using the stoichiometry of the reaction.

$\begin{array}{cccccc}& {\text{C}}_5{\text{H}}_5\text{N}& +& \text{HCl}& \rightleftharpoons & {\text{C}}_5{\text{H}}_5\text{NHCl}\\ \text{Number}\text{of}\text{Moles}\text{before}\text{the}\text{reaction}& 0.0025& & 0.0008& & 0\\ \text{Number}\text{of}\text{Moles}\text{after}\text{the}\text{reaction}& 0.0017& & 0& & 0.0008\end{array}$

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

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

Where,

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

The standard value of $p{K}_{\text{a}}$ for pyridine is $5.2$.

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

$\begin{array}{l}\text{pH}=5.2+\text{log}\left[\frac{0.0017}{0.0008}\right]\\ \text{pH}=5.53\end{array}$

At $\text{12}\text{.5}\text{mL}$, addition of $\text{HCl}$ to the $25\text{mL}\text{of}\text{0}\text{.100}\text{M}\text{pyridine}$,the concentration of pyridinium ion at equilibrium is calculated by using the stoichiometry of the reaction.

$\begin{array}{cccccc}& {\text{C}}_5{\text{H}}_5\text{N}& +& \text{HCl}& \rightleftharpoons & {\text{C}}_5{\text{H}}_5\text{NHCl}\\ \text{Number}\text{of}\text{Moles}\text{before}\text{the}\text{reaction}& 0.0025& & 0.00125& & 0\\ \text{Number}\text{of}\text{Moles}\text{after}\text{the}\text{reaction}& 0.00125& & 0& & 0.00125\end{array}$

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

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

Where,

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

The standard value of $p{K}_{\text{a}}$ for pyridine is $5.2$.

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

$\begin{array}{l}\text{pH}=5.2+\text{log}\left[\frac{0.00125}{0.00125}\right]\\ \text{pH}=5.2\end{array}$

At $\text{20}\text{.0}\text{mL}$ addition of $\text{HCl}$ to the $25\text{mL}\text{of}\text{0}\text{.100}\text{M}\text{pyridine}$,the concentration of pyridinium ion at equilibrium is calculated by using the stoichiometry of the reaction.

$\begin{array}{cccccc}& {\text{C}}_5{\text{H}}_5\text{N}& +& \text{HCl}& \rightleftharpoons & {\text{C}}_5{\text{H}}_5\text{NHCl}\\ \text{Number}\text{of}\text{Moles}\text{before}\text{the}\text{reaction}& 0.0025& & 0.0020& & 0\\ \text{Number}\text{of}\text{Moles}\text{after}\text{the}\text{reaction}& 0.005& & 0& & 0.0020\end{array}$

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

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

Where,

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

The standard value of $p{K}_{\text{a}}$ for pyridine is $5.2$.

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

$\begin{array}{l}\text{pH}=5.2+\text{log}\left[\frac{0.0005}{0.002}\right]\\ \text{pH}=4.6\end{array}$

At $\text{24}\text{.5}\text{mL}$ addition of $\text{HCl}$ to the $25\text{mL}\text{of}\text{0}\text{.100}\text{M}\text{pyridine}$,the concentration of pyridinium ion at equilibrium is calculated by using the stoichiometry of the reaction.

$\begin{array}{cccccc}& {\text{C}}_5{\text{H}}_5\text{N}& +& \text{HCl}& \rightleftharpoons & {\text{C}}_5{\text{H}}_5\text{NHCl}\\ \text{Number}\text{of}\text{Moles}\text{before}\text{the}\text{reaction}& 0.0025& & 0.0024& & 0\\ \text{Number}\text{of}\text{Moles}\text{after}\text{the}\text{reaction}& 0.0001& & 0& & 0.0024\end{array}$

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

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

Where,

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

The standard value of $p{K}_{\text{a}}$ for pyridine is $5.2$.

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

$\begin{array}{l}\text{pH}=5.2+\text{log}\left[\frac{0.001}{0.0025}\right]\\ \text{pH}=3.82\end{array}$

At $\text{25}\text{.0}\text{mL}$ addition of $\text{HCl}$ to the $25\text{mL}\text{of}\text{0}\text{.100}\text{M}\text{pyridine}$,the concentration of pyridinium ion at equilibrium is calculated by using the stoichiometry of the reaction.

$\begin{array}{cccccc}& {\text{C}}_5{\text{H}}_5\text{N}& +& \text{HCl}& \rightleftharpoons & {\text{C}}_5{\text{H}}_5\text{NHCl}\\ \text{Number}\text{of}\text{Moles}\text{before}\text{the}\text{reaction}& 0.0025& & 0.0025& & 0\\ \text{Number}\text{of}\text{Moles}\text{after}\text{the}\text{reaction}& 0& & 0& & 0.0025\end{array}$

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

${\text{C}}_5{\text{H}}_5\text{NHCl}\rightleftharpoons {\text{C}}_5{\text{H}}_5{\text{NH}}^{+}+{\text{Cl}}^{-}$

The concentration of  pyridium ion $\left({\text{C}}_5{\text{H}}_{\varphi 5}{\text{NH}}^{+}\right)$$=\frac{\text{Moles}\text{of}\text{pyridium}\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 pyridium ion.

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

The value of 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,.

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

The relation,

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

The value of $K_{\text{b}}$ is $1.7\times {10}^{-9}$.

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

$\begin{array}{l}{K}_{\text{a}}=\frac{1.0\times {10}^{-14}}{1.7\times {10}^{-9}}\\ K_{\text{a}}=5.88\times {10}^{-6}\end{array}$

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

$\begin{array}{c}5.88\times {10}^{-6}=\frac{x\times x}{0.05-x}\\ x^2+5.88\times {10}^{-6}-2.94\times {10}^{-7}=0\\ x=5.42\times {10}^{-4}\end{array}$

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

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.

$\begin{array}{l}\text{pH}=-\log \left[5.42\times {10}^{-4}\right]\\ \text{pH}=\underset{\_}{3.3}\end{array}$

At $\text{26}\text{.0}\text{mL}$ addition of $\text{HCl}$ to the $25\text{mL}\text{of}\text{0}\text{.100}\text{M}\text{pyridine}$,the concentration of pyridinium ion at equilibrium is calculated by using the stoichiometry of the reaction.

$\begin{array}{cccccc}& {\text{C}}_5{\text{H}}_5\text{N}& +& \text{HCl}& \rightleftharpoons & {\text{C}}_5{\text{H}}_5\text{NHCl}\\ \text{Number}\text{of}\text{Moles}\text{before}\text{the}\text{reaction}& 0.0025& & 0.0026& & 0\\ \text{Number}\text{of}\text{Moles}\text{after}\text{the}\text{reaction}& 0& & 0.0001& & 0.0025\end{array}$

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

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.

$\begin{array}{l}\text{pH}=-\log \left[0.00196\right]\\ \text{pH}=\underset{\_}{2.71}\end{array}$

At $\text{28}\text{.0}\text{mL}$ addition of $\text{HCl}$ to the $25\text{mL}\text{of}\text{0}\text{.100}\text{M}\text{pyridine}$,the concentration of pyridinium ion at equilibrium is calculated by using reaction stoichiometric coefficients.

$\begin{array}{cccccc}& {\text{C}}_5{\text{H}}_5\text{N}& +& \text{HCl}& \rightleftharpoons & {\text{C}}_5{\text{H}}_5\text{NHCl}\\ \text{Number}\text{of}\text{Moles}\text{before}\text{the}\text{reaction}& 0.0025& & 0.0028& & 0\\ \text{Number}\text{of}\text{Moles}\text{after}\text{the}\text{reaction}& 0& & 0.0003& & 0.0025\end{array}$

The $\left[{\text{H}}^{+}\right]$ value $\begin{array}{l}=\frac{0.0003}{0.025+0.026}\\ =0.00566\text{M}\end{array}$

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.

$\begin{array}{l}\text{pH}=-\log \left[0.00566\right]\\ \text{pH}=\underset{\_}{2.24}\end{array}$

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

$\begin{array}{cccccc}& {\text{C}}_5{\text{H}}_5\text{N}& +& \text{HCl}& \rightleftharpoons & {\text{C}}_5{\text{H}}_5\text{NHCl}\\ \text{Number}\text{of}\text{Moles}\text{before}\text{the}\text{reaction}& 0.0025& & 0.030& & 0\\ \text{Number}\text{of}\text{Moles}\text{after}\text{the}\text{reaction}& 0& & 0.0005& & 0.0025\end{array}$

The $\left[{\text{H}}^{+}\right]$ value $\begin{array}{l}=\frac{0.0005}{0.025+0.030}\\ =0.009M\end{array}$

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.

$\begin{array}{l}\text{pH}=-\log \left[0.009\right]\\ \text{pH}=\underset{\_}{2.04}\end{array}$

The graph between $\text{pH}$ and the $\text{volume}\text{of}\text{HCl}\text{used}$ is,

Conclusion

Conclusion

• The $\text{pH}$ when $0.0\text{mL}\text{of}0.100\text{M}\text{HCl}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}\text{pyridine}$ is $\underset{\_}{9.11}$.
• The $\text{pH}$ when $4.0\text{mL}\text{of}0.100\text{M}\text{HCl}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}\text{pyridine}$ is $\underset{\_}{5.92}$.
• The $\text{pH}$ when $8.0\text{mL}\text{of}0.100\text{M}\text{HCl}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}\text{pyridine}$ is $\underset{\_}{5.53}$.
• The $\text{pH}$ when $12.5\text{mL}\text{of}0.100\text{M}\text{HCl}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}\text{pyridine}$ is $\underset{\_}{5.2}$.
• The $\text{pH}$ when $20.0\text{mL}\text{of}0.100\text{M}\text{HCl}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}\text{pyridine}$ is $\underset{\_}{4.6}$.
• The $\text{pH}$ when $24.5\text{mL}\text{of}0.100\text{M}\text{HCl}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}\text{pyridine}$ is $\underset{\_}{3.82}$.
• The $\text{pH}$ when $25.0\text{mL}\text{of}0.100\text{M}\text{HCl}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}\text{pyridine}$ is $\underset{\_}{0.05}$.
• The $\text{pH}$ when $26.0\text{mL}\text{of}0.100\text{M}\text{HCl}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}\text{pyridine}$ is $\underset{\_}{2.71}$.
• The $\text{pH}$ when $28.0\text{mL}\text{of}0.100\text{M}\text{HCl}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}\text{pyridine}$ is $\underset{\_}{2.24}$.
• The $\text{pH}$ when $30.0\text{mL}\text{of}0.100\text{M}\text{HCl}$ is added to $25\text{mL}\text{of}\text{0}\text{.100}\text{M}\text{pyridine}$ is $\underset{\_}{2.04}$.