Chapter 11, Problem 11.25P

Interpretation Introduction

Interpretation:

The $pH$ at $10mL$ intervals( from $0mL\text{ to 100 mL}$) in titration of $40.0mL$ of piperazine with $0.100M\text{ HCl}$ has to be calculated and a graph of $pH{\text{ vs V}}_a$ has to be drawn where $V_a$ is volume of acid added.

Concept introduction:

Acid-base titration is titration between acid and base. It is also known as neutralization reaction.

There are primarily four types of acid-base titrations –

• Strong base vs strong acid
• Strong base vs weak acid
• Weak base vs strong acid
• Weak base vs weak acid

Equivalence point ensures the titration reaction becomes completed and at this point the number of moles of titrant and the number of moles of analyte becomes equal.

Explanation of Solution

Piperazine is a weak base and $HCl$ is a strong acid.  So the titration of piperazine with $HCl$ is titration of weak base with strong acid.  The reaction of piperazine with $HCl$ is written as,

Piperazine has two N-atoms and each N-atom has a pair of non-bonding electrons available for bonding which makes the molecule basic. The protons from $HCl$ protonates piperazine.  The $pH$ of the titration varies at each different stage of titration depending upon the volume of the base and acid present.  Let ‘B’ be the term that denotes base here.

Given data:

$\begin{array}{l}volume\text{ of piperazine}\text{= 40}\text{.0 mL}\\ \text{Molarity of piperazine}\text{= 0}\text{.100M}\\ \text{Molarity of HCl}\text{= 0}\text{.100M}\end{array}$

When volume of acid added $V_a$ is $0mL,$ only base is present in aqueous medium.  The base gets hydrolyzed by water as,

$B+H_{2}O\stackrel{}{\rightleftharpoons }B{H}^{+}O{H}^{-}$

Then, $[B{H}^{+}]=[O{H}^{-}]=x,[B]=100-x$

The base has two $K_a$ values as there are two basic N-atoms.  One of the $K_a$ value is $1.86\times {10}^{-10}$ and another one is $4.65\times {10}^{-6}$ .  The equation of $K_b$ for the above reaction is,

$\begin{array}{l}{K}_b=\frac{x^2}{0.100-x}\\ \frac{K_w}{K_{a2}}=\frac{1.0\times {10}^{-14}}{1.86\times {10}^{-10}}=\frac{x^2}{0.100-x}\end{array}$

Solving for ‘x’,

$x=2.29\times {10}^{-3}M=[O{H}^{-}]$

Therefore $pH$ is calculated as,

$\begin{array}{l}pH=\text{ -log}\left(\frac{K_w}{[O{H}^{-}]}\right)\\ =\text{ -log}\left(\frac{1\times {10}^{-14}}{2.29\times {10}^{-3}M}\right)=11.36\end{array}$

$V_a=10.0mL$

$10mL$ of acid consumes $10mL$ of base and so volume of base remaining unreacted is $30mL.$

Therefore $pH$ is calculated as,

$\begin{array}{l}pH={\text{ pK}}_2\text{+log}\left(\frac{[B]}{[B{H}^{+}]}\right)\\ =9.73\text{+log}\left(\frac{30}{10}\right)=10.21\end{array}$

$V_a=20.0mL$

$20mL$ of acid consumes $20mL$ of base and so volume of base remaining unreacted is $20mL.$

Therefore $pH$ is calculated as,

$\begin{array}{l}pH={\text{ pK}}_2\text{+log}\left(\frac{[B]}{[B{H}^{+}]}\right)\\ =9.73\text{+log}\left(\frac{20}{20}\right)\\ =\text{ 9}\text{.73+0 = 9}\text{.73}\end{array}$

$V_a=30.0mL$

$30mL$ of acid consumes $30mL$ of base and so volume of base remaining unreacted is $10mL.$

Therefore $pH$ is calculated as,

$\begin{array}{l}pH={\text{ pK}}_2\text{+log}\left(\frac{[B]}{[B{H}^{+}]}\right)\\ =9.73\text{+log}\left(\frac{30}{10}\right)\\ =\text{ 9}\text{.25}\end{array}$

$V_a=40.0mL$

$40mL$ of acid consumes $40mL$ of base and so no volume of base is remaining unreacted.  Thus B is converted to BH⁺ at a formal concentration of,

$\begin{array}{l}F=\left(\frac{40.0mL}{40.0mL+40.0mL}\right)(0.100M)\\ =\text{ 0}\text{.0500M}\end{array}$

$[H^{+}]$ is calculated as,

$\begin{array}{l}[H^{+}]=\sqrt{\frac{K_1{K}_{2}F+K_1{K}_W}{K_1+F}}\\ =\sqrt{\frac{\left(4.65\times {10}^{-6}\times 1.86\times {10}^{-10}\times 0.0500M\right)+\left(4.65\times {10}^{-6}\times 1.0\times {10}^{-14}\right)}{4.65\times {10}^{-6}+0.0500M}}\end{array}$

Solving the above equation for $[H^{+}]$ and calculating $pH$ using the formula $pH=-\log [H^{+}],$

$pH=7.53$

$V_a=50.0mL$

$50mL$ of acid consumes $20mL$ of BH⁺ and volume of BH⁺ remaining unreacted is $50mL-20mL=30mL$ and so volume of $B{H}_2{}^{2+}$ formed is $10mL$

Therefore $pH$ is calculated as,

$\begin{array}{l}pH={\text{ pK}}_1\text{+log}\left(\frac{[B{H}^{+}]}{[B{H}_2{}^{2+}]}\right)\\ =\text{ 5}.33\text{+log}\left(\frac{30}{10}\right)\\ =\text{ 5}\text{.81}\end{array}$

$V_a=60.0mL$

$60mL$ of acid consumes $30mL$ of BH⁺ and so volume of $B{H}_2{}^{2+}$ formed is $60-30=30mL$ and volume of BH⁺ remaining unreacted is $30mL.$

Therefore $pH$ is calculated as,

$\begin{array}{l}pH={\text{ pK}}_1\text{+log}\left(\frac{[B{H}^{+}]}{[B{H}_2{}^{2+}]}\right)\\ =\text{ 5}.33\text{+log}\left(\frac{30}{30}\right)\\ =\text{ 5}\text{.33}\end{array}$

$V_a=70.0mL$

$70mL$ of acid consumes $40mL$ of BH⁺ and so volume of $B{H}_2{}^{2+}$ formed is $70-40=30mL$ and volume of BH⁺ remaining unreacted is $10mL.$

Therefore $pH$ is calculated as,

$\begin{array}{l}pH={\text{ pK}}_1\text{+log}\left(\frac{[B{H}^{+}]}{[B{H}_2{}^{2+}]}\right)\\ =\text{ 5}.33\text{+log}\left(\frac{10}{30}\right)\\ =\text{ 4}\text{.86}\end{array}$

$V_a=80.0mL$

All of B is converted to $B{H}_2{}^{2+}$ at formal concentration,

$F=\left(\frac{40mL}{40mL+80mL}\right)\left(0.100M\right)=0.0333M$

Therefore $pH$ is determined by dissociation of $B{H}_2{}^{2+}$

$B{H}_2{}^{2+}\stackrel{}{\rightleftharpoons }B{H}^{+}{H}^{+}$

Then, $[B{H}^{+}]=[H^{+}]=x,{\text{ [BH}}_2^{2+}\text{]=0}\text{.0333-x}$

Dissociation constant Ka is,

$\begin{array}{l}{K}_a=4.65\times {10}^{-6}\\ K_a=\frac{x^2}{0.0333-x}\\ \frac{x^2}{0.0333-x}=4.65\times {10}^{-6}\end{array}$

Solving for ‘x’,

$x=3.91\times {10}^{-4}M=[H^{+}]$

Therefore $pH$ is calculated as,

$\begin{array}{l}pH={\text{ -log[H}}^{+}\text{]}\\ =\text{ -log}\left(3.91\times {10}^{-4}M\right)\\ =\text{ 3}\text{.4 }\end{array}$

$V_a=90.0mL$

$\begin{array}{l}[H^{+}]=\left(\frac{100mL-90mL}{90mL+40mL}\right)\times 0.100M\\ =\text{ 0}\text{.0077M}\end{array}$

Therefore $pH$ is calculated as,

$\begin{array}{l}pH={\text{ -log[H}}^{+}\text{]}\\ =\text{ -log}\left(0.0077\right)\\ =\text{ 2}\text{.11 }\end{array}$

$V_a=100.0mL$

$\begin{array}{l}[H^{+}]=\left(\frac{100mL-80mL}{100mL+40mL}\right)\times 0.100M\\ =\text{ 0}\text{.0143M}\end{array}$

Therefore $pH$ is calculated as,

$\begin{array}{l}pH={\text{ -log[H}}^{+}\text{]}\\ =\text{ -log}\left(0.0143\right)\\ =\text{ 1}\text{.84 }\end{array}$

For each volume of acid added corresponding $pH$ is calculated.  The graph of $pH{\text{ vs V}}_a$ is plotted as,

Figure 1

Conclusion

The $pH$ at $10mL$ intervals( from $0mL\text{ to 100 mL}$) in titration of $40.0mL$ of piperazine with $0.100M\text{ HCl}$ was calculated and a graph of $pH{\text{ vs V}}_a$ was drawn.