Chapter 11, Problem 11.CE

Interpretation Introduction

Interpretation:

The $pH$ values of the reaction medium during titration of $0.100cocaine-0.200MHN{O}_3$ system has to be calculated for the given volumes of $HN{O}_3.$ A graph of $pH{\text{ vs V}}_a$ has to be drawn.

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

The term $pH$ refers to concentration of $H^{+}$ion in a solution.

Answer to Problem 11.CE

The $pH$ values of the reaction medium during titration of $0.100cocaine-0.200MHN{O}_3$ system are calculated for the given volumes of $HN{O}_3$ as,

S.No	Volume of $HN{O}_3$, $V_a$ $(mL)$	$pH$
1.	$0.0$	$10.71$
2.	$10.0$	$9.01$
3.	$20.0$	$8.59$
4.	$25.0$	$8.41$
5.	$30.0$	$8.23$
6.	$40.0$	$7.81$
7.	$49.0$	$6.72$
8.	$49.9$	$5.71$
9.	$50.0$	$4.80$
10.	$50.1$	$3.89$
11.	$51.0$	$2.88$
12.	$60.0$	$1.48$

A graph of $pH{\text{ vs V}}_a$ is drawn as –

Explanation of Solution

Given that volume of acid, $HN{O}_3$ and it is denoted by $V_a.$ Given the strength ( $M_1$) and volume of cocaine solution ( $V_1$) and strength of $HN{O}_3$ ( $M_2$), the volume of $HN{O}_3$ at equivalence point ( $V_2$)is calculated as,

$\begin{array}{l}{V}_1{M}_1={\text{ V}}_2{\text{M}}_2\\ 100\text{ mL}\times \text{ 0}\text{.100 M}{\text{= V}}_2\times 0.200M\\ V_2=\frac{\text{100 mL}\times \text{ 0}\text{.100 M}}{0.200M}\\ =\text{ 50 mL}\end{array}$

The titration reaction is

$B+H^{+}\stackrel{}{\to }B{H}^{+}$

Where B represents cocaine and $B{H}^{+}$ represents conjugate acid formed by protonation of cocaine by proton from nitric acid.

When volume of acid added is $0.00mL$, the reaction is,

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

Molarity of cocaine is $0.100M$.  Assume, $[B]{\text{ = 100-x, [BH}}^{+}{\text{] = [OH}}^{-}\text{] = x}$

$K_b$ for this reaction is given as,

$K_b\text{ = }\frac{{\text{[BH}}^{+}\text{] }{\text{. [OH}}^{-}\text{]}}{[B]}\text{ = 2}\text{.6}\times {\text{10}}^{-6}$

Therefore,

$\frac{x^2}{0.100-x}\text{ = 2}\text{.6}\times {\text{10}}^{-6}$

Solving for ‘x’,

$x\text{ = 5}\text{.09}\times {\text{10}}^{-4}M$

Then $pH$ of reaction medium when volume of acid added is $0.0mL$ is,

$\begin{array}{l}pH\text{= - }\log \left[\frac{K_w}{\left[O{H}^{-}\right]}\right]\\ =\text{ -log}\left[\frac{1\times {10}^{-14}}{5.09\times {10}^{-4}M}\right]\\ \text{= 10}\text{.71}\end{array}$

When volume of acid added is $10.0mL$, the reaction is,

$B+H^{+}\stackrel{}{\to }B{H}^{+}$

Thus the fraction of Cocaine titrated when $10.0\text{ mL}$ of $HN{O}_3$ used is – $10.0\text{ mL}/50.0\text{ mL = 0}\text{.2}\text{.}$ It is denoted by $\left[B{H}^{+}\right]$. Then the remaining fraction of Cocaine is $1-0.2=0.8$ and it is denoted by $\left[B\right].$ $pH$ of the titration medium at this point is calculated as,

$pH{\text{= pK}}_a+\log \left[\frac{\left[B\right]}{{\left[BH\right]}^{+}}\right]$

For protonated cocaine, $B{H}^{+}$ $p{K}_a$ value is $8.41$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH\text{= 8}\text{.41}+\log \left[\frac{0.8}{0.2}\right]\\ =\text{ 9}\text{.01}\end{array}$

When volume of acid added is $20.0mL$, the reaction is,

$B+H^{+}\stackrel{}{\to }B{H}^{+}$

Thus the fraction of Cocaine titrated when $20.0mL$ of $HN{O}_3$ used is – $20.0\text{ mL}/50.0\text{ mL = 0}\text{.4}\text{.}$ It is denoted by   $\left[B{H}^{+}\right]$. Then the remaining fraction of Cocaine is $1-0.4=0.6$ and it is denoted by $\left[B\right].$ $pH$ of the titration medium at this point is calculated as,

$pH{\text{= pK}}_a+\log \left[\frac{\left[B\right]}{{\left[BH\right]}^{+}}\right]$

For protonated cocaine, $B{H}^{+}$ $p{K}_a$ value is $8.41$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH\text{= 8}\text{.41}+\log \left[\frac{0.6}{0.4}\right]\\ =\text{ 8}\text{.59}\end{array}$

When volume of acid added is $25.0mL$, the reaction is,

$B+H^{+}\stackrel{}{\to }B{H}^{+}$

Thus the fraction of Cocaine titrated when $25.0mL$ of $HN{O}_3$ used is – $25.0\text{ mL}/50.0\text{ mL = 0}\text{.5}\text{.}$ It is denoted by   $\left[B{H}^{+}\right]$. Then the remaining fraction of Cocaine is $1-0.5=0.5$ and it is denoted by $\left[B\right].$ $pH$ of the titration medium at this point is calculated as,

$pH{\text{= pK}}_a+\log \left[\frac{\left[B\right]}{{\left[BH\right]}^{+}}\right]$

For protonated cocaine, $B{H}^{+}$ $p{K}_a$ value is $8.41$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH\text{= 8}\text{.41}+\log \left[\frac{0.5}{0.5}\right]\\ =\text{ 8}\text{.41}\end{array}$

When volume of acid added is $30.0mL$, the reaction is,

$B+H^{+}\stackrel{}{\to }B{H}^{+}$

Thus the fraction of Cocaine titrated when $30.0mL$ of $HN{O}_3$ used is – $30.0\text{ mL}/50.0\text{ mL = 0}\text{.6}\text{.}$ It is denoted by   $\left[B{H}^{+}\right]$. Then the remaining fraction of Cocaine is $1-0.6=0.4$ and it is denoted by $\left[B\right].$ $pH$ of the titration medium at this point is calculated as,

$pH{\text{= pK}}_a+\log \left[\frac{\left[B\right]}{{\left[BH\right]}^{+}}\right]$

For protonated cocaine, $B{H}^{+}$ $p{K}_a$ value is $8.41$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH\text{= 8}\text{.41}+\log \left[\frac{0.4}{0.6}\right]\\ =\text{ 8}\text{.23}\end{array}$

When volume of acid added is $40.0mL$, the reaction is,

$B+H^{+}\stackrel{}{\to }B{H}^{+}$

Thus the fraction of Cocaine titrated when $40.0mL$ of $HN{O}_3$ used is – $40.0\text{ mL}/50.0\text{ mL = 0}\text{.8}\text{.}$ It is denoted by   $\left[B{H}^{+}\right]$. Then the remaining fraction of Cocaine is $1-0.8=0.2$ and it is denoted by $\left[B\right].$ $pH$ of the titration medium at this point is calculated as,

$pH{\text{= pK}}_a+\log \left[\frac{\left[B\right]}{{\left[BH\right]}^{+}}\right]$

For protonated cocaine, $B{H}^{+}$ $p{K}_a$ value is $8.41$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH\text{= 8}\text{.41}+\log \left[\frac{0.2}{0.8}\right]\\ =\text{ 7}\text{.81}\end{array}$

When volume of acid added is $49.0mL$, the reaction is,

$B+H^{+}\stackrel{}{\to }B{H}^{+}$

Thus the fraction of Cocaine titrated when $49.0mL$ of $HN{O}_3$ used is – $49.0\text{ mL}/50.0\text{ mL = 0}\text{.98}\text{.}$ It is denoted by   $\left[B{H}^{+}\right]$. Then the remaining fraction of Cocaine is $1-0.98=0.02$ and it is denoted by $\left[B\right].$ $pH$ of the titration medium at this point is calculated as,

$pH{\text{= pK}}_a+\log \left[\frac{\left[B\right]}{{\left[BH\right]}^{+}}\right]$

For protonated cocaine, $B{H}^{+}$ $p{K}_a$ value is $8.41$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH\text{= 8}\text{.41}+\log \left[\frac{0.02}{0.98}\right]\\ =\text{ 6}\text{.72}\end{array}$

When volume of acid added is $49.9mL$, the reaction is,

$B+H^{+}\stackrel{}{\to }B{H}^{+}$

Thus the fraction of Cocaine titrated when $49.9mL$ of $HN{O}_3$ used is – $49.9\text{ mL}/50.0\text{ mL = 0}\text{.998}\text{.}$ It is denoted by   $\left[B{H}^{+}\right]$. Then the remaining fraction of Cocaine is $1-0.998=0.002$ and it is denoted by $\left[B\right].$ $pH$ of the titration medium at this point is calculated as,

$pH{\text{= pK}}_a+\log \left[\frac{\left[B\right]}{{\left[BH\right]}^{+}}\right]$

For protonated cocaine, $B{H}^{+}$ $p{K}_a$ value is $8.41$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH\text{= 8}\text{.41}+\log \left[\frac{0.002}{0.998}\right]\\ =\text{ 5}\text{.71}\end{array}$

When volume of acid added is $49.9mL$, the reaction is,

$B+H^{+}\stackrel{}{\to }B{H}^{+}$

When volume of acid added is $50.0mL$, all of the cocaine is converted to its conjugate acid that formal concentration of $B{H}^{+}$ would be $\frac{100mL}{150mL}\times 0.100M=0.0667M$.

$B{H}^{+}$ dissociates as,

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

Assume, ${\text{[BH}}^{+}\text{] = 0}\text{.0667-x, }[B]{\text{= [H}}^{+}\text{] = x}$

$K_a$ for this reaction is given as,

$\begin{array}{l}{K}_a\text{= }\frac{[B]{\text{. [H}}^{+}\text{]}}{[B{H}^{+}]}\\ \frac{K_w}{K_b}=\frac{x^2}{0.0667-x}\end{array}$

Solving for ‘x’,

$x\text{ = 1}\text{.60}\times {\text{10}}^{-5}M$

Then $pH$ of reaction medium when volume of acid added is $50.00mL$ is,

$\begin{array}{l}pH\text{= - }\log [H^{+}]\\ =\text{ -log(1}\text{.60}\times {\text{10}}^{-5}M\text{)}\\ \text{= 4}\text{.80}\end{array}$

After reaching the equivalence point, continual addition of acid increases the concentration of acid so that we need to calculate $[H^{+}]$ to determine $pH.$ As the concentration of $[H^{+}]$ increases upon each addition of acid, $pH$ of the reaction medium decreases.

When volume of acid added is $50.1mL$,

$\begin{array}{l}\left[H^{+}\right]=\left(\frac{0.1\text{ mL}}{\text{150}\text{.01 mL}}\right)\times 0.200M\\ =0.00013M\end{array}$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH={\text{-log[H}}^{+}\text{]}\\ \text{= -log [}0.00013\text{] =3}\text{.89}\end{array}$

When volume of acid added is $51.0mL$,

$\begin{array}{l}\left[H^{+}\right]=\left(\frac{1.0\text{ mL}}{151.0mL}\right)\times 0.200M\\ =0.00132M\end{array}$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH={\text{-log[H}}^{+}\text{]}\\ \text{= -log [}0.00132\text{] = 2}\text{.88}\end{array}$

When volume of acid added is $60.0mL$,

$\begin{array}{l}\left[H^{+}\right]=\left(\frac{10.0\text{ mL}}{160.0mL}\right)\times 0.200M\\ =0.0333M\end{array}$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH={\text{-log[H}}^{+}\text{]}\\ \text{= -log [}0.0333\text{] = 1}\text{.48}\end{array}$

A graph of $pH{\text{ vs V}}_a$ is drawn as –

Conclusion

The $pH$ values of the reaction medium during titration of $0.100cocaine-0.200MHN{O}_3$ system was calculated using relevant formula.