Chapter 11, Problem 11.17P

Interpretation Introduction

Interpretation:

The $pH$ values of the reaction medium during titration of $0.0319Mbenzylamine-0.0500MHCl$ system has to be calculated for the given volumes of $HCl$.

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 11.17P

The $pH$ values of the reaction medium during titration of $0.0100MNaOH-0.100MHCl$ system are calculated for the given volumes of $HCl$ as,

S.No	Volume of $HCl$, $V_a$ $(mL)$	$pH$
1.	$0.00$	$10.92$
2.	$12.0$	$9.57$
3.	$1/2{\text{ V}}_e$(half the volume at equivalence point - $15.95$)	$9.35$
4.	$30.0$	$8.15$
5.	$V_e(31.9)$	$5.53$
6.	$35.0$	$2.74$

Explanation of Solution

Given that volume of acid, $HCl$ and it is denoted by $V_a.$ Given the strength ( $M_1$) and volume of benzylamine solution ( $V_1$) and strength of $HCl$( $M_2$), the volume of $HCl$ at equivalence point ( $V_e$)is calculated as,

$\begin{array}{l}{V}_1{M}_1={\text{ V}}_e{\text{M}}_2\\ 50\text{ mL}\times \text{ 0}\text{.0319 M}{\text{= V}}_e\times 0.0500M\\ V_e=\frac{50\text{ mL}\times \text{ 0}\text{.0319 M}}{0.0500M}\\ =\text{ 31}\text{.9 mL}\end{array}$

Proton from $HCl$ protonates the base benzylamine (B) and forms conjugate acid (BH⁺).

But when no acid is added (V_a is $0.00mL$,), the base initially reacts with water in the reaction medium.

The equation for this reaction is written as,

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

Thus, $[B{H}^{+}]=[O{H}^{-}]=x,[B]=0.0319-x$

Dissociation constant $K_b$ for the above reaction is known to be $2.2\times {10}^{-5}$ and hence,

$\begin{array}{l}{K}_b=\left[\frac{[B{H}^{+}][O{H}^{-}]}{[B]}\right]=2.2\times {10}^{-5}\\ 2.2\times {10}^{-5}=\frac{x^2}{0.0319-x}\end{array}$

Solving for ‘x’

$x=[O{H}^{-}]=8.31\times {10}^{-4}M$

Determining $pH$,

$\begin{array}{l}pH+pOH=14\\ pOH=-\log [O{H}^{-}]\\ =-\text{log}\left(8.31\times {10}^{-4}\right)=3.08\end{array}$

Then as we know,

$\begin{array}{l}pH+pOH=14\\ pH\text{ + }3.08=14\\ pH=\text{14-3}\text{.08 = 10}\text{.92}\end{array}$

When volume of acid added is $12.0mL$,

$\begin{array}{l}\text{Reaction: }B+H^{+}\stackrel{}{\to }B{H}^{+}\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ Initial\text{ conc}\text{. 31}\text{.9 12}\text{.0 0}\\ \text{Final conc}\text{. 31}\text{.9-12}\text{.0 0 12}\text{.0}\end{array}$

$p{K}_a$ for benzylamine is known to be $\mathrm{9.35.}$ Substitute the values known to determine $pH.$

$\begin{array}{l}pH{\text{= pK}}_a\text{+ }\log \frac{[B]}{[B{H}^{+}]}\\ =\text{ 9}\text{.35+log}\left[\frac{19.9}{12.0}\right]\\ \text{= 9}\text{.57}\end{array}$

When volume of acid added is half the volume of equivalence point, $1/2{\text{ V}}_e=\frac{31.9mL}{2}=15.95mL,$

$\begin{array}{l}\text{Reaction: }B+H^{+}\stackrel{}{\to }B{H}^{+}\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ Initial\text{ conc}\text{. 31}\text{.9 15}\text{.95 0}\\ \text{Final conc}\text{. 31}\text{.9-15}\text{.95 0 15}\text{.95}\end{array}$

$p{K}_a$ for benzylamine is known to be $\mathrm{9.35.}$ Substitute the values known to determine $pH.$

$\begin{array}{l}pH{\text{= pK}}_a\text{+ }\log \frac{[B]}{[B{H}^{+}]}\\ =\text{ 9}\text{.35+log}\left[\frac{15.95}{15.95}\right]\\ \text{= 9}\text{.35+0 = 9}\text{.35}\end{array}$

When volume of acid added is $30.0mL$,

$\begin{array}{l}\text{Reaction: }B+H^{+}\stackrel{}{\to }B{H}^{+}\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ Initial\text{ conc}\text{. 31}\text{.9 30}\text{.0 0}\\ \text{Final conc}\text{. 31}\text{.9-30}\text{.0 0 30}\text{.0}\end{array}$

$p{K}_a$ for benzylamine is known to be $\mathrm{9.35.}$ Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH{\text{= pK}}_a\text{+ }\log \frac{[B]}{[B{H}^{+}]}\\ =\text{ 9}\text{.35+log}\left[\frac{1.9}{30}\right]\\ \text{= 9}\text{.35-1}\text{.20 = 8}\text{.15}\end{array}$

When volume of acid added is $31.9mL$ which is the volume of acid at equivalence point of the titration, at this stage the base benzylamine is completely converted to its conjugate acid BH⁺.  Concentration of BH⁺ at equivalence point is,

$\frac{50.0mL}{50.0+31.9mL}\times 0.0319=0.0195M$

The equation at this stage is written as,

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

Thus, $[H^{+}]=[B]=x,[B{H}^{+}]=0.0195-x$

Dissociation constant $K_a$ for the above reaction is known to be $4.7\times {10}^{-10}$ and hence,

$\begin{array}{l}{K}_a=\left[\frac{[H^{+}][B]}{[B{H}^{+}]}\right]=4.7\times {10}^{-10}\\ 4.7\times {10}^{-10}=\frac{x^2}{0.0195-x}\end{array}$

Solving for ‘x’

$x=[H^{+}]=2.96\times {10}^{-6}M$

Determining $pH$,

$\begin{array}{l}pH=\text{ -log}\left[H^{+}\right]\\ =\text{ -log(}2.96\times {10}^{-6}M\text{)}\\ \text{= 5}\text{.53}\end{array}$

$\text{35}\text{.0 mL}$ of $HCl$ is beyond the equivalence point which is $\text{31}\text{.9 mL}\text{.}$ thus volume of $HCl$ at this stage is $\text{35}\text{.0 mL-31}\text{.9 mL = 3}\text{.1 mL}\text{.}$ $pH$ of the solution has to be determined not exactly at equivalence point but after reaching the equivalence point – that is post-equivalence region.  At this stage the concentration of base is nil.  $pH$ is determined by measuring the concentration of excess acid added beyond the equivalence point. Therefore,

$\begin{array}{l}{\text{[H]}}^{+}=\left(\frac{3.1mL}{50.0mL+35.0mL}\right)\times 0.0500M\\ =\text{ 0}\text{.00182 M}\end{array}$

Determining $pH$,

$\begin{array}{l}pH=-\log [H^{+}]\\ =-\log \left(0.00182\right)\\ =\text{ 2}\text{.74}\end{array}$

Conclusion

The $pH$ values of the reaction medium during titration of $0.0319Mbenzylamine-0.0500MHCl$ system was calculated using relevant formula.