Chapter 11, Problem 11.AE

Interpretation Introduction

Interpretation:

The $pH$ values of the reaction medium during titration of $0.0100MNaOH-0.100MHCl$ system has to be calculated for the given volumes of $HCl$.  A graph of $pH$ versus volume of $HCl$ added 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.AE

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.	$0.00$	$12$
2.	$1.00$	$11.89$
3.	$2.00$	$11.76$
4.	$3.00$	$11.58$
5.	$4.00$	$11.27$
6.	$4.50$	$10.96$
7.	$4.90$	$10.26$
8.	$4.99$	$9.26$
9.	$5.00$	$7.00$
10.	$5.01$	$4.74$
11.	$5.10$	$3.74$
12.	$5.50$	$3.05$
13.	$6.00$	$2.74$
14.	$8.00$	$2.30$
15.	$10.00$	$2.09$

The graph of $pH$ versus volume of $HCl$ is plotted as,

Figure 1

Explanation of Solution

Given that volume of acid, $HCl$ and it is denoted by $V_a.$ Given the strength ( $M_1$) and volume of $NaOH$ solution ( $V_1$) and strength of $HCl$( $M_2$), the volume of $HCl$ at equivalence point ( $V_2$)is calculated as,

$\begin{array}{l}{V}_1{M}_1={\text{ V}}_2{\text{M}}_2\\ 50\text{ mL}\times \text{ 0}\text{.0100 M}{\text{= V}}_2\times 0.100M\\ V_2=\frac{50\text{ mL}\times \text{ 0}\text{.0100 M}}{0.100M}\\ =\text{ 5}\text{.00 mL}\end{array}$

When volume of acid added is $0.00mL$,

$\begin{array}{l}\left[O{H}^{-}\right]=\left(\frac{5.00\text{ mL}}{5.00\text{ mL}}\right)\times 0.0100M\times \left(\frac{50.00mL}{50.00mL}\right)\\ =\text{ 1}\times \text{0}\text{.0100M}\times \text{1 = 0}\text{.0100M}\end{array}$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH\text{= - }\log \left[\frac{K_w}{\left[O{H}^{-}\right]}\right]\\ =\text{ -log}\left[\frac{1\times {10}^{-14}}{0.0100M}\right]\\ \text{= 12}\end{array}$

When volume of acid added is $1.00mL$,

$\begin{array}{l}\left[O{H}^{-}\right]=\left(\frac{4.00\text{ mL}}{5.00\text{ mL}}\right)\times 0.0100M\times \left(\frac{50.00mL}{51.00mL}\right)\\ =\text{ 0}\text{.00784M}\end{array}$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH\text{= - }\log \left[\frac{K_w}{\left[O{H}^{-}\right]}\right]\\ =\text{ -log}\left[\frac{1\times {10}^{-14}}{0.000784M}\right]\\ \text{= 11}\text{.89}\end{array}$

When volume of acid added is $2.00mL$,

$\begin{array}{l}\left[O{H}^{-}\right]=\left(\frac{3.00\text{ mL}}{5.00\text{ mL}}\right)\times 0.0100M\times \left(\frac{50.00mL}{52.00mL}\right)\\ =\text{ 0}\text{.6}\times \text{0}\text{.0100M}\times \text{0}\text{.96 = 0}\text{.00576 M}\end{array}$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH\text{= - }\log \left[\frac{K_w}{\left[O{H}^{-}\right]}\right]\\ =\text{ -log}\left[\frac{1\times {10}^{-14}}{0.00576M}\right]\\ \text{= 11}\text{.76}\end{array}$

When volume of acid added is $3.00mL$,

$\begin{array}{l}\left[O{H}^{-}\right]=\left(\frac{2.00\text{ mL}}{5.00\text{ mL}}\right)\times 0.0100M\times \left(\frac{50.00mL}{53.00mL}\right)\\ =\text{ 0}\text{.4}\times \text{0}\text{.0100M}\times \text{0}\text{.94 = 0}\text{.00376 M}\end{array}$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH\text{= - }\log \left[\frac{K_w}{\left[O{H}^{-}\right]}\right]\\ =\text{ -log}\left[\frac{1\times {10}^{-14}}{0.00376M}\right]\\ \text{= 11}\text{.58}\end{array}$

When volume of acid added is $4.00mL$,

$\begin{array}{l}\left[O{H}^{-}\right]=\left(\frac{1.00\text{ mL}}{5.00\text{ mL}}\right)\times 0.0100M\times \left(\frac{50.00mL}{54.00mL}\right)\\ =\text{ 0}\text{.2}\times \text{0}\text{.0100M}\times \text{0}\text{.92 = 0}\text{.00184 M}\end{array}$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH\text{= - }\log \left[\frac{K_w}{\left[O{H}^{-}\right]}\right]\\ =\text{ -log}\left[\frac{1\times {10}^{-14}}{0.00184M}\right]\\ \text{= 11}\text{.27}\end{array}$

When volume of acid added is $4.50mL$,

$\begin{array}{l}\left[O{H}^{-}\right]=\left(\frac{0.50\text{ mL}}{5.00\text{ mL}}\right)\times 0.0100M\times \left(\frac{50.00mL}{54.50mL}\right)\\ =\text{ 0}\text{.1}\times \text{0}\text{.0100M}\times \text{0}\text{.92 = 0}\text{.00092 M}\end{array}$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH\text{= - }\log \left[\frac{K_w}{\left[O{H}^{-}\right]}\right]\\ =\text{ -log}\left[\frac{1\times {10}^{-14}}{0.00092M}\right]\\ \text{= 10}\text{.96}\end{array}$

When volume of acid added is $4.90mL$,

$\begin{array}{l}\left[O{H}^{-}\right]=\left(\frac{0.10\text{ mL}}{5.00\text{ mL}}\right)\times 0.0100M\times \left(\frac{50.00mL}{54.90mL}\right)\\ =\text{ 0}\text{.02}\times \text{0}\text{.0100M}\times \text{0}\text{.91 = 0}\text{.000182 M}\end{array}$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH\text{= - }\log \left[\frac{K_w}{\left[O{H}^{-}\right]}\right]\\ =\text{ -log}\left[\frac{1\times {10}^{-14}}{0.000182M}\right]\\ \text{= 10}\text{.26}\end{array}$

When volume of acid added is $4.99mL$,

$\begin{array}{l}\left[O{H}^{-}\right]=\left(\frac{0.01\text{ mL}}{5.00\text{ mL}}\right)\times 0.0100M\times \left(\frac{50.00mL}{54.99mL}\right)\\ =\text{ 0}\text{.002}\times \text{0}\text{.0100M}\times \text{0}\text{.91 = 0}\text{.0000182 M}\end{array}$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH\text{= - }\log \left[\frac{K_w}{\left[O{H}^{-}\right]}\right]\\ =\text{ -log}\left[\frac{1\times {10}^{-14}}{0.0000182M}\right]\\ \text{= 9}\text{.26}\end{array}$

When volume of acid added is $5mL$, the neutralization reaction is in equilibrium,

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

Let ‘x’ be the concentration of $H^{+}$ and $O{H}^{-}$ ions, so that,

$\begin{array}{l}{x}^2{\text{= K}}_w\text{ = 1}\times {\text{10}}^{-14}\\ x={\text{ 10}}^{-7}\end{array}$

Then,

$\begin{array}{l}pH={\text{-log[H}}^{+}\text{]}\\ {\text{= -log [10}}^{-7}\text{] = 7}\end{array}$

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 $5.01mL$,

$\begin{array}{l}\left[H^{+}\right]=\left(\frac{0.01\text{ mL}}{55.01\text{ mL}}\right)\times 0.100M\\ =\text{ 1}\text{.82}\times {\text{10}}^{-5}M\end{array}$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH={\text{-log[H}}^{+}\text{]}\\ \text{= -log [}1.82\times {10}^{-5}\text{] = 4}\text{.74}\end{array}$

When volume of acid added is $5.10mL$,

$\begin{array}{l}\left[H^{+}\right]=\left(\frac{0.1\text{ mL}}{55.1\text{ mL}}\right)\times 0.100M\\ =0.00018M\end{array}$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH={\text{-log[H}}^{+}\text{]}\\ \text{= -log [}0.00018\text{] = 3}\text{.74}\end{array}$

When volume of acid added is $5.50mL$,

$\begin{array}{l}\left[H^{+}\right]=\left(\frac{0.5\text{ mL}}{55.5\text{ mL}}\right)\times 0.100M\\ =0.0009M\end{array}$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH={\text{-log[H}}^{+}\text{]}\\ \text{= -log [}0.0009\text{] = 3}\text{.05}\end{array}$

When volume of acid added is $6.00mL$,

$\begin{array}{l}\left[H^{+}\right]=\left(\frac{1.00\text{ mL}}{56\text{ mL}}\right)\times 0.100M\\ =0.0018M\end{array}$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH={\text{-log[H}}^{+}\text{]}\\ \text{= -log [}0.0018\text{] = 2}\text{.74}\end{array}$

When volume of acid added is $8.00mL$,

$\begin{array}{l}\left[H^{+}\right]=\left(\frac{3.00\text{ mL}}{58\text{ mL}}\right)\times 0.100M\\ =0.005M\end{array}$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH={\text{-log[H}}^{+}\text{]}\\ \text{= -log [}0.005\text{] = 2}\text{.30}\end{array}$

When volume of acid added is $10.00mL$,

$\begin{array}{l}\left[H^{+}\right]=\left(\frac{5.00\text{ mL}}{\text{60 mL}}\right)\times 0.100M\\ =0.008M\end{array}$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH={\text{-log[H}}^{+}\text{]}\\ \text{= -log [}0.008\text{] = 2}\text{.09}\end{array}$

The graph of $pH$ versus volume of $HCl$ is plotted as,

Figure 1

Conclusion

The $pH$ values of the reaction medium during titration of $0.0100MNaOH-0.100MHCl$ system was calculated using relevant formula.