Chapter 11, Problem 11.BE

Interpretation Introduction

Interpretation:

The $pH$ values of the reaction medium during titration of $0.0500MKOH-0.0500MHCOOH$ system has to be calculated for the given volumes of $KOH$.  A graph of $pH$ versus volume of $KOH$ 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.BE

The $pH$ values of the reaction medium during titration of $0.0500MKOH-0.0500MHCOOH$ system are  calculated for the given volumes of $KOH$ as,

S.No	Volume of $KOH$, $V_b$ $(mL)$	$pH$
1.	$0.00$	$2.54$
2.	$10.0$	$3.14$
3.	$20.0$	$3.57$
4.	$25.0$	$3.74$
5.	$30.0$	$3.92$
6.	$40.0$	$4.35$
7.	$45.0$	$4.70$
8.	$48.0$	$5.12$
9.	$49.0$	$5.43$
10.	$49.5$	$5.74$
11.	$50.0$	$8.07$
12.	$50.5$	$10.40$
13.	$51.0$	$10.69$
14.	$52.0$	$10.99$
15.	$55.0$	$11.38$
16.	$60.0$	$11.65$

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

Explanation of Solution

Given that volume of base, $KOH$ and it is denoted by $V_b.$ Given the strength ( $M_1$) and volume of formic acid $HCOOH$ solution ( $V_1$) and strength of $KOH$( $M_2$), the volume of $KOH$ at equivalence point ( $V_2$)is calculated as,

$\begin{array}{l}{V}_1{M}_1={\text{ V}}_2{\text{M}}_2\\ 50.0\text{ mL}\times \text{ 0}\text{.0500 M}{\text{= V}}_2\times 0.0500M\\ V_2=\frac{50\text{ mL}\times \text{ 0}\text{.0500 M}}{0.0500M}\\ =\text{ 50}\text{.0 mL}\end{array}$

The reaction occurring in titration of formic acid with $KOH$ is represented as,

$HCOOH+O{H}^{-}\stackrel{}{\to }HCO{O}^{-}+H_{2}O$

When volume of base added is $0.00mL$, that is in absence of base, the $pH$ depends upon the dissociation of formic acid.  Since it is a weak acid it doesn’t dissociate completely into individual ions.

$HA\stackrel{}{\rightleftharpoons }{H}^{+}+A^{-}$

$[H^{+}]=[A^{-}]=x,[HA]=0.0500-x$ and $K_a\text{ of HCOOH is 1}\text{.8}\times {\text{10}}^{-4}$

$K_a$ for the above reaction is,

$\begin{array}{l}{K}_a=\frac{x^2}{0.0500-x}\\ \text{1}\text{.8}\times {\text{10}}^{-4}=\frac{x^2}{0.0500-x}\end{array}$

Solving for ‘x’,

$x=2.91\times {10}^{-3}M=[H^{+}]$

$pH$ is determined as,

$\begin{array}{l}pH\text{= - }\log \left[H^{+}\right]\\ =\text{ -log}\left[2.91\times {10}^{-3}\right]\\ \text{= 2}\text{.54}\end{array}$

When volume of $KOH$ added is $10.0mL$, the base begins to react with acid.  Upon reaction with base the acid forms conjugate base.  The titrating mixture then becomes buffer.  The concentration of conjugate base formed increases with addition of base at each step. Using Henderson-Hasselbalch equation $pH$ can be determined.

The titration reaction at this instant is,

$HA{\text{ + OH}}^{-}\stackrel{}{\to }{\text{ A}}^{-}+H_{2}O$

Their respective concentration is,

$\begin{array}{l}HA{\text{ + OH}}^{-}\stackrel{}{\to }{\text{A}}^{-}+H_{2}O\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ Initial:\text{ 50 mL 10 mL - -}\\ \text{Final: 50-10 mL - 10 mL -}\end{array}$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH{\text{= pK}}_a\text{+ }\log \frac{[A^{-}]}{[HA]}\\ =\text{ 3}\text{.744 +log}\left(\frac{10}{40}\right)\\ \text{= 3}\text{.14}\end{array}$

When volume of $KOH$ added is $20.0mL$, the base begins to react with acid.  Upon reaction with base the acid forms conjugate base.  The titrating mixture then becomes buffer.  The concentration of conjugate base formed. Using Henderson-Hasselbalch equation $pH$ can be determined.

The titration reaction at this instant is,

$HA{\text{ + OH}}^{-}\stackrel{}{\to }{\text{ A}}^{-}+H_{2}O$

Their respective concentration is,

$\begin{array}{l}HA{\text{ + OH}}^{-}\stackrel{}{\to }{\text{A}}^{-}+H_{2}O\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ Initial:\text{ 50 mL 20 mL - -}\\ \text{Final: 50-20 mL - 20 mL -}\end{array}$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH{\text{= pK}}_a\text{+ }\log \frac{[A^{-}]}{[HA]}\\ =\text{ 3}\text{.744 +log}\left(\frac{20}{30}\right)\\ \text{= 3}\text{.57}\end{array}$

When volume of $KOH$ added is $25.0mL$, the base begins to react with acid.  Upon reaction with base the acid forms conjugate base.  The titrating mixture then becomes buffer.  The concentration of conjugate base formed. Using Henderson-Hasselbalch equation $pH$ can be determined.

The titration reaction at this instant is,

$HA{\text{ + OH}}^{-}\stackrel{}{\to }{\text{ A}}^{-}+H_{2}O$

Their respective concentration is,

$\begin{array}{l}HA{\text{ + OH}}^{-}\stackrel{}{\to }{\text{A}}^{-}+H_{2}O\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ Initial:\text{ 50 mL 25 mL - -}\\ \text{Final: 50-25 mL - 25 mL -}\end{array}$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH{\text{= pK}}_a\text{+ }\log \frac{[A^{-}]}{[HA]}\\ =\text{ 3}\text{.744 +log}\left(\frac{25}{25}\right)\\ \text{= 3}\text{.74}\end{array}$

When volume of $KOH$ added is $30.0mL$, the base begins to react with acid.  Upon reaction with base the acid forms conjugate base.  The titrating mixture then becomes buffer.  The concentration of conjugate base formed. Using Henderson-Hasselbalch equation $pH$ can be determined.

The titration reaction at this instant is,

$HA{\text{ + OH}}^{-}\stackrel{}{\to }{\text{ A}}^{-}+H_{2}O$

Their respective concentration is,

$\begin{array}{l}HA{\text{ + OH}}^{-}\stackrel{}{\to }{\text{A}}^{-}+H_{2}O\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ Initial:\text{ 50 mL 30 mL - -}\\ \text{Final: 50-30 mL - 30 mL -}\end{array}$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH{\text{= pK}}_a\text{+ }\log \frac{[A^{-}]}{[HA]}\\ =\text{ 3}\text{.744 +log}\left(\frac{30}{20}\right)\\ \text{= 3}\text{.92}\end{array}$

When volume of $KOH$ added is $40.0mL$, the base begins to react with acid.  Upon reaction with base the acid forms conjugate base.  The titrating mixture then becomes buffer.  The concentration of conjugate base formed. Using Henderson-Hasselbalch equation $pH$ can be determined.

The titration reaction at this instant is,

$HA{\text{ + OH}}^{-}\stackrel{}{\to }{\text{ A}}^{-}+H_{2}O$

Their respective concentration is,

$\begin{array}{l}HA{\text{ + OH}}^{-}\stackrel{}{\to }{\text{A}}^{-}+H_{2}O\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ Initial:\text{ 50 mL 40 mL - -}\\ \text{Final: 50-40 mL - 40 mL -}\end{array}$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH{\text{= pK}}_a\text{+ }\log \frac{[A^{-}]}{[HA]}\\ =\text{ 3}\text{.744 +log}\left(\frac{40}{10}\right)\\ \text{= 4}\text{.35}\end{array}$

When volume of $KOH$ added is $45.0mL$, the base begins to react with acid.  Upon reaction with base the acid forms conjugate base.  The titrating mixture then becomes buffer.  The concentration of conjugate base formed. Using Henderson-Hasselbalch equation $pH$ can be determined.

The titration reaction at this instant is,

$HA{\text{ + OH}}^{-}\stackrel{}{\to }{\text{ A}}^{-}+H_{2}O$

Their respective concentration is,

$\begin{array}{l}HA{\text{ + OH}}^{-}\stackrel{}{\to }{\text{A}}^{-}+H_{2}O\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ Initial:\text{ 50 mL 45 mL - -}\\ \text{Final: 50-45 mL - 45 mL -}\end{array}$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH{\text{= pK}}_a\text{+ }\log \frac{[A^{-}]}{[HA]}\\ =\text{ 3}\text{.744 +log}\left(\frac{45}{5}\right)\\ \text{= 4}\text{.70}\end{array}$

When volume of $KOH$ added is $48.0mL$, the base begins to react with acid.  Upon reaction with base the acid forms conjugate base.  The titrating mixture then becomes buffer.  The concentration of conjugate base formed. Using Henderson-Hasselbalch equation $pH$ can be determined.

The titration reaction at this instant is,

$HA{\text{ + OH}}^{-}\stackrel{}{\to }{\text{ A}}^{-}+H_{2}O$

Their respective concentration is,

$\begin{array}{l}HA{\text{ + OH}}^{-}\stackrel{}{\to }{\text{A}}^{-}+H_{2}O\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ Initial:\text{ 50 mL 48 mL - -}\\ \text{Final: 50-48 mL - 2 mL -}\end{array}$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH{\text{= pK}}_a\text{+ }\log \frac{[A^{-}]}{[HA]}\\ =\text{ 3}\text{.744 +log}\left(\frac{48}{2}\right)\\ \text{= 5}\text{.12}\end{array}$

When volume of $KOH$ added is $50.0mL$, all of the acid has been converted to its conjugate base as the equivalence point is reached.  Hydrolysis of conjugate base determines the $pH.$

The titration reaction at this instant is,

$A^{-}{\text{ + H}}_2\text{O}\stackrel{}{\to }\text{ HA}+O{H}^{-}$

formal concentration of formic acid at this instant is,

$F\text{ = }\left(\frac{50mL}{50mL+50mL}\right)\times 0.05M=0.025M$

After reaction of Hydrolysis of conjugate base,

$[HA]=[O{H}^{-}]=x,[A^{-}]=0.025-x$

$K_b$ for the above reaction is known to be $5.56\times {10}^{-11}$ and hence,

$\begin{array}{l}{K}_b=\frac{x^2}{0.025-x}\\ 5.56\times {10}^{-11}=\frac{x^2}{0.025-x}\end{array}$

Solving for ‘x’,

$x=[O{H}^{-}]=1.18\times {10}^{-6}M$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH\text{= -}\log \frac{K_w}{[O{H}^{-}]}\\ =\text{ -log}\left(\frac{1\times {10}^{-14}}{1.18\times {10}^{-6}}\right)\\ \text{= 8}\text{.07}\end{array}$

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

When volume of base added is $50.5mL$,

$\begin{array}{l}\left[O{H}^{-}\right]=\left(\frac{50.5-50\text{ mL}}{\text{100}\text{.5 mL}}\right)\times 0.0500M\\ =\text{ 2}\text{.5}\times {\text{10}}^{-4}M\end{array}$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH+pOH=\text{ 14}\\ pH=\text{ 14 - }pOH\\ \text{= 14-}\left(\text{-log [}O{H}^{-}\text{]}\right)\\ \text{= 14-}\left(\text{-log(2}\text{.5}\times {\text{10}}^{-4}\text{)}\right)\\ \text{= 14 - 3}\text{.60 = 10}\text{.4}\end{array}$

When volume of base added is $51mL$,

$\begin{array}{l}\left[O{H}^{-}\right]=\left(\frac{51-50\text{ mL}}{\text{101 mL}}\right)\times 0.0500M\\ =\text{ 4}\text{.95}\times {\text{10}}^{-4}M\end{array}$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH+pOH=\text{ 14}\\ pH=\text{ 14 - }pOH\\ \text{= 14-}\left(\text{-log [}O{H}^{-}\text{]}\right)\\ \text{= 14-}\left(\text{-log(4}\text{.95}\times {\text{10}}^{-4}\text{)}\right)\\ \text{= 14 - 3}\text{.31 = 10}\text{.69}\end{array}$

When volume of acid added is $52mL$,

$\begin{array}{l}\left[O{H}^{-}\right]=\left(\frac{52-50\text{ mL}}{\text{102 mL}}\right)\times 0.0500M\\ =\text{ 9}\text{.80}\times {\text{10}}^{-4}M\end{array}$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH+pOH=\text{ 14}\\ pH=\text{ 14 - }pOH\\ \text{= 14-}\left(\text{-log [}O{H}^{-}\text{]}\right)\\ \text{= 14-}\left(\text{-log(9}\text{.80}\times {\text{10}}^{-4}\text{)}\right)\\ \text{= 14 - 3}\text{.009 = 10}\text{.99}\end{array}$

When volume of acid added is $55mL$,

$\begin{array}{l}\left[O{H}^{-}\right]=\left(\frac{55-50\text{ mL}}{\text{105 mL}}\right)\times 0.0500M\\ =0.0024M\end{array}$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH+pOH=\text{ 14}\\ pH=\text{ 14 - }pOH\\ \text{= 14-}\left(\text{-log [}O{H}^{-}\text{]}\right)\\ \text{= 14-}\left(\text{-log(0}\text{.0024)}\right)\\ \text{= 14 - 2}\text{.62 = 11}\text{.38}\end{array}$

When volume of acid added is $60mL$,

$\begin{array}{l}\left[O{H}^{-}\right]=\left(\frac{60-50\text{ mL}}{\text{110 mL}}\right)\times 0.0500M\\ =0.0045M\end{array}$

Substitute the values known to determine $pH$ ,

$\begin{array}{l}pH+pOH=\text{ 14}\\ pH=\text{ 14 - }pOH\\ \text{= 14-}\left(\text{-log [}O{H}^{-}\text{]}\right)\\ \text{= 14-}\left(\text{-log(0}\text{.0045)}\right)\\ \text{= 14 - 2}\text{.35 = 11}\text{.65}\end{array}$

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

Conclusion

The $pH$ values of the reaction medium during titration of $0.0500MKOH-0.0500MHCOOH$ system was calculated using relevant formula and a graph of $pH$ versus volume of $KOH$ was drawn.