Chapter 19, Problem 19.106P

Interpretation Introduction

Interpretation:

The $pH$ buffer made from $475mL$ of $0.200M$ benzoic acid and $25mL$ of $2.00MNaOH$ with make $500mL$ of $0.200MHCOOH$ and $2.00MNaOH$ volume has to be calculated.

Concept Introduction:

pH definition:

The concentration of hydrogen ion is measured using $\text{pH}$ scale.  The acidity of aqueous solution is expressed by $\text{pH}$ scale.

The $\text{pH}$ of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

  $\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$

Buffer solution:

• Buffer solution is defined as a solution that oppose changes in $\text{pH}$ while adding little amount of either an acid or a base. In general, addition of acid or base does not affect the $\text{pH}$ in buffer solution but if it is more than amount of conjugate base or conjugate acid, then buffer loses its buffering capacity.
• Buffer solution is a combination of a weak acid and its conjugate base or a weak base and its conjugate acid.

Henderson-Hasselbalch equation:

The $\text{pH}$ at an intermediate stage of the titration is calculated by using the following equation,

  $\begin{array}{l}\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}{\frac{\left[\text{base}\right]}{\left[\text{acid}\right]}}_{}^{}\\ \text{where,}\\ \text{pH}\text{=}\text{-}\text{log}\left[{\text{H}}^{\text{+}}\right]\\ {\text{pK}}_{\text{a}}\text{=}\text{-}\text{log}{\text{K}}_{\text{a}}\\ \left({\text{K}}_{\text{a}}\text{-}\text{acid}\text{dissociation}\text{constant}\right)\end{array}$

This equation is known as Henderson-Hasselbalch equation.

Answer to Problem 19.106P

The required volume of the base $NaOH$ is $36mL$ and the volume of the formic acid $HCOOH$ is $0.464=464mL$.

Explanation of Solution

Consider the following chemical reaction,

  $C_6{H}_{5}COO{H}_{\left(aq\right)}+NaO{H}_{\left(aq\right)}\stackrel{}{\rightleftarrows }N{a}^{2+}{}_{\left(aq\right)}+C_6{H}_{5}CO{O}^{-}{}_{\left(aq\right)}+H_2{O}_{\left(l\right)}$

Record the given data,

Concentration of $H-COOH=0.200M$

Concentration of $NaOH=2.00M$

Volume of solution $V=500.0mL$

Concentration of benzoic acid $=0.200M$

Calculate the number of mole of benzoic acid present in the solutions,

  $\begin{array}{c}Molesof{C}_6{H}_{5}COOH=\left(\frac{0.200mol{C}_6{H}_{5}COOH}{\cancel{L}}\right)\left(\frac{{10}^{-3}\cancel{L}}{1\cancel{mL}}\right)\left(475\cancel{mL}\right)\\ \\ =0.0950molof{C}_6{H}_{5}COOH\end{array}$

Calculate the number of mole of $NaOH$ present in the solution,

  $\begin{array}{c}MolesofNaOH=\left(\frac{2.00molNaOH}{L}\right)\left(\frac{{10}^{-3}L}{1mL}\right)\left(25mL\right)\\ \\ =0.050molNaOH\end{array}$

$NaOH$ is the limiting reagent,

Write the balance equation of the reaction,

$\begin{array}{l}{C}_6{H}_{5}COO{H}_{\left(aq\right)}+NaO{H}_{\left(aq\right)}\stackrel{}{\rightleftarrows }N{a}^{2+}{}_{\left(aq\right)}+C_6{H}_{5}CO{O}^{-}{}_{\left(aq\right)}+H_2{O}_{\left(l\right)}\\ \\ \text{Initial}\text{(M):}0.0950mol0.050mol-0-\\ \\ \text{Change}\text{(M):}-0.050mol-0.050mol-+0.050mol-\\ \\ ----------------------------------------------\\ \text{Equilibrium}\text{(M):}0.045mol0mol-0.050mol-\end{array}$

The concentrations after the reaction are,

  $\begin{array}{c}\left[C_6{H}_{5}COOH\right]=\left(\frac{0.045mol{C}_6{H}_{5}COOH}{\left(475+25\right)mL}\right)\left(\frac{1mL}{{10}^{-3}L}\right)\\ \\ =0.090M{C}_6{H}_{5}COOH\end{array}$

  $\begin{array}{c}\left[C_6{H}_{5}CO{O}^{-}\right]=\left(\frac{0.050mol{C}_6{H}_{5}CO{O}^{-}}{\left(475+25\right)mL}\right)\left(\frac{1mL}{{10}^{-3}L}\right)\\ \\ =0.10M{C}_6{H}_{5}CO{O}^{-}\end{array}$

Calculate the $pH$ of the reaction

Henderson-Hasselbalch equation is,

  $\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}{\frac{\left[\text{base}\right]}{\left[\text{acid}\right]}}_{}^{}$

  $pH=p{K}_a+\log \left(\frac{C_6{H}_{5}CO{O}^{-}}{C_6{H}_{5}COOH}\right)$

The $K_a$ for benzoic acid is $6.3\times {10}^{-5}$ (this value referred from Appendix table).

The $p{K}_a$ is $-\log \left(6.3\times {10}^{-5}\right)=4.201$

The $p{K}_a$ value is plugging above equation, to solve $pH$ of the reaction,

  $\begin{array}{c}pH=4.201+\log \left(\frac{0.10}{0.090}\right)\\ =4.201+\log \left(1.1111\right)\\ =4.201+0.045757\\ =4.24675\\ \\ pH=4.2\end{array}$

Calculate the formic acid $\left(H-COOH\right)$and use the Henderson-Hasselbalch equation,

  $\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}{\frac{\left[\text{base}\right]}{\left[\text{acid}\right]}}_{}^{}$

The $K_a$ for formic acid is $1.8\times {10}^{-4}$ (this value referred from Appendix table).

  The $p{K}_a$ is $-\log \left(1.8\times {10}^{-4}\right)=3.7447$

  $pH=p{K}_a+\log \left(\frac{HCO{O}^{-}}{HCOOH}\right)$

Benzoic acid $pH$ value is  $4.24675$

Formic acid $p{K}_a$ value is $3.7447$ 

The $pH$ and $p{K}_a$ values are plugging above equation,

  $\begin{array}{l}\Rightarrow 4.24676=3.7447+\log \left(\frac{HCO{O}^{-}}{HCOOH}\right)\\ \\ \Rightarrow 4.24676-3.7447=\log \left(\frac{HCO{O}^{-}}{HCOOH}\right)\end{array}$

Hence,

  $\begin{array}{l}\log \left(\frac{HCO{O}^{-}}{HCOOH}\right)=0.50206\\ \\ p{K}_{a}valueforformicacidis3.177313\\ \\ \left(\frac{HCO{O}^{-}}{HCOOH}\right)=3.177313\\ \\ \left(HCO{O}^{-}\right)=3.177313\left(HCOOH\right)\end{array}$

Since the conjugate acid and the conjugate base are in the same volume, the mole ratio and the molarity ratios are identical.

  $MolesofHCO{O}^{-}=3.177313molHCOOH$

The total volume of the solution is,

  $\begin{array}{l}=\left(500mL\right)\times \left(\frac{{10}^{-3}L}{1mL}\right)\\ =0.500L\end{array}$

Let,

  $V_a=$ Volume of acid solution added and

  $V_b=$ Volume of base solution added

Thus,

    $V_a+V_b=0.500L$

The reaction between the formic acid and the sodium hydroxide is,

  $HOO{H}_{\left(aq\right)}+NaO{H}_{\left(aq\right)}\stackrel{}{\to }HCOON{a}_{\left(aq\right)}+H_2{O}_{\left(l\right)}$

The moles of $NaOH$ added equal the moles of $HCOOH$ reacted and the moles of $HCOONa$ formed.

  $\begin{array}{l}MolesNaOH=\left(2.00molNaOH/L\right)\left(V_b\right)=2.00V_{b}mol\\ \\ TotalmolesHCOOH=\left(0.200molHCOOH/L\right)\left(V_a\right)=2.00V_{a}mol\end{array}$  

The stoichiometric ratios in this reaction are 1:1 ratio,

  $\begin{array}{l}MolesHCOOHremainingaftertheraction=\left(0.200V_a-2.00V_b\right)mol\\ \\ MolesHCO{O}^{-}=molesHCOONa=molesofNaOH=2.00V_b\end{array}$

Using these moles and the mole ratio determined for the buffer solution gives,

  $\begin{array}{l}MolesHCO{O}^{-}=3.177313molHCOOH\\ \\ 2.00V_{b}mol=3.177313\left(0.200V_a-2.00V_b\right)mol\\ \\ 2.00V_b=0.6354626V_a-6.354626V_b\\ \\ \underset{\_}{addedfor{V}_b}\\ \\ 8.354626V_b=0.6354626V_a\end{array}$ 

The volume relationship given above calculation,

  $\begin{array}{l}{V}_a+V_b=0.500L\\ \\ V_a=\left(0.500-V_b\right)L\end{array}$

Calculate for the volume $\left(V\right)$

  $\begin{array}{l}8.354626V_b=0.6354626V_a\left(0.500-V_b\right)\\ \\ 8.354626V_b=0.3177313-0.6354626V_b\end{array}$

Added for volume $\left(V_b\right)$

  $\begin{array}{c}8.9900886V_b=0.3177313\\ V_b=\frac{0.3177313}{8.9900886}\\ V_b=0.0353424\\ \\ V_b=0.035LNaOH\end{array}$

Next calculate for volume $\left(V_a\right)$

  $\begin{array}{c}{V}_a=0.500-0.0353424\\ =0.4646576\\ \\ =0.465LHCOOH\end{array}$

Hence, the volume of $NaOH$ is $0.500-0.464=0.036L=36mL$ and the volume of the formic acid $HCOOH$ is $0.464=464mL$.