Chapter 14, Problem ISP

Lactic acid, CH₃CH(OH)COOH, is a weak monoprotic acid with a melting point of 53 °C. It exists as two enantiomers (Sec. 7-2f) that have slightly different K_a values. The D form has a K_a of 1.5 × 10⁻⁴ and the L form has a K_a of 1.6 × 10⁻⁴. The D form is synthesized by some bacteria. The L form is produced in muscle cells during anaerobic metabolism in which glucose molecules are broken down into lactic acid and molecules of adenosine triphosphate (ATP) are formed. When lactic acid builds up too rapidly in muscle tissue, severe pain results.

1. (a) Which form of lactic acid (D or L) is the stronger acid? Explain your answer.
2. (b) Determine the pK_a that would be measured for a 50:50 mixture of the two forms of lactic acid in aqueous solution, pK_a = −log K_a
3. (c) A solution of D-lactic acid is prepared. Use HL as a general formula for lactic acid, and write the equation for the ionization of lactic acid in water.
4. (d) If 0.100-M solutions of these two acids (D and L) were prepared, calculate what the pH of each solution would be.
5. (e) Before any lactic acid dissolves in the water, what reaction determines the pH?
6. (f) Calculate the pH of a solution made by dissolving 4.46 g D-lactic acid in 500. mL of water.
7. (g) Calculate the volume (mL) of 1.15-M NaOH(aq) required to completely neutralize 4.46 g of pure lactic acid.
8. (h) Calculate the pH of the solution when exactly enough NaOH was added to neutralize all of the lactic acid for (i) the D form; (ii) the L form; and (iii) a 50:50 mixture of the two forms.

Interpretation Introduction

Interpretation:

The stronger acid has to be chosen between D- form of lactic acid and L- form of lactic acid.

Explanation of Solution

The acid having larger $K_a$ value is known as strong acid.

The $K_a$ values of L- and D- form of lactic acid are $1.6\times {10}^{-4}and1.5\times {10}^{-4}$ respectively.  The $K_a$ value of L- form of lactic acid is slightly larger than that of D- form of lactic acid.  Thus, L- form of lactic acid is stronger acid.

Interpretation Introduction

Interpretation:

The value of $p{K}_a$ that would be measured for a $50:50$ mixture of the two forms of lactic acid in aqueous solution has to be determined.

Answer to Problem ISP

The value of $p{K}_a$ of the mixture is $3.81.$

Explanation of Solution

As the solution contains a mixture of two acids, the observed value of the $K_a$ is the combination of the two $K_a\text{'}s.$

When the reactions are added, $K_a\text{'}s$ are multiplied as they are logarithmic.

  $\begin{array}{l}H{L}_D\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons L_D{}^{-}\left(aq\right)+H_3{O}^{+}\left(aq\right)K_{a,D}\\ \\ \underset{\_}{H{L}_L\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons L_L{}^{-}\left(aq\right)+H_3{O}^{+}\left(aq\right)K_{a,L}}\\ H{L}_D\left(aq\right)+H{L}_L\left(aq\right)+2H_{2}O\left(l\right)\rightleftharpoons L_D{}^{-}\left(aq\right)+L_L{}^{-}\left(aq\right)+2H_3{O}^{+}\left(aq\right)K=K_{a,D}.K_{a,L}\end{array}$

It can be further written as given below.

  $\begin{array}{c}\frac{1}{2}H{L}_D\left(aq\right)+\frac{1}{2}H{L}_L\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons \frac{1}{2}{L}_D{}^{-}\left(aq\right)+\frac{1}{2}{L}_L{}^{-}\left(aq\right)+H_3{O}^{+}\left(aq\right)K_a={\left(K_{a,D}.K_{a,L}\right)}^{1/2}\\ \\ p{K}_a=-\log K_a=-\log \left[{\left(K_{a,D}.K_{a,L}\right)}^{1/2}\right]\end{array}$

On simplification,

  $\begin{array}{c}p{K}_a=\frac{1}{2}\left\{-\log \left[\left(K_{a,D}.K_{a,L}\right)\right]\right\}=\frac{1}{2}\left[\left(-\log K_{a,D}\right)-\left(-\log K_{a,L}\right)\right]\\ \\ p{K}_a=\frac{1}{2}\left(p{K}_{a,D}+p{K}_{a,L}\right)\end{array}$

Now,

  $\begin{array}{c}p{K}_{a,L}=-\log K_{a,L}=-\log \left(1.6\times {10}^{-4}\right)=3.80\\ \\ p{K}_{a,D}=-\log K_{a,D}=-\log \left(1.5\times {10}^{-4}\right)=3.82\\ \\ p{K}_a=\frac{1}{2}\left(p{K}_{a,D}+p{K}_{a,L}\right)=\frac{1}{2}\left(3.80+3.82\right)=3.81.\end{array}$

Therefore, the value of $p{K}_a$ of the mixture is $3.81.$

Interpretation Introduction

Interpretation:

The equation for the ionization of lactic acid in water has to be written.

Explanation of Solution

The equation for the ionization of lactic acid in water is given below.  The ionization causes the transfer of the carboxyl hydrogen ion from lactic acid to water.

  $HL\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons L^{-}\left(aq\right)+H_3{O}^{+}\left(aq\right)$

Interpretation Introduction

Interpretation:

The $pH$ of $0.100M$ solutions of D and L- lactic acids has to be determined.

Answer to Problem ISP

The $pH$ of $0.100M$ solutions of D and L- lactic acids are $2.41and2.40$ respectively.

Explanation of Solution

The equilibrium reaction is given below.

  $HL\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons L^{-}\left(aq\right)+H_3{O}^{+}\left(aq\right)$

The acid ionization constant can be written as given below.

  $K_b=\frac{\left[L^{-}\right]\left[H_3{O}^{+}\right]}{\left[HL\right]}$

A table can be set up as shown below.

  $\begin{array}{cccc}& HL& L^{-}& H_3{O}^{+}\\ Initialconc.\left(M\right)& 0.10& 0& 1\times {10}^{-7}\\ Changeinconc.\left(M\right)& -x& +x& +x\\ Equilibriumconc.\left(M\right)& 0.10-x& x& x\end{array}$

The concentration of $H_3{O}^{+}$ comes from its autoionization that is $1.0\times {10}^{-7}.$  It is assumed to be small compared to the change.

At equilibrium, $K_a=\frac{\left(x\right)\left(x\right)}{\left(0.10-x\right)}$

Assuming x is very small, it can be written as $0.10-x\cong \mathrm{0.10.}$

Then,

For L- form of the acid:

  $\begin{array}{l}{x}^2=\left(K_a\right)\left(0.10\right)\\ \\ x=\left(1.6\times {10}^{-4}\right)\left(0.10\right)\\ \\ x=\text{4}{\text{.0×10}}^{\text{-3}}\text{M=}\left[{\text{H}}_3{\text{O}}^{+}\right]\end{array}$

The $pH$ of the solution can be calculated as shown below.

  $pH=-\log \left[H_3{O}^{+}\right]=-\log \left(4.0\times {10}^{-3}\right)=2.40.$

For D- form of the acid:

  $\begin{array}{l}{x}^2=\left(K_a\right)\left(0.10\right)\\ \\ x=\left(1.5\times {10}^{-4}\right)\left(0.10\right)\\ \\ x=3.{\text{9×10}}^{\text{-3}}\text{M=}\left[{\text{H}}_3{\text{O}}^{+}\right]\end{array}$

The $pH$ of the solution can be calculated as shown below.

  $pH=-\log \left[H_3{O}^{+}\right]=-\log \left(3.9\times {10}^{-3}\right)=2.41.$

Therefore, the $pH$ of $0.100M$ solutions of D and L- lactic acids are $2.41and2.40$ respectively.

Interpretation Introduction

Interpretation:

The reaction which determines the $pH$ before any lactic acid dissolves in the water has to be written.

Explanation of Solution

The $pH$ is determined by the autoionization of water.  The reaction is given below.

  $2H_{2}O\left(l\right)\rightleftharpoons O{H}^{-}\left(aq\right)+H_3{O}^{+}\left(aq\right)$

Interpretation Introduction

Interpretation:

The $pH$ of a solution made by dissolving $4.46g$ D- lactic acid in $500mL$ of water has to be calculated.

Answer to Problem ISP

The $pH$ of solutions of D-lactic acid is $2.41.$

Explanation of Solution

Calculation of initial concentration of D- lactic acid:

The number of moles of D-HL can be calculated as given below.

  $Moles=\frac{Mass}{Molarmass}=\frac{4.46g}{90.0777g/mol}=0.0495mol.$

Then, the concentration of D-HL is given below.

  $\begin{array}{l}Molarity=\frac{Moles}{VolumeofsolutioninL}=\frac{0.0495mol}{500\times {10}^{-3}L}=0.0990M.\\ \\ Where,1L=1000mL.\end{array}$

Therefore, the initial concentration of D- lactic acid is $0.0990M.$

The equilibrium reaction is given below.

  $HL\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons L^{-}\left(aq\right)+H_3{O}^{+}\left(aq\right)$

The acid ionization constant can be written as given below.

  $K_b=\frac{\left[L^{-}\right]\left[H_3{O}^{+}\right]}{\left[HL\right]}$

A table can be set up as shown below.

  $\begin{array}{cccc}& HL& L^{-}& H_3{O}^{+}\\ Initialconc.\left(M\right)& 0.0990& 0& 1\times {10}^{-7}\\ Changeinconc.\left(M\right)& -x& +x& +x\\ Equilibriumconc.\left(M\right)& 0.0990-x& x& x\end{array}$

The concentration of $H_3{O}^{+}$ comes from its autoionization that is $1.0\times {10}^{-7}.$  It is assumed to be small compared to the change.

At equilibrium, $K_a=\frac{\left(x\right)\left(x\right)}{\left(0.0990-x\right)}=1.5\times {10}^{-4}$

Assuming x is very small, it can be written as $0.0990-x\cong \mathrm{0.0990.}$

Then,

  $\begin{array}{l}{x}^2=\left(1.6\times {10}^{-4}\right)\left(0.10\right)\\ \\ x=3.9{\text{×10}}^{\text{-3}}\text{M=}\left[{\text{H}}_3{\text{O}}^{+}\right]\end{array}$

The $pH$ of the solution can be calculated as shown below.

  $pH=-\log \left[H_3{O}^{+}\right]=-\log \left(3.9\times {10}^{-3}\right)=2.41.$

Therefore, the $pH$ of solutions of D-lactic acid is $2.41.$

Interpretation Introduction

Interpretation:

The volume of $1.15MNaOH\left(aq\right)$ required to completely neutralize $4.46g$ of pure lactic acid has to be calculated.

Answer to Problem ISP

The volume of $1.15MNaOH\left(aq\right)$ required to completely neutralize $4.46g$ of pure lactic acid is $43.0mL.$

Explanation of Solution

The neutralization reaction is given below.

  $HL\left(aq\right)+O{H}^{-}\left(aq\right)\rightleftharpoons L^{-}\left(aq\right)+H_{2}O\left(l\right)$

The stoichiometry of acid and the base is $1:1.$

The number of moles of D-HL can be calculated as given below.

  $Moles=\frac{Mass}{Molarmass}=\frac{4.46g}{90.0777g/mol}=0.0495mol.$

Hence, the number of moles of $NaOH$ required is $0.0495mol.$

Then, the volume of $NaOH$ required can be calculated as given below.

  $\begin{array}{c}Molarity=\frac{Moles}{VolumeofsolutioninL}\\ \\ Volume=\frac{Moles}{Molarity}\\ \\ =\frac{0.0495mol}{1.15mol/L}\\ \\ =0.0430L=0.0430\times {10}^{3}mL=43.0mL.\\ \\ Where,1L=1000mL.\end{array}$

Therefore, the volume of $1.15MNaOH\left(aq\right)$ required to completely neutralize $4.46g$ of pure lactic acid is $43.0mL.$

Interpretation Introduction

Interpretation:

The $pH$ of the solution when exactly enough $NaOH$ was added to neutralize all of the lactic acid for $\left(i\right)$ the D form, $\left(ii\right)$ the L form and $\left(iii\right)$ a $50:50$ mixture of the two forms has to be calculated.

Explanation of Solution

The equilibrium reaction is given below.

  $L^{-}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons HL\left(aq\right)+O{H}^{-}\left(aq\right)$

The equilibrium constant for the above reaction can be written as given below.

  $K_b=\frac{\left[HL\right]\left[O{H}^{-}\right]}{\left[L^{-}\right]}$

There is relationship exist as given below.

  $\begin{array}{l}{K}_w=K_a{K}_b\\ \\ K_b=\frac{K_w}{K_a}\end{array}$

Now,

For D form:

  $K_{b,D}=\frac{1.0\times {10}^{-14}}{1.5\times {10}^{-4}}=6.7\times {10}^{-11}$

For L form:

  $K_{b,L}=\frac{1.0\times {10}^{-14}}{1.6\times {10}^{-4}}=6.3\times {10}^{-11}$

For $50:50$ mixture of the two forms:

  $\begin{array}{l}p{K}_{b,L}=-\log K_{b,L}=-\log \left(6.7\times {10}^{-11}\right)=10.17\\ \\ p{K}_{b,D}=-\log K_{b,D}=-\log \left(6.3\times {10}^{-11}\right)=10.20\\ \\ p{K}_b=\frac{1}{2}\left(p{K}_{b,L}+p{K}_{b,D}\right)=\frac{1}{2}\left(10.17+10.20\right)=10.19\\ \\ K_b-{10}^{-p{K}_b}={10}^{-10.19}=6.5\times {10}^{-11}.\end{array}$

Calculation of concentration of lactate ion:

The number of moles of HL can be calculated as given below.

  $Moles=\frac{Mass}{Molarmass}=\frac{4.46g}{90.0777g/mol}=0.0495mol.$

Then,

  $\begin{array}{c}Molarity=\frac{Moles}{VolumeofsolutioninL}\\ \\ =\frac{0.0495mol}{43.0\times {10}^{-3}L}\\ \\ =1.15M.\end{array}$

Therefore, the concentration of lactate ion is $1.15M.$

A table can be set up as shown below.

  $\begin{array}{cccc}& L^{-}\left(aq\right)& HL\left(aq\right)& O{H}^{-}\left(aq\right)\\ Initialconc.\left(M\right)& 1.15& 0& 1\times {10}^{-7}\\ Changeinconc.\left(M\right)& -x& +x& +x\\ Equilibriumconc.\left(M\right)& 1.15-x& x& x\end{array}$

The concentration of $O{H}^{-}$ comes from its autoionization that is $1.0\times {10}^{-7}.$  It is assumed to be small compared to the change.

At equilibrium, $K_b=\frac{\left(x\right)\left(x\right)}{\left(1.15-x\right)}$

Assuming x is very small, it can be written as $1.15-x\cong \mathrm{1.15.}$

Then,

(i) For D form:

  $\begin{array}{l}{x}^2=\left(6.7\times {10}^{-11}\right)\left(1.15\right)\\ \\ x=8.8{\text{×10}}^{\text{-6}}\text{M=}\left[O{H}^{-}\right]\end{array}$

The $pH$ of the solution can be calculated as shown below.

  $\begin{array}{l}pOH=-\log \left[O{H}^{-}\right]=-\log \left(8.8\times {10}^{-6}\right)=5.06\\ \\ pH=14-pOH=14-5.06=8.94.\end{array}$

Therefore, the $pH$ of solutions of D-lactic acid is $8.94.$

(ii) For L form:

  $\begin{array}{l}{x}^2=\left(6.3\times {10}^{-11}\right)\left(1.15\right)\\ \\ x=8.5{\text{×10}}^{\text{-6}}\text{M=}\left[O{H}^{-}\right]\end{array}$

The $pH$ of the solution can be calculated as shown below.

  $\begin{array}{l}pOH=-\log \left[O{H}^{-}\right]=-\log \left(8.5\times {10}^{-6}\right)=5.07\\ \\ pH=14-pOH=14-5.07=8.93.\end{array}$

Therefore, the $pH$ of solutions of L-lactic acid is $8.93.$

(iii) For $50:50$ mixture of the two forms:

  $\begin{array}{l}{x}^2=\left(6.5\times {10}^{-11}\right)\left(1.15\right)\\ \\ x=8.6{\text{×10}}^{\text{-6}}\text{M=}\left[O{H}^{-}\right]\end{array}$

The $pH$ of the solution can be calculated as shown below.

  $\begin{array}{l}pOH=-\log \left[O{H}^{-}\right]=-\log \left(8.6\times {10}^{-6}\right)=5.06\\ \\ pH=14-pOH=14-5.06=8.94.\end{array}$

Therefore, the $pH$ of $50:50$ mixture of the two forms is $8.94.$