Chapter 15, Problem 15.51QA

Interpretation Introduction

To find:

The value of $K_a$ for lactic acid

Answer to Problem 15.51QA

Solution:

$K_a=8.91 \times {10}^{-4}$

Explanation of Solution

1) Concept:

We are asked to determine the $K_a$ value for lactic acid from known initial concentration of the acid and given percent ionization. Lactic acid $\left(C{H}_{3}CH\left(OH\right)COOH\right)$ indicates that only one H atom per molecule is ionisable in aqueous solution.

The ionization reaction of butanoic acid can be written as

$C{H}_{3}CH\left(OH\right)COOH \left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons C{H}_{3}CH\left(OH\right)CO{O}^{-}\left(aq\right)+H_3{O}^{+}\left(aq\right)$

And the equilibrium constant expression can be written as

$K_a=\frac{\left[C{H}_{3}CH\left(OH\right)CO{O}^{-}\right]\left[H_3{O}^{+}\right]}{\left[C{H}_{3}CH\left(OH\right)COOH \right]}$

2) Formulae:

i) $Percent ionization=\frac{{\left[A^{-}\right]}_{equilibrium}}{{\left[HA\right]}_{initial}} \times 100$

ii) $K_a=\frac{\left[C{H}_{3}CH\left(OH\right)CO{O}^{-}\right]\left[H_3{O}^{+}\right]}{\left[C{H}_{3}CH\left(OH\right)COOH\right]}$

3) Given:

i) Initial concentration of butanoic acid is $1.00M$

ii) Percent ionization of the acid is $2.94\%$

4) Calculations:

From the given percent dissociation, we can find the ${\left[C{H}_{3}CH\left(OH\right)CO{O}^{-}\right]}_{equilibrium}$ using the percent dissociation formula as

$Percent ionization=\frac{{\left[C{H}_{3}CH\left(OH\right)CO{O}^{-}\right]}_{equilibrium}}{{\left[C{H}_{3}CH\left(OH\right)COOH\right]}_{initial}} \times 100$

$2.94\%=\frac{{\left[C{H}_{3}CH\left(OH\right)CO{O}^{-}\right]}_{equilibrium}}{1.00 M} \times 100$

$0.0294=\frac{{\left[C{H}_{3}CH\left(OH\right)CO{O}^{-}\right]}_{equilibrium}}{1.00 M}$

${\left[C_3{H}_{7}CO{O}^{-}\right]}_{equilibrium}=0.0294 \times 1.00 M=0.0294 M$

The equilibrium concentration of $C{H}_{3}CH\left(OH\right)CO{O}^{-}$ is calculated as $0.0294M$.

We can set up the RICE table to solve this problem.

Reaction	$C{H}_{3}CH\left(OH\right)COOH \left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons C{H}_{3}CH\left(OH\right)CO{O}^{-}\left(aq\right)+H_3{O}^{+}\left(aq\right)$ $\left[C{H}_{3}CH\left(OH\right)COOH\right] \left(M\right)$ $\left[C{H}_{3}CH\left(OH\right)CO{O}^{-}\right]\left(M\right)$ $\left[H_3{O}^{+}\right] \left(M\right)$
Initial (I)	$1.00$	$0.00$	$0.00$
Change (C)	$-x$	$+x$	$+x$
Equilibrium (E)	$\left(1.00-x\right)$	$x$	$x$

At equilibrium, $\left[H_3{O}^{+}\right]=\left[C{H}_{3}CH\left(OH\right)CO{O}^{-}\right]=x=2.94\times {10}^{-2}M$

The equilibrium concentration for the acid is

${\left[C{H}_{3}CH\left(OH\right)COOH\right]}_{equilibrium}={\left[C{H}_{3}CH\left(OH\right)COOH\right]}_{initial}-{\left[H_3{O}^{+}\right]}_{equilibrium}$

${\left[C{H}_{3}CH\left(OH\right)COOH\right]}_{equilibrium}=\left(1.00-2.94\times {10}^{-2}\right)M= 0.9706M$

Inserting the equilibrium values for all the species in the $K_a$ expression, we get

$K_a=\frac{\left[C{H}_{3}CH\left(OH\right)CO{O}^{-}\right]\left[H_3{O}^{+}\right]}{\left[C{H}_{3}CH\left(OH\right)COOH\right]}$

$K_a=\frac{\left(2.94\times {10}^{-2}\right)\left(2.94\times {10}^{-2}\right)}{0.9706 }=8.905 \times {10}^{-4}$

Thus, the acid dissociation constant value for lactic acid is $8.91 \times {10}^{-4}$

Conclusion:

$K_a$ value is calculated from the initial concentration and percent ionization of lactic acid.