Chapter 15, Problem 15.54QA

Interpretation Introduction

To find:

The acid dissociation constant value ${(K}_{a)}$ for butanoic acid.

Answer to Problem 15.54QA

Solution:

The acid dissociation constant value $\left(K_a\right)$ for butanoic acid is $1.53 \times {10}^{-5}$

Explanation of Solution

1) Concept:

We are asked to determine the $K_a$ value for the butanoic acid $(C_3{H}_{7}COO-H)$ from known initial concentration of the acid and percent ionization. Structural formula for butanoic acid indicates that only one H atom per molecule is ionisable in aqueous solution.

The ionization reaction of butanoic acid can be written as

$C_3{H}_{7}COOH \left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons C_3{H}_{7}CO{O}^{-}\left(aq\right)+H_3{O}^{+}\left(aq\right)$

And the equilibrium constant expression can be written as

$K_a=\frac{\left[C_3{H}_{7}CO{O}^{-}\right]\left[H_3{O}^{+}\right]}{\left[C_3{H}_{7}COOH\right]}$

2) Formulae:

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

ii) $K_a=\frac{\left[C_3{H}_{7}CO{O}^{-}\right]\left[H_3{O}^{+}\right]}{\left[C_3{H}_{7}COOH\right]}$

3) Given:

i) Initial concentration of butanoic acid is $0.100M$

ii) Percent ionization of the acid is $1.23\%$

4) Calculations:

From the given percent dissociation, we can find the ${\left[C_3{H}_{7}CO{O}^{-}\right]}_{equilibrium}$ using the percent dissociation formula as

$Percent ionization=\frac{{\left[C_3{H}_{7}CO{O}^{-}\right]}_{equilibrium}}{{\left[C_3{H}_{7}COOH\right]}_{initial}} \times 100$

$1.23\%=\frac{{\left[C_3{H}_{7}CO{O}^{-}\right]}_{equilibrium}}{0.100 M} \times 100$

$0.0123=\frac{{\left[C_3{H}_{7}CO{O}^{-}\right]}_{equilibrium}}{0.100 M}$

${\left[C_3{H}_{7}CO{O}^{-}\right]}_{equilibrium}=0.0123 \times 0.100 M=0.00123M=1.23\times {10}^{-3}M$

The equilibrium concentration of $C_3{H}_{7}CO{O}^{-}$ is calculated as $1.23\times {10}^{-3}M$

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

Reaction	$C_3{H}_{7}COOH \left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons C_3{H}_{7}CO{O}^{-}\left(aq\right)+H_3{O}^{+}\left(aq\right)$ $\left[C_3{H}_{7}COOH \right] \left(M\right)$ $\left[C_3{H}_{7}CO{O}^{-}\right]\left(M\right)$ $\left[H_3{O}^{+}\right] \left(M\right)$
Initial (I)	$0.100$	$0.00$	$0.00$
Change (C)	$-x$	$+x$	$+x$
Equilibrium (E)	$\left(0.100-x\right)$	$x$	$x$

At equilibrium, $\left[H_3{O}^{+}\right]=\left[C_3{H}_{7}CO{O}^{-}\right]=1.23\times {10}^{-3}M$

The equilibrium concentration for the acid is

${\left[C_3{H}_{7}COOH\right]}_{equilibrium}={\left[C_3{H}_{7}COOH\right]}_{initial}-{\left[H_3{O}^{+}\right]}_{equilibrium}$

${\left[C_3{H}_{7}COOH\right]}_{equilibrium}=\left(0.100-1.23\times {10}^{-3}\right)M= 0.09877M$

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

$K_a=\frac{\left[C_3{H}_{7}CO{O}^{-}\right]\left[H_3{O}^{+}\right]}{\left[C_3{H}_{7}COOH\right]}$

$K_a=\frac{\left(1.23\times {10}^{-3}\right)\left(1.23\times {10}^{-3}\right)}{0.09877 }=1.53 \times {10}^{-5}$

Thus, the acid dissociation constant value for butanoic acid is $1.53 \times {10}^{-5}$

Conclusion:

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