Chapter 16, Problem 16.134QA

Interpretation Introduction

To find:

a) Which acid yields a solution with the lower pH if $0.100 M$ solutions are prepared?

b) Which acid will dissociate to a greater degree?

c) What is the pH of a solution that contains $15.8 mM$ of acetoacetic acid and $10.8 mM$ sodium acetoacetate?

d) What is the $pH$ of a solution of $90 mg/L$ β-hydroxybutyric acid and $90 mg/L$ β-hydroxybutyrate anion after 100 µL of $0.100 M$ HCl has been added.

Answer to Problem 16.134QA

Solution:

a) Acetoacetic acid

b) Acetoacetic acid

c) $pH = 3.41$

d) $pH = 4.62$

Explanation of Solution

1) Concept:

We are asked to find which acid yields a solution with the lower $pH$ if $0.100 M$ solutions of β-hydroxybutyric acid and acetoacetic acid are prepared. To calculate the $pH$, we must first determine the equilibrium concentration of $H^{+}$ ions and then convert it to $pH$. For this step, we need to draw the RICE table. From the equilibrium concentrations of acid, we will find out the acid that will dissociate to a greater degree. We are given the $p{K}_a$ values. The $pH$ of a solution of a weak acid and its conjugate base can be calculated using the Henderson-Hasselbalch equation.

2) Formula:

i) $K_a= \frac{\left[A^{-}\right] \left[H_3{O}^{+}\right]}{\left[HA\right]}$

where, $K_a$ is acid ionization constant

$\left[A^{-}\right]$= Concentration of conjugate base

$\left[HA\right]=$ Concentration of acid

ii) $p{K}_a= -\mathrm{l}\mathrm{o}\mathrm{g}\left[K_a\right]$

iii) $pH=p{K}_a+\log \frac{\left[A^{-}\right]}{\left[HA\right]}$

3) Given:

β-hydroxybutyric acid, $p{K}_a=4.72$

Acetoacetic acid, $p{K}_a=3.58$

For part a:

Concentration of acetoacetic acid $= 0.100 M$

Concentration of β-hydroxybutyric acid $= 0.100 M$

For part c:

Concentration of acetoacetic acid $= 15.8 mM$

Concentration of sodium acetoacetate $= 10.8 mM$

For part d:

Concentration of β-hydroxybutyric acid $= 90 mg/L$

Concentration of β-hydroxybutyrate anion $= 90 mg/L$

Concentration of $HCl= 0.100 M$

Volume of $HCl=100\mu L=1.00\times {10}^{-4}L$

Molar mass of β-hydroxybutyric acid $=104.1 g/mol$

Molar mass of β-hydroxybutyrate anion $=126.1 g/mol$

4) Calculation:

a) Which acid yields a solution with the lower pH if $0.100 M$ solutions are prepared?

The $p{K}_a$ value of acetoacetic acid is lower than $p{K}_a$ value of β-hydroxybutyric acid. Therefore, acetoacetic acid will dissociate more and give more ${[H}_3{O}^{+}]$. So acetoacetic acid yields a solution with the lower $pH$.

The dissociation reaction for β-hydroxybutyric acid can be written as

	[β — hydroxybutyric acid] (M)	[β — hydroxybutyric acid] (M)	[H3O+] (M)
Initial (I)	$0.100$	$0$	$0$
Change (C)	$-x$	$+x$	$+x$
Equilibrium (E)	$0.100-x$	$x$	$x$

$p{K}_a= -\mathrm{l}\mathrm{o}\mathrm{g}\left[K_a\right]$

$K_a={10}^{(-p{K}_a)}$

$K_a={10}^{(-4.72)}= 1.905 \times {10}^{-5}$

$K_a= \frac{[-hydroxybutyrate anion]\left[H_3{O}^{+}\right]}{\left[-hydroxybutyric acid\right]}$

$1.905 \times {10}^{-5}= \frac{\left[x\right]\left[x\right]}{[0.100-x]}$

Assuming $x$ is very small and $0.100-x\approx 0.100$,

$1.905 \times {10}^{-5}= \frac{\left[x\right]\left[x\right]}{\left[0.100\right]}$

${\left[x\right]}^2=1.905 \times {10}^{-6}$

$\left[x\right]=\left[H_3{O}^{+}\right]=0.00138$

$pH= -\log \left[H_3{O}^{+}\right]= -\mathrm{l}\mathrm{o}\mathrm{g}\left(0.00138\right)$

$pH=2.86$

The dissociation reaction for acetoacetic acid can be written as

	[Acetoacetic acid] (M)	[Acetoacetate anion] (M)	[H3O+] (M)
Initial (I)	$0.100$	$0$	$0$
Change ( C)	$-x$	$+x$	$+x$
Equilibrium ( E)	$0.100-x$	$x$	$x$

$p{K}_a= -\mathrm{l}\mathrm{o}\mathrm{g}\left[K_a\right]$

$K_a={10}^{(-p{K}_a)}$

$K_a={10}^{(-3.58)}= 2.63\times {10}^{-4}$

$K_a= \frac{\left[Acetoacetate anion\right]\left[H_3{O}^{+}\right]}{\left[Acetoacetic acid\right]}$

$2.63\times {10}^{-4}= \frac{\left[x\right]\left[x\right]}{[0.100-x]}$

Assuming $x$ is very small and $0.100-x\approx 0.100$,

$2.63\times {10}^{-4}= \frac{\left[x\right]\left[x\right]}{\left[0.100\right]}$

${\left[x\right]}^2=2.63\times {10}^{-5}$

$\left[x\right]=\left[H_3{O}^{+}\right]=0.00138$

$pH= -\log \left[H_3{O}^{+}\right]= -\mathrm{l}\mathrm{o}\mathrm{g}\left(0.00513\right)$

$pH=2.29$

b) Which acid will dissociate to a greater degree?

Lower the $p{K}_a$ value, higher will be the degree of dissociation. The $p{K}_a$ value of acetoacetic acid is lower than $p{K}_a$ value of β-hydroxybutyric acid. Therefore, acetoacetic acid will dissociate to a greater degree.

c) What is the pH of a solution that contains $15.8 mM$ of acetoacetic acid and $10.8 mM$ sodium acetoacetate?

$pH=p{K}_a+\log \frac{\left[A^{-}\right]}{\left[HA\right]}$

$pH=3.58+\log \left(\frac{\left[10.8\right]}{\left[15.8\right]}\right)$

$pH=3.58+(-0.1652)$

$pH=3.415$

The $pH$ of a solution that contains $15.8 mM$ of acetoacetic acid and $10.8 mM$ sodium acetoacetate is $3.41$.

d) What is the $pH$ of a solution of $90 mg/L$ β-hydroxybutyric acid and $90 mg/L$ β-hydroxybutyrate anion after 100 µL of $0.100 M$ HCl has been added?

First we will convert the given concentration from $mg/L$ to $mol/L$.

$90\frac{mg}{L} \times \frac{1 g}{1000 mg} \times \frac{1 mol }{104.1 g}=8.6455 \times {10}^{-4} M$

$90\frac{mg}{L} \times \frac{1 g}{1000 mg} \times \frac{1 mol }{126.1 g}=7.1372 \times {10}^{-4} M$

We assume that the volume of buffer solution is $1L$.

$1L \times 8.6455 \times {10}^{-4}\frac{mol}{L}= 8.6455 \times {10}^{-4} mol$

$1L \times 7.1372 \times {10}^{-4}\frac{mol}{L}= 7.1372 \times {10}^{-4} mol$

Calculation of mole of $HCl$:

$100 \mu L \times \frac{1 L}{{10}^{6 }\mu L} \times 0.100\frac{mol}{L}= 1.00\times {10}^{-5} mol$

	[A−] (mol)	[HCL] (mol)	[HA] (mol)
Initial	$8.6455 \times {10}^{-4}$	$1.00\times {10}^{-5}$	$7.1372 \times {10}^{-4}$
Change	$-1.00\times {10}^{-5}$	-$1.00\times {10}^{-5}$	$+1.00\times {10}^{-5}$
Final	$7.0372\times {10}^{-4}$	$0$	$8.7455\times {10}^{-4}$

$pH=p{K}_a+\log \frac{\left[A^{-}\right]}{\left[HA\right]}$

$pH=4.72+\log \frac{(7.0372\times {10}^{-4})}{(8.7455\times {10}^{-4})}$

$pH=4.72+(-0.09439)$

$pH=4.6256$

The $pH$ of a solution of $90 mg/L$ β-hydroxybutyric acid and $90 mg/L$ β-hydroxybutyrate anion after 100 µL of $0.100 M$$HCl$ has been added is $4.62$.

Conclusion:

Lower the $p{K}_a$ value, higher will be the degree of dissociation. Using Henderson-Hasselbalch equation, the $pH$ of solution after addition of $HCl$ acid is calculated.