Chapter 13, Problem 52QAP

Penicillin $\left(\text{MM}=356\text{g}/\text{mol}\right)$, an antibiotic often used to treat bacterial infections, is a weak acid. Its K_a is $1.7\times {10}^{-3}$. Calculate [H⁺] in solutions prepared by adding enough water to the following to make 725 mL.

(a) 0.187 mol

(b) 127 g

Interpretation Introduction

(a)

Interpretation:

The concentration of hydrogen ion in the given solution should be calculated.

Concept introduction:

The dissociation reaction of a weak acid is represented as follows:

$\text{HB}\rightleftarrows {\text{H}}^{+}+{\text{B}}^{-}$

The expression for the acid dissociation constant will be as follows:

$K_a=\frac{\left[{\text{H}}^{+}\right]\left[{\text{B}}^{-}\right]}{\left[\text{HB}\right]}$

Here, $\left[{\text{H}}^{+}\right]$ is equilibrium concentration of hydrogen ion, $\left[{\text{B}}^{-}\right]$ is equilibrium concentration of conjugate base and $\left[\text{HB}\right]$ is the equilibrium concentration of weak acid.

Answer to Problem 52QAP

0.020 M

Explanation of Solution

The acid dissociation constant of penicillin is $1.7\times {10}^{-3}$. The volume of solution is 725 mL and number of moles of the acid is 0.187 mol.

The molarity of penicillin can be calculated as follows:

$M=\frac{n}{V\left(L\right)}$

Putting the values,

$M=\frac{0.187\text{ mol}}{725\text{ mL}\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)}=0.258\text{ M}$

Thus, the concentration of penicillin is 0.258 M.

The molecular formula of penicillin is ${\text{C}}_{\text{16}}{\text{H}}_{\text{18}}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{S}$.

The concentration all the species can be calculated using the ICE table as follows:

$\begin{array}{l}{\text{ C}}_{\text{16}}{\text{H}}_{\text{18}}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{S}\rightleftharpoons {\text{C}}_{\text{16}}{\text{H}}_{\text{17}}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}{\text{S}}^{-}+{\text{H}}^{+}\\ I\text{ 0}\text{.258 - -}\\ C-xxx\\ E0.258-xxx\end{array}$

The expression for $K_a$ is as follows:

$K_a=\frac{\left[{\text{C}}_4{\text{H}}_3{\text{N}}_{\text{2}}{\text{O}}_3{}^{-}\right]\left[{\text{H}}^{+}\right]}{\left[{\text{HC}}_4{\text{H}}_3{\text{N}}_{\text{2}}{\text{O}}_3\right]}$

Putting the values,

$1.7\times {10}^{-3}=\frac{\left(x\right)\left(x\right)}{\left(0.258-x\right)}$

On rearranging,

$\begin{array}{l}1.7\times {10}^{-3}\left(0.258\right)-1.7\times {10}^{-3}x=x^2\\ 4.386\times {10}^{-4}-1.7\times {10}^{-3}x=x^2\end{array}$

Or,

$x^2+1.7\times {10}^{-3}x-4.386\times {10}^{-4}=0$

Comparing this with the general quadratic equation as follows:

$a{x}^2+bx-c=0$

Solving the quadratic equation,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Putting the values,

$\begin{array}{l}x=\frac{-1.7\times {10}^{-3}\pm \sqrt{{\left(1.7\times {10}^{-3}\right)}^2-4\left(1\right)\left(-4.386\times {10}^{-4}\right)}}{2\left(1\right)}\\ =\frac{-1.7\times {10}^{-3}\pm \sqrt{2.89\times {10}^{-6}+1.75\times {10}^{-3}}}{2\left(1\right)}\\ =\frac{-1.7\times {10}^{-3}\pm 0.042}{2}\end{array}$

$x=0.020,-0.022$

Since, the value of x cannot be negative thus, the value of x will be 0.020.

From the ICE table, it is equal to the concentration of hydrogen ion thus,

$\left[{\text{H}}^{+}\right]=0.020$

Therefore, the concentration of hydrogen ion is 0.020 M.

Interpretation Introduction

(b)

Interpretation:

The concentration of hydrogen ion in the given solution should be calculated.

Concept introduction:

The dissociation reaction of a weak acid is represented as follows:

$\text{HB}\rightleftarrows {\text{H}}^{+}+{\text{B}}^{-}$

The expression for the acid dissociation constant will be as follows:

$K_a=\frac{\left[{\text{H}}^{+}\right]\left[{\text{B}}^{-}\right]}{\left[\text{HB}\right]}$

Here, $\left[{\text{H}}^{+}\right]$ is equilibrium concentration of hydrogen ion, $\left[{\text{B}}^{-}\right]$ is equilibrium concentration of conjugate base and $\left[\text{HB}\right]$ is the equilibrium concentration of weak acid.

Answer to Problem 52QAP

0.02805 M

Explanation of Solution

The mass of penicillin is given as 127 g. The molecular formula of the penicillin is ${\text{C}}_{\text{16}}{\text{H}}_{\text{18}}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{S}$ thus, its molar mass is 356 g/mol.

Now, from mass and molar mass of the ${\text{C}}_{\text{16}}{\text{H}}_{\text{18}}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{S}$, its number of moles can be calculated as follows:

$n=\frac{m}{M}=\frac{127\text{ g}}{356\text{ g/mol}}=0.3567\text{ mol}$

Now, the molarity of ${\text{C}}_{\text{16}}{\text{H}}_{\text{18}}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{S}$ can be calculated as follows:

$M=\frac{n}{V\left(L\right)}$

Putting the values,

$M=\frac{0.3567\text{ mol}}{\text{725 mL}\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)}=0.492\text{ M}$

Thus, the concentration of ${\text{C}}_{\text{16}}{\text{H}}_{\text{18}}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{S}$ is 0.492 M.

The concentration all the species can be calculated using the ICE table as follows:

$\begin{array}{l}{\text{ C}}_{\text{16}}{\text{H}}_{\text{18}}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{S}\rightleftharpoons {\text{C}}_{\text{16}}{\text{H}}_{\text{17}}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}{\text{S}}^{-}+{\text{H}}^{+}\\ I\text{ 0}\text{.492 - -}\\ C-xxx\\ E0.492-xxx\end{array}$

The expression for $K_a$ is as follows:

$K_a=\frac{\left[{\text{C}}_4{\text{H}}_3{\text{N}}_{\text{2}}{\text{O}}_3{}^{-}\right]\left[{\text{H}}^{+}\right]}{\left[{\text{HC}}_4{\text{H}}_3{\text{N}}_{\text{2}}{\text{O}}_3\right]}$

Putting the values,

$1.7\times {10}^{-3}=\frac{\left(x\right)\left(x\right)}{\left(0.492-x\right)}$

On rearranging,

$\begin{array}{l}1.7\times {10}^{-3}\left(0.492\right)-1.7\times {10}^{-3}x=x^2\\ 8.364\times {10}^{-4}-1.7\times {10}^{-3}x=x^2\end{array}$

Or,

$x^2+1.7\times {10}^{-3}x-8.364\times {10}^{-4}=0$

Comparing this with the general quadratic equation as follows:

$a{x}^2+bx-c=0$

Solving the quadratic equation,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Putting the values,

$\begin{array}{l}x=\frac{-1.7\times {10}^{-3}\pm \sqrt{{\left(1.7\times {10}^{-3}\right)}^2-4\left(1\right)\left(-8.364\times {10}^{-4}\right)}}{2\left(1\right)}\\ =\frac{-1.7\times {10}^{-3}\pm \sqrt{2.89\times {10}^{-6}+3.3456\times {10}^{-3}}}{2\left(1\right)}\\ =\frac{-1.7\times {10}^{-3}\pm 0.0578}{2}\end{array}$

$x=0.02805,-0.03$

Since, the value of x cannot be negative thus, the value of x will be 0.02805.

From the ICE table, it is equal to the concentration of hydrogen ion thus,

$\left[{\text{H}}^{+}\right]=0.02805$

Therefore, the concentration of hydrogen ion is 0.02805 M.