Chapter 13, Problem 51QAP

Barbituric acid, HC₄H₃N₂O₃, is used to prepare barbiturates, a class of drugs used as sedatives. Its K_a is $9.8\times {10}^{-5}$. Calculate [H⁺] in solutions prepared by adding enough water to the following to make 1.45 L.

(a) 0.344 mol

(b) 28.9 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 51QAP

$4.84\times {10}^{-3}\text{ M}$

Explanation of Solution

The acid dissociation constant of barbituric acid is $9.8\times {10}^{-5}$. The volume of solution is 1.45 L and number of moles of the acid is 0.344 mol.

The molarity of ${\text{HC}}_4{\text{H}}_3{\text{N}}_{\text{2}}{\text{O}}_3$ can be calculated as follows:

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

Putting the values,

$M=\frac{0.344\text{ mol}}{1.45\text{ L}}=0.24\text{ M}$

Thus, the concentration of ${\text{HC}}_4{\text{H}}_3{\text{N}}_{\text{2}}{\text{O}}_3$ is 0.24 M.

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

$\begin{array}{l}{\text{ HC}}_4{\text{H}}_3{\text{N}}_{\text{2}}{\text{O}}_3\rightleftharpoons {\text{C}}_4{\text{H}}_3{\text{N}}_{\text{2}}{\text{O}}_3{}^{-}+{\text{H}}^{+}\\ I\text{ 0}\text{.24 - -}\\ C-xxx\\ E0.24-xxx\end{array}$

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,

$9.8\times {10}^{-5}=\frac{\left(x\right)\left(x\right)}{\left(0.24-x\right)}$

On rearranging,

$\begin{array}{l}9.8\times {10}^{-5}\left(0.24\right)-9.8\times {10}^{-5}x=x^2\\ 2.352\times {10}^{-5}-9.8\times {10}^{-5}x=x^2\end{array}$

Or,

$x^2+9.8\times {10}^{-5}x-2.352\times {10}^{-5}=0$

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,

$\begin{array}{l}x=\frac{-9.8\times {10}^{-5}\pm \sqrt{{\left(9.8\times {10}^{-5}\right)}^2-4\left(1\right)\left(-2.352\times {10}^{-5}\right)}}{2\left(1\right)}\\ =\frac{-1.2\times {10}^{-5}\pm \sqrt{9.604\times {10}^{-9}+9.41\times {10}^{-5}}}{2\left(1\right)}\\ =\frac{-1.2\times {10}^{-5}\pm 9.7\times {10}^{-3}}{2}\end{array}$

$x=4.84\times {10}^{-3},-4.856\times {10}^{-3}$

Since, the value of x cannot be negative thus, the value of x will be $4.84\times {10}^{-3}$.

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

$\left[{\text{H}}^{+}\right]=4.84\times {10}^{-3}$

Therefore, the concentration of hydrogen ion is $4.84\times {10}^{-3}\text{ M}$.

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 51QAP

$3.86\times {10}^{-3}\text{ M}$

Explanation of Solution

The mass of barbituric acid is given as 28.9 g. The molecular formula of the barbituric acid is ${\text{HC}}_4{\text{H}}_3{\text{N}}_{\text{2}}{\text{O}}_3$ thus, its molar mass is 128.1 g/mol.

Now, from mass and molar mass of the ${\text{HC}}_4{\text{H}}_3{\text{N}}_{\text{2}}{\text{O}}_3$, its number of moles can be calculated as follows:

$n=\frac{m}{M}=\frac{28.9\text{ g}}{128.1\text{ g/mol}}=0.2256\text{ mol}$

Now, the molarity of ${\text{HC}}_4{\text{H}}_3{\text{N}}_{\text{2}}{\text{O}}_3$ can be calculated as follows:

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

Putting the values,

$M=\frac{0.2256\text{ mol}}{1.45\text{ L}}=0.156\text{ M}$

Thus, the concentration of ${\text{HC}}_4{\text{H}}_3{\text{N}}_{\text{2}}{\text{O}}_3$ is 0.156 M.

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

$\begin{array}{l}{\text{ HC}}_4{\text{H}}_3{\text{N}}_{\text{2}}{\text{O}}_3\rightleftharpoons {\text{C}}_4{\text{H}}_3{\text{N}}_{\text{2}}{\text{O}}_3{}^{-}+{\text{H}}^{+}\\ I\text{ 0}\text{.156 - -}\\ C-xxx\\ E0.156-xxx\end{array}$

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,

$K_a=\frac{\left(x\right)\left(x\right)}{\left(0.156-x\right)}=9.8\times {10}^{-5}$

On rearranging,

$\begin{array}{l}9.8\times {10}^{-5}\left(0.156\right)-9.8\times {10}^{-5}x=x^2\\ 1.53\times {10}^{-5}-9.8\times {10}^{-5}x=x^2\end{array}$

Or,

$x^2+9.8\times {10}^{-5}x-1.53\times {10}^{-5}=0$

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,

$\begin{array}{l}x=\frac{-9.8\times {10}^{-5}\pm \sqrt{{\left(9.8\times {10}^{-5}\right)}^2-4\left(1\right)\left(-1.53\times {10}^{-5}\right)}}{2\left(1\right)}\\ =\frac{-9.8\times {10}^{-5}\pm \sqrt{9.604\times {10}^{-9}+6.12\times {10}^{-5}}}{2\left(1\right)}\\ =\frac{-9.8\times {10}^{-5}\pm 7.82\times {10}^{-3}}{2}\end{array}$

$x=3.86\times {10}^{-3},-3.96\times {10}^{-3}$

Since, the value of x cannot be negative thus, the value of x will be $3.86\times {10}^{-3}$.

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

$\left[{\text{H}}^{+}\right]=3.86\times {10}^{-3}$

Therefore, the concentration of hydrogen ion is $3.86\times {10}^{-3}\text{ M}$.