Chapter 15, Problem 15.54QA

Interpretation Introduction

To find:

The $K_a$ value of the acid $HN{O}_3$

Answer to Problem 15.54QA

Solution:

The $K_a$ value for $7.5M$ concentration aqueous solution of nitric acid is $1.2$.

Explanation of Solution

1) Concept:

We are asked to determine the $K_a$ value for the nitric acid from known initial concentration of the acid and percent ionization. The value for the ${\left[N{O}_3^{-}\right]}_{equilibrium}$ can be calculated using the percent dissociation formula. At equilibrium,$\left[H_3{O}^{+}\right]=\left[{NO}_3^{-}\right]$, and the equilibrium concentration of the acid is calculated from the initial value minus the equilibrium concentration of $N{O}_3^{-}$ ions.

The acid ionization reaction for $HN{O}_3$ is written as

$HN{O}_3\left(aq\right)+H_{2}O \left(l\right)\rightleftharpoons N{O}_3^{-}\left(aq\right)+H_3{O}^{+}\left(aq\right)$

The dissociation constant expression for the acid is written as

$K_a=\frac{\left[H_3{O}^{+}\right]\left[{NO}_3^{-}\right]}{\left[HN{O}_3\right]}$

2) Formula:

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

ii) $K_a=\frac{\left[H_3{O}^{+}\right]\left[{NO}_3^{-}\right]}{\left[HN{O}_3\right]}$

3) Given:

i) A $50\%$ solution of $7.5M$ nitric acid; only $33\%$ of the molecules dissociate into $H^{+}$ and $N{O}_3^{-}$.

ii) Mass percent of solution is given as $50\%$

iii) Ionization is given as $33\%$

iv) Initial concentration of the nitric acid is given as $7.5 M$

4) Calculations:

From the given percent dissociation, we can find the ${\left[N{O}_3^{-}\right]}_{equilibrium}$ using the percent dissociation formula as:

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

$33\%=\frac{{\left[{NO}_3^{-}\right]}_{equilibrium}}{7.5 M} \times 100$

$0.33=\frac{{\left[{NO}_3^{-}\right]}_{equilibrium}}{7.5 M}$

${\left[{NO}_3^{-}\right]}_{equilibrium}=0.33 \times 7.5 M=2.475M$

The equilibrium concentration of $N{O}_3^{-}$ is calculated as $2.475 M.$

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

Reaction	$HN{O}_3\left(aq\right)+H_{2}O \left(l\right)\rightleftharpoons N{O}_3^{-}\left(aq\right)+H_3{O}^{+}\left(aq\right)$ $\left[HN{O}_3\right] \left(M\right)$ $\left[N{O}_3^{-}\right]\left(M\right)$ $\left[H_3{O}^{+}\right] \left(M\right)$
Initial (I)	$7.5$	$0.00$	$0.00$
Change (C)	$-x$	$+x$	$+x$
Equilibrium (E)	$\left(7.5-x\right)$	$x$	$x$

At equilibrium, $\left[H_3{O}^{+}\right]=\left[{NO}_3^{-}\right]=x=2.475$

The equilibrium concentration for the acid is

${\left[HN{O}_3\right]}_{equilibrium}={\left[HN{O}_3\right]}_{initial}-{\left[H_3{O}^{+}\right]}_{equilibrium}$

${\left[HN{O}_3\right]}_{equilibrium}=\left(7.5-2.475\right)M= 5.025M$

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

$K_a=\frac{\left[H_3{O}^{+}\right]\left[{NO}_3^{-}\right]}{\left[HN{O}_3\right]}$

$K_a=\frac{\left(2.475\right)\left(2.475\right)}{5.025 }=1.219$

Thus, the acid dissociation constant value for the strong acid $HN{O}_3$ is 1.2.

Conclusion:

The Ka value is calculated from the initial concentration and percent dissociation by using RICE table.