Question

Calculate the percent ionization of nitrous acid in a solution that is 0.139 M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50 ⋅ 10-4.

Answer 1

The percentage dissociation of nitrous acid is $\boxed{{\text{5}}{\text{.85\%}}}$ .

Further explanation:

Chemical equilibrium is the state in which concentration of reactants and products become constant and do not change with time. This is because the rate of forward and backward direction becomes equal. The general equilibrium reaction is as follows:

${\text{A(g)}}+{\text{B(g)}}\rightleftharpoons{\text{C(g)}}+{\text{D(g)}}$

Equilibrium constant is the constant that relates the concentration of product and reactant at equilibrium. The formula to calculate the equilibrium constant for general reaction is as follows:

${\text{K}}=\frac{{\left[{\text{D}}\right]\left[{\text{C}}\right]}}{{\left[{\text{A}}\right]\left[ {\text{B}}\right]}}$

Here, K is the equilibrium constant.

The equilibrium constant for the dissociation of acid is known as ${{\text{K}}_{\text{a}}}$  and equilibrium constant for the dissociation of base is known as ${{\text{K}}_{\text{b}}}$ .

Percentage ionization:

When weak acid or base dissolves into the water, it partially dissociates into its ions. The amount of weak acid exists as ions in the solution is known as percentage ionization.

The formula for percentage ionization for a weak acid is,

$\% {\text{ ionization}}=\frac{{\left[ {{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}}\right]}}{{\left[{{\text{HA}}}\right]}}\times100$

Here, $\left[{{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}}\right]$  is the concentration of hydronium ion and $\left[{{\text{HA}}}\right]$  is the concentration of acid.

The nitrous acid is a weak acid that dissociates in water to form ${\text{NO}}_2^ -$  and hydronium ion.

${\text{HN}}{{\text{O}}_{\text{2}}}\left({aq}\right)+{{\text{H}}_2}{\text{O}}\left(l\right)\rightleftharpoons{\text{NO}}_2^-\left({aq}\right)+{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}\left({aq}\right)$

The expression of ${{\text{K}}_{\text{a}}}$ for the above reaction is as follows:  

[tex{{\text{K}}_{\text{a}}}=\frac{{\left[{{\text{NO}}_2^-}\right]\left[{{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}}\right]}}{{\left[{{\text{HN}}{{\text{O}}_{\text{2}}}}\right]}}[/tex]                   …... (1)

For nitrous acid, percentage dissociation can be calculated as,

$\% {\text{ ionization}}=\frac{{\left[{{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}}\right]}}{{\left[{{\text{HN}}{{\text{O}}_2}}\right]}}\times100$                    …… (2)

The value of ${{\text{K}}_{\text{a}}}$ is ${\text{4}}{\text{.50}}\times{\text{1}}{{\text{0}}^{-4}}$ .

The initial concentration of nitrous acid is 0.139 M.

Let the change in concentration at equilibrium is x. Therefore, the concentration of ${\text{HN}}{{\text{O}}_2}$  becomes 0139-x at equilibrium. The concentration of ${\text{NO}}_2^-$ and ${{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}$  ion becomes x at equilibrium.

Substitute x for $\left[{{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}}\right]$, x for $\left[{{\text{NO}}_2^-}\right]$  and 0.139-x for ${\text{HN}}{{\text{O}}_2}$ in equation (1).

${{\text{K}}_{\text{a}}}=\frac{{{\text{x}}\times{\text{x}}}}{{0.139-{\text{x}}}}$                    …… (3)

Rearrange the equation (3) and substitute ${\text{4}}{\text{.50}}\times{\text{1}}{{\text{0}}^{-4}}$ for ${{\text{K}}_{\text{a}}}$ to calculate value of x.

${{\text{x}}^2}=\left({{\text{4}}{\text{.50}}\times{\text{1}}{{\text{0}}^{-4}}}\right)\left({{\text{0}}{\text{.139}}-{\text{x}}}\right)$

The final quadric equation is,

${{\text{x}}^2}-\left({{\text{4}}{\text{.50}}\times{\text{1}}{{\text{0}}^{-4}}}\right){\text{x}}-6.255\times{10^{-5}}=0$

After solving the quadratic equation the value of x obtained is,

${\mathbf{x=0}}{\mathbf{.008137}}$

The concentration of ${{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}$ is equal to x and therefore the concentration of ${{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}$ is 0.008137 M.

Substitute 0.008137 M for $\left[{{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}}\right]$  and 0.139 M for $\left[{{\text{HN}}{{\text{O}}_2}}\right]$  in equation (2).

$\begin{aligned}\%{\text{ ionization}}&=\frac{{0.008137{\text{ M}}}}{{0.139{\text{ M}}}}\times100\\&={\mathbf{5}}{\mathbf{.85\%}}\\\end{aligned}$

Learn more:

1. Calculation of equilibrium constant of pure water at 25°c: brainly.com/question/3467841

2. Complete equation for the dissociation of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ (aq): brainly.com/question/5425813

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Chemical equilibrium

Keywords: pH, 0.139 M, solution of nitrous acid, hno2, pOH, equilibrium, percentage dissociation, nitrous acid, 0.008137, and 5.85%.

Answer 2

Ok first, we have to create a balanced equation for the dissolution of nitrous acid.

HNO2 <-> H(+) + NO2(-)

Next, create an ICE table

           HNO2   <-->  H+        NO2-
[]i        0.139M          0M       0M
Δ[]      -x                   +x         +x
[]f        0.139-x          x           x

Then, using the concentration equation, you get

4.5x10^-4 = [H+][NO2-]/[HNO2]

4.5x10^-4 = x*x / .139 - x

However, because the Ka value for nitrous acid is lower than 10^-3, we can assume the amount it dissociates is negligable, 

assume 0.139-x ≈ 0.139

4.5x10^-4 = x^2/0.139

Then, we solve for x by first multiplying both sides by 0.139 and then taking the square root of both sides.

We get the final concentrations of [H+] and [NO2-] to be x, which equals 0.007M.

Then to find percent dissociation, you do final concentration/initial concentration.

0.007M/0.139M = .0503 or 

≈5.03% dissociation.