Chemistry, 05.10.2019 18:30, sportygirlscand

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.

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answered: soonerlady19

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).