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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Answer from: 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).

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

Learn more:

1. Calculation of equilibrium constant of pure water at 25°c:

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

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 from: Quest

explanation:

the largest database of organic compounds lists about 10 million substances, which include compounds originating from living organisms and those synthesized by chemists. the number of potential organic compounds has been estimated at 1060—an astronomically high number. the existence of so many organic molecules is a consequence of the ability of carbon atoms to form up to four strong bonds to other carbon atoms, resulting in chains and rings of many different sizes, shapes, and complexities. the simplest organic compounds contain only the elements carbon and hydrogen, and are called hydrocarbons. even though they are composed of only two types of atoms, there is a wide variety of hydrocarbons because they may consist of varying lengths of chains, branched chains, and rings of carbon atoms, or combinations of these structures. in addition, hydrocarbons may differ in the types of carbon-carbon bonds present in their molecules. many hydrocarbons are found in plants, animals, and their fossils; other hydrocarbons have been prepared in the laboratory. we use hydrocarbons every day, mainly as fuels, such as natural gas, acetylene, propane, butane, and the principal components of gasoline, diesel fuel, and heating oil. the familiar plastics polyethylene, polypropylene, and polystyrene are also hydrocarbons. we can distinguish several types of hydrocarbons by differences in the bonding between carbon atoms. this leads to differences in geometries and in the hybridization of the carbon orbitals.

Answer from: Quest

It’s the energy source of stars

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Calculate the percent ionization of nitrous acid in a solution that is 0.139 m in nitrous acid. the...

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