Question

Nitrous acid,${\mathbf{\text{HNO}}}_{\mathbf{\text{2}}}$, has a${\mathbf{\text{K}}}_{\mathbf{\text{a}}}$of${\mathbf{\text{7.1×10}}}^{\mathbf{\text{-4}}}$. What are${\mathbf{\text{[H}}}_{\mathbf{\text{3}}}{\mathbf{\text{O}}}^{\mathbf{\text{+}}}{\mathbf{\text{],[NO}}}_{\mathbf{\text{2}}}^{\mathbf{\text{-}}}\mathbf{\text{],}}$andin${\mathbf{\text{0.60MHNO}}}_{\mathbf{\text{2}}}$?

Short answer

The values for${\text{K}}_{\text{a}}$${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{],[NO}}_{\text{2}}^{\text{-}}\text{]}$ and${\text{[OH}}^{\text{-}}\text{]}$ are –

${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{]=2.1×10}}^{\text{-2}}$

${\text{[NO}}_{\text{2}}^{\text{-}}{\text{]=2.1×10}}^{\text{-2}}$

${\text{[OH}}^{\text{-}}{\text{]=4.8×10}}^{\text{-13}}$

Step-by-step solution

Concept Introduction

When a chemical process reaches equilibrium, the equilibrium constant (typically indicated by the symbol) offers information on the relationship between the products and reactants.

Calculation for the Equation

$\text{0}$The information provided is –

$\begin{array}{c}{\text{K}}_{\text{a}}{\text{=7.1×10}}^{\text{-4}}\\ {\text{[HNO}}_{\text{2}}\text{]=0.60M}\end{array}$

The reaction for the dissociation of${\text{HNO}}_{\text{2}}$is –

${\text{HNO}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{+NO}}_{\text{2}}^{\text{-}}$

Construct the ICE table to obtain the equation for ${\text{K}}_{\text{a}}$.

	${\text{HNO}}_2$	${\mathrm{H}}_2\mathrm{O}$	${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$	${\text{NO}}_{\text{2}}^{\text{-}}$
Initial	$\text{0.60M}$	-	$\text{0}$	role="math" localid="1657281398175" $\text{0}$
Change	$\text{-x}$	-	$\text{+x}$	$+x$
Equilibrium	$\text{0.60M-x}$	-	$x$	$x$

Write the expression for the equilibrium constant of the reaction in terms of concentration –

$\begin{array}{l}{\text{K}}_a\text{=}\frac{\text{Products}}{\text{Reactants}}\\ {\text{K}}_a\text{=}\frac{{\text{[NO}}_{\text{2}}^{\text{-}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{{\text{[HNO}}_{\text{2}}\text{]}}\end{array}$

Substitute the equilibrium equations from the reaction table to solve for –

$\begin{array}{l}{\text{K}}_{\text{a}}\text{=}\frac{{\text{[NO}}_{\text{2}}^{\text{-}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{{\text{[HNO}}_{\text{2}}\text{]}}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{x}}^{{}_{\text{2}}}}{\text{0.60-x}}\end{array}$

Substitute the equilibrium equations from the reaction table to solve for –

$\begin{array}{l}{\text{K}}_{\text{a}}\text{=}\frac{{\text{[NO}}_{\text{2}}^{\text{-}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{{\text{[HNO}}_{\text{2}}\text{]}}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{x}}^{{}_{\text{2}}}}{\text{0.60-x}}\end{array}$

Calculation for the Equilibrium Constant Concentrations

It is known that that ${\text{x=[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{]=[NO}}_{\text{2}}^{\text{-}}\text{]}$. Since ${\text{HNO}}_{\text{2}}$is a weak acid, it’s ${\text{K}}_{\text{a}}$must be very small. So, assume that the $x$has no effect on $\text{0.60M}$in the denominator. Then substitute the ${\text{K}}_{\text{a}}$to solve for $x$.

$\begin{array}{c}{\text{K}}_{\text{a}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0.60}}\\ {\text{x}}^{\text{2}}\text{=}\left({\text{K}}_{\text{a}}\right)\text{(0.60)}\\ \text{x=}\sqrt{\left({\text{K}}_{\text{a}}\right)\text{(0.60)}}\\ \text{=}\sqrt{\left({\text{7.1×10}}^{\text{-4}}\right)\text{(0.60)}}\\ \text{x=0.021}\\ {\text{=2.1×10}}^{\text{-2}}\end{array}$

Since${\text{x=[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{]=[NO}}_{\text{2}}^{\text{-}}\text{]}$then ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{]=[NO}}_{\text{2}}^{\text{-}}{\text{]=2.1×10}}^{\text{-2}}$.

Now, solve for ${\text{[OH}}^{\text{-}}\text{]}$using ${\text{K}}_{\text{w}}{\text{=1.0×10}}^{\text{-14}}$.

$\begin{array}{c}{\text{K}}_{\text{w}}\text{=}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]\\ \left[{\text{OH}}^{\text{-}}\right]\text{=}\frac{{\text{K}}_{\text{w}}}{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}\\ \text{=}\frac{{\text{1.0×10}}^{\text{-14}}}{{\text{2.1×10}}^{\text{-2}}}\\ \left[{\text{OH}}^{\text{-}}\right]{\text{=4.8×10}}^{\text{-13}}\end{array}$

Therefore, the values of concentrations of${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{],[NO}}_{\text{2}}^{\text{-}}\text{]}$ and${\text{[OH}}^{\text{-}}\text{]}$ are${\text{2.1×10}}^{\text{-2}}{\text{,2.1×10}}^{\text{-2}}$ and${\text{4.8×10}}^{\text{-13}}$ respectively.