Q18.65P - Chemistry: Molecular Nature Of Matter And Change

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 ${\mathbf{\text{[H}}}_{\mathbf{\text{3}}}{\mathbf{\text{O}}}^{\mathbf{\text{+}}}{\mathbf{\text{], [NO}}}_{\mathbf{\text{2}}}^{\mathbf{\text{-}}}\mathbf{\text{],}}$ and in ${\mathbf{\text{0.60 M HNO}}}_{\mathbf{\text{2}}}$?

Step-by-Step Solution

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 –

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

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

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

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 –

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

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	$\text{0.60 M}$	-	$\text{0}$	$\text{0}$
Change	$\text{-x}$	-	$\text{+x}$	$+x$
Equilibrium	$\text{0.60 M-x}$	-	$x$	$x$

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 $\text{0.60 M}$ 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.