Question

Sodium azide \(\left(\mathrm{NaN}_{3}\right)\) is sometimes added to water to kill bacteria. Calculate the concentration of all species in a \(0.010-M\) solution of \(\mathrm{NaN}_{3} .\) The \(K_{\mathrm{a}}\) value for hydrazoic acid (HN_) is \(1.9 \times 10^{-5} .\)

Short answer

In a \(0.010-M\) solution of sodium azide \(\left(\mathrm{NaN}_{3}\right)\), the equilibrium concentrations of the species are as follows: \([HN] \approx 1.37 \times 10^{-3} M\), \([OH^-] \approx 1.37 \times 10^{-3} M\), and \([N_{3}^-] \approx 0.00863 M\).

Step-by-step solution

Write the balanced chemical equation

The dissociation reaction of the azide ion can be written as follows (ignoring the sodium because it is a spectator ion): \[ \mathrm{N}_{3}^- (aq) + H2O(l) \rightleftharpoons HN(aq) + OH^- (aq) \]

Write the expression for the acidity constant

Since the balanced chemical equation gives us the stoichiometric coefficients, we can write the expression for the acidity constant (\(K_{a}\)) of hydrazoic acid: \[ K_{a} = \frac{[HN][OH^-]}{[N_{3}^-]} \]

Create the initial and final concentration table – an ICE table

Now, we will create an ICE (Initial, Change, Equilibrium) table to represent the initial concentrations, the change during the reaction and the final equilibrium concentrations: | | N\(_{3}^-\) | HN | OH\(^-\) | |-----------|-----------|----|--------| | Initial | 0.010-M | 0 | 0 | | Change | -x | x | x | | Equilibrium | 0.010-M - x | x | x |

Substitute equilibrium concentrations in the \(K_{a}\) expression

Now, we will substitute our equilibrium values from the ICE table into the \(K_{a}\) expression: \[ 1.9 \times 10^{-5} = \frac{x^2}{0.010 - x} \]

Simplify and solve for x

Because \(1.9 \times 10^{-5}\) is small, we can approximate that \(x<<0.010\) and thus \(0.010-x \approx 0.010\). This allows us to simplify the equation and solve for x directly: \[ 1.9 \times 10^{-5} \approx \frac{x^2}{0.010} \] Divide both sides by \(0.010\) and take the square root to find x: \[ x \approx \sqrt{1.9 \times 10^{-5} \times 0.010} \approx 1.37 \times 10^{-3} \]

Calculate equilibrium concentrations

Now we can find the equilibrium concentrations of all species in the solution: \[ [HN] = x = 1.37 \times 10^{-3} M \] \[ [OH^-] = x = 1.37 \times 10^{-3} M \] \[ [N_{3}^-] = 0.010 - x \approx 0.010 - 1.37 \times 10^{-3} = 0.00863 M \] Thus, the concentrations of all species in the \(0.010-M\) solution of sodium azide are as follows: \([HN] \approx 1.37 \times 10^{-3} M\) \([OH^-] \approx 1.37 \times 10^{-3} M\) \([N_{3}^-] \approx 0.00863 M\)

Key concepts

ICE Table

An ICE table is a crucial tool in chemistry to visualize how reactant and product concentrations change over time, particularly during equilibrium reactions. ICE stands for Initial, Change, and Equilibrium. This approach helps organize data and simplifies solving for unknowns.
Initially, list what you know: the starting concentration of the reactants and products. In the sodium azide example, the initial concentration of \(N_3^-\) is given as 0.010 M, while \(HN\) and \(OH^-\) start at 0 M.
Next, identify the changes, usually represented by \'x\', that occur as the reaction progresses towards equilibrium. In equilibrium conditions, these changes result in final concentrations, which are the sum of the initial concentration and the change. Use the balanced chemical equation to determine these changes.

• Initial: concentrations before any reaction.
• Change: modifications as the system moves to equilibrium.
• Equilibrium: concentrations when the reaction stabilizes.

This organized format allows you to neatly solve for unknowns using the given equilibrium expressions.

Dissociation Reaction

A dissociation reaction describes how a compound separates into simpler products when dissolved. It involves breaking chemical bonds and results in ions in a solution.
For sodium azide, the dissociation focuses on the azide ion \(N_3^-\) reacting with water to form hydrazoic acid (HN) and hydroxide ions (OH^-). This step excludes the sodium ion because it doesn't actively participate in the equilibrium.
It's key to note that dissociation reactions are often reversible, represented by the double arrow \(\rightleftharpoons\). These reactions reach a state of balance where the rate of forward reaction equals the rate of the reverse reaction. Understanding dissociation will also guide us in understanding acidity constants and calculating equilibrium concentrations, putting everything in perspective.

Acidity Constant

The acidity constant (\(K_a\)) is a vital measure indicating the strength of an acid in solution. It quantifies the extent to which an acid dissociates into its ions in an aqueous solution.
In our example, the hydrazoic acid forms through the dissociation of \(N_3^-\), with its acidity constant \(K_a = 1.9 \times 10^{-5}\). This small number shows it is a weak acid—it doesn't fully dissociate in water.
The expression for \(K_a\) is derived directly from the balanced chemical equation and helps to calculate equilibrium concentrations:
\[ K_a = \frac{[HN][OH^-]}{[N_3^-]} \]
This expression represents the ratio of the product concentrations to the reactant concentration. Higher \(K_a\) values imply stronger acids as they dissociate more extensively. For weak acids, the assumption that the change \'x\' is small simplifies calculations.

Equilibrium Concentration

The equilibrium concentration is the amount of each species present in a reaction mixture once the system reaches equilibrium.
Using the ICE table and the \(K_a\) expression, we solved for \'x\', representing the change in concentration. This helped us calculate equilibrium amounts of each ion.

• For \(HN\) and \(OH^-\), \[ [HN] = [OH^-] = x = 1.37 \times 10^{-3} \, M \]
• For \(N_3^-\), \[ [N_3^-] = 0.010 - x = 0.00863 \, M \]

These values reflect how much of each component is present when the reaction has settled. Knowing equilibrium concentrations allows chemists to predict the behavior of solutions and make informed decisions in various applications like titration or pH adjustment. It's a foundational aspect of understanding chemical dynamics.