Problem 121 - Chemistry An Atoms First Approach

Question

Sodium azide (NaN, ) 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 \(\left(\mathrm{HN}_{3}\right)\) is \(1.9 \times 10^{-5}\)

Step-by-Step Solution

Answer

The concentrations of all species in the 0.010 M solution of sodium azide are: [Na+] = 0.010 M, [N3-] = 8.63 x 10^-3 M, [HN3] = 1.37 x 10^-3 M, and [OH-] = 1.37 x 10^-3 M.

Step 1: Write the dissociation reactions for sodium azide (NaN3) and hydrazoic acid (HN3)

When sodium azide dissolves in water, it dissociates into sodium ions (Na+) and azide ions (N3-). Hydrazoic acid (HN3) can then form through the reaction between azide ions and water: 1) NaN3 -> Na+ + N3- 2) N3- + H2O <-> HN3 + OH-

Step 2: Determine the initial concentration of each species

We know that the initial concentration of sodium azide is 0.010 M. When it dissociates completely, it will produce an equal concentration of sodium ions and azide ions: [Na+] = [N3-] = 0.010 M [HN3] and [OH-] are initially 0 M.

Step 3: Write the Ka expression for hydrazoic acid and its related equilibrium expression

The Ka expression for hydrazoic acid (HN3) is given as follows, and we are given the value of Ka as 1.9 x 10^-5: Ka = \(\frac{[HN_3][OH^-]}{[N_3^-]}\) Since we are considering changes in equilibrium concentrations, let x represent the change in [HN3] and [OH-]: Increase in [HN3] and [OH-] = x Decrease in [N3-] = x Equilibrium concentrations: [HN3] = x [OH-] = x [N3-] = 0.010 - x

Step 4: Solve for x using Ka expression

Substitute the equilibrium concentrations into the Ka expression: \(1.9 \times 10^{-5} = \frac{x \times x}{0.010 - x}\) Since Ka is very small, we can assume that x is negligible compared to 0.010: \(1.9 \times 10^{-5} = \frac{x^2}{0.012}\) Solve for x: \(x = 1.37 \times 10^{-3}\)

Step 5: Find the equilibrium concentrations of all species

Now that we have the value of x, we can find the equilibrium concentrations of each species: [HN3] = x = 1.37 x 10^-3 M [OH-] = x = 1.37 x 10^-3 M [N3-] = 0.010 - x = 0.010 - 1.37 x 10^-3 = 8.63 x 10^-3 M [Na+] = 0.010 M (since it doesn't participate in the acid-base reaction) The concentrations of all species in the 0.010 M solution of sodium azide are: [Na+] = 0.010 M [N3-] = 8.63 x 10^-3 M [HN3] = 1.37 x 10^-3 M [OH-] = 1.37 x 10^-3 M

Key Concepts

Acid-Base Reactions

Acid-base reactions are interactions where an acid donates a proton (H\(^+\)) to a base. In the given exercise, the focus is on the reaction between azide ions \(\text{N}_3^−\) and water, which results in the formation of hydrazoic acid \(\text{HN}_3\) and hydroxide ions \(\text{OH}^-\). This is an example of how bases can accept protons from water:

• The azide ion acts as a base and accepts a hydrogen ion from water.
• The water acts as an acid by donating a hydroxide ion as it releases a hydrogen ion to form \(\text{HN}_3\).

These interactions and the concept of acids and bases exchanging hydrogen ions are central to understanding and predicting the outcome of the reactions, especially in determining the concentrations of the reaction species involved.

Dissociation in Water

Dissociation is the process by which an ionic compound separates into its ions in solution. Sodium azide (NaN\(_3\)) dissociates completely when it is dissolved in water:

• Sodium azide dissociates into sodium ions (Na\(^+\)) and azide ions (N\(_3^−\)).

This is represented by the chemical equation:\[\text{NaN}_3 \rightarrow \text{Na}^+ + \text{N}_3^-\]After dissociation, the azide ion can further engage in reactions. In a balanced and well-mixed solution, the ions are evenly distributed within the solvent, allowing secondary reactions like the formation of hydrazoic acid \(\text{HN}_3\) to occur. The complete dissociation is key to calculating the initial concentrations of ions before any acid-base reaction takes place.

Equilibrium Concentrations

Equilibrium in a chemical reaction is the state where the concentrations of reactants and products remain constant over time. For the dissociation of hydrazoic acid in water, we need to find the concentrations of various species at equilibrium. Given the weak acid \(\text{HN}_3\), which does not dissociate completely in water, we calculate the concentrations using changes in initial and equilibrium states:

• [N\(_3^−\)] starts at 0.010 M, but decreases by x as it forms \(\text{HN}_3\).
• [HN\(_3\)] and [OH\(^-\)] increase by x, as they are products of the reaction.

Therefore, the equilibrium concentrations are:

• [HN\(_3\)] = x = 1.37 x 10\(^{-3}\) M
• [OH\(^-\)] = x = 1.37 x 10\(^{-3}\) M
• [N\(_3^−\)] = 0.010 - x = 8.63 x 10\(^{-3}\) M
These concentrations help to further understand the dynamics and extent of the reaction within the solution.

Ka Expression

The acid dissociation constant \(K_a\) provides insight into the strength of a weak acid in solution. It specifically measures the extent of dissociation of an acid into its ionic components. The formula for \(K_a\) is given by:\[K_a = \frac{[\text{HN}_3][\text{OH}^-]}{[\text{N}_3^-]}\]Here, \(K_a\) is used to determine the change in concentrations at equilibrium. In practice:

• The given \(K_a\) value helps us set up an equilibrium expression based on the change "\(x\)" which represents the concentration of products formed.
• Since \(K_a\) is quite small (1.9 x 10\(^{-5}\)), we assume \(x\) is negligible compared to 0.010, simplifying calculations.

Understanding \(K_a\) and its application in calculating equilibrium concentrations is crucial for predicting how much reactants remain un-dissociated and how much product is formed in a given solution.