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Chapter 13: Problem 121

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 \(1.9 \times 10^{-5}\)

Short Answer

Expert verified
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 by step solution

01

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-
02

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.
03

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
04

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: \(x = 1.37 \times 10^{-3}\)
05

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

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