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

NaN3 is a salt containing the azide ion (N3-). When it dissolves in water, it breaks down into sodium ions (Na+) and azide ions (N3-). Additionally, the azide ions can react with water to form hydrazoic acid (HN3) and hydroxide ions (OH-). So, the species present in the solution will be Na+, N3-, HN3, and OH-.

The dissociation equation for azide ions (N3-) reacting with water (H2O) can be written as: \[N^-_ 3 (aq) + H_2 O (l) \rightleftharpoons HN_3 (aq) + OH^- (aq)\] The corresponding equilibrium constant expression can be written as: \(K_b = \frac{[HN_3][OH^-]}{[N_3^-]}\) Since we know the Ka for hydrazoic acid, we can determine the Kb for azide ions using the relationship: \(K_a \times K_b = K_w\) where \(K_w = 1.0 \times 10^{-14}\), the ion product of water. Thus, the Kb for azide ions can be calculated as: \(K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.9 \times 10^{-5}}\)

Before the reaction, we have the following initial concentrations: [N3-] = 0.010 M (from the given NaN3 concentration) [HN3] = 0 M [OH-] = 0 M Let x be the change in concentration of the species as the reaction proceeds: [N3-] = 0.010 - x [HN3] = x [OH-] = x Now, substitute these concentrations into the Kb expression: \(K_b = \frac{x^2}{0.010 - x}\) Solve for x: \(x^2 = K_b (0.010 - x)\) When the value for Kb is small, x will also be small. Hence, we can omit x in the denominator: \(x^2 = K_b \times 0.010\) Now, solve for x: \(x = \sqrt{K_b \times 0.010}\) \(x = \sqrt{\frac{1.0 \times 10^{-14}}{1.9 \times 10^{-5}} \times 0.010}\) Now that we've found x, we can calculate the final concentrations of all species: [OH-] = x (as x represents the concentration of OH- ions) [HN3] = x (since x also represents the concentration of HN3) [N3-] = 0.010 - x [Na+] = 0.010 M (it remains the same as the initial concentration since it does not participate in the reaction) These are the concentrations of all the species in the 0.010 M solution of NaN3.