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Chapter 14: Problem 121

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

Expert verified
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

01

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): \[ \mathrm{N}_{3}^- (aq) + H2O(l) \rightleftharpoons HN(aq) + OH^- (aq) \]
02

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: \[ K_{a} = \frac{[HN][OH^-]}{[N_{3}^-]} \]
03

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 |
04

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

Now, we will substitute our equilibrium values from the ICE table into the \(K_{a}\) expression: \[ 1.9 \times 10^{-5} = \frac{x^2}{0.010 - x} \]
05

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: \[ x \approx \sqrt{1.9 \times 10^{-5} \times 0.010} \approx 1.37 \times 10^{-3} \]
06

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: \([HN] \approx 1.37 \times 10^{-3} M\) \([OH^-] \approx 1.37 \times 10^{-3} M\) \([N_{3}^-] \approx 0.00863 M\)

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Most popular questions from this chapter

Place the species in each of the following groups in order of increasing acid strength. Explain the order you chose for each group. a. \(\mathrm{HIO}_{3}, \mathrm{HBrO}_{3}\) b. \(\mathrm{HNO}_{2}, \mathrm{HNO}_{3}\) c. HOCl, HOI d. \(\mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{H}_{3} \mathrm{PO}_{3}\)

For propanoic acid $\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}, K_{\mathrm{a}}=1.3 \times 10^{-5}\right),$ determine the concentration of all species present, the \(\mathrm{pH},\) and the percent dissociation of a \(0.100-M\) solution.

A typical vitamin \(\mathrm{C}\) tablet (containing pure ascorbic acid, \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{6}\) ) weighs $500 . \mathrm{mg}\( . One vitamin \)\mathrm{C}$ tablet is dissolved in enough water to make 200.0 \(\mathrm{mL}\) of solution. Calculate the \(\mathrm{pH}\) of this solution. Ascorbic acid is a diprotic acid.

Saccharin, a sugar substitute, has the formula $\mathrm{HC}_{7} \mathrm{H}_{4} \mathrm{NSO}_{3}$ and is a weak acid with \(K_{\mathrm{a}}=2.0 \times 10^{-12} .\) If 100.0 \(\mathrm{g}\) of saccharin is dissolved in enough water to make 340 \(\mathrm{mL}\) of solution, calculate the \(\mathrm{pH}\) of the resulting solution.

Which of the following represent conjugate acid-base pairs? For those pairs that are not conjugates, write the correct conjugate acid or base for each species in the pair. a. \(\mathrm{H}_{2} \mathrm{O}, \mathrm{OH}^{-}\) b. \(\mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{SO}_{4}^{2-}\) c. \(\mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) d. $\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}, \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}$
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