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Chapter 43: Problem 4

Write the \(K_{\mathrm{a}}\) expression for \(\mathrm{HNO}_{3}\).

Short Answer

Expert verified
The \(K_{\mathrm{a}}\) expression for \(\mathrm{HNO}_{3}\) is \(K_{\mathrm{a}} = [\mathrm{H}^{+}][\mathrm{NO}_{3}^{-}].\)

Step by step solution

01

Write the Ionization Reaction of HNO3 in Water

First, write the ionization reaction of \(\mathrm{HNO}_{3}\) in water. Since \(\mathrm{HNO}_{3}\) is a strong acid, it ionizes completely in water: \(\mathrm{HNO}_{3}(aq)\) \(\rightarrow\) \(\mathrm{H}^{+}(aq)\) + \(\mathrm{NO}_{3}^- (aq).\)
02

Write the Ka Expression for the Reaction

Next, write the \(K_{\mathrm{a}}\) expression for the reaction. The \(K_{\mathrm{a}}\) for a reaction can be represented as the ratio of the product of the concentrations of the products, each raised to the power of its stoichiometric coefficient, to the product of the concentrations of the reactants, each raised to the power of its stoichiometric coefficient: \(K_{\mathrm{a}} = \frac{[\mathrm{H}^{+}][\mathrm{NO}_{3}^{-}]}{[\mathrm{HNO}_{3}]}.\)
03

Simplify the Ka Expression

Finally, simplify the \(K_{\mathrm{a}}\) expression for \(\mathrm{HNO}_{3}\). Since \(\mathrm{HNO}_{3}\) is a strong acid and ionizes completely in water, the concentration of \(\mathrm{HNO}_{3}\) in the denominator of the \(K_{\mathrm{a}}\) expression will be practically zero. Hence, this simplifies the \(K_{\mathrm{a}} \) expression to: \(K_{\mathrm{a}} = [\mathrm{H}^{+}][\mathrm{NO}_{3}^{-}].\)

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Most popular questions from this chapter

The acids in Table 2 are divided into two groups: those with \(K_{\mathrm{a}}>1\) and those with \(\mathrm{K}_{\mathrm{a}}<1\). Describe each of these two groups in terms of percent dissociation.

Use the value of \(K_{\mathrm{w}}\) to calculate the hydronium ion concentration and the hydroxide ion concentration in pure water.
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