The \(K_{a}\) of nitrous acid, \(\mathrm{HNO}_{2},\) is \(6.76 \times 10^{-4} .\) Write the equation describing the equilibrium established when \(\mathrm{HNO}_{2}\) reacts with \(\mathrm{NH}_{3} .\) Use unequal arrows to indicate whether reactants or products are favored.

Short Answer

Expert verified
The reaction of nitrous acid with ammonia is given by: \(\mathrm{HNO}_{2} + \mathrm{NH}_{3} \longleftarrow \mathrm{NH}_{4}^{+} + \mathrm{NO}_{2}^{-}\). Since the \(K_{a}\) value is \(6.76 \times 10^{-4}\), less than 1, the equilibrium favors the reactants, represented by a longer arrow pointing to the left.

Step by step solution

01

Write the acid-base reaction

First, we model the reaction of nitrous acid (\(\mathrm{HNO}_{2}\)) with ammonia (\(\mathrm{NH}_{3}\)). In this reaction, the nitrous acid donates a proton (H+) to the ammonia, resulting in ammonium (\(\mathrm{NH}_{4}^{+}\)) and nitrite ion (\(\mathrm{NO}_{2}^{-}\)). The general form of the reaction is: \[\mathrm{HNO}_{2} + \mathrm{NH}_{3} \leftrightharpoons \mathrm{NH}_{4}^{+} + \mathrm{NO}_{2}^{-}\]
02

Analyze the reaction based on \(K_{a}\) value

Next, recall that the equilibrium constant \(K_{a}\) for nitrous acid is given as \(6.76 \times 10^{-4}\), which is less than 1. This means that the equilibrium favors the reactants more than the products. As more reactants tend to get converted into products, the reaction shifts back to reactants to maintain the equilibrium. In other words, the reactants \(\mathrm{HNO}_{2}\) and \(\mathrm{NH}_{3}\) are favored over the products \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{NO}_{2}^{-}\). Therefore, the reaction proceeds largely to the left and can be represented by unequal arrows: \[\mathrm{HNO}_{2} + \mathrm{NH}_{3} \longleftarrow \mathrm{NH}_{4}^{+} + \mathrm{NO}_{2}^{-}\]

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