Question

What are \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) in a \(0.344 \mathrm{M}\) solution of \(\mathrm{HNO}_{3}\) ?

Short answer

\([\mathrm{H}^{+}] = 0.344 \,\text{M}\) and \([\mathrm{OH}^{-}] = 2.91 \times 10^{-14} \,\text{M}\).

Step-by-step solution

Identify the Nature of Nitric Acid

Nitric acid (\(\mathrm{HNO}_3\)) is a strong acid, meaning it completely dissociates in water. The dissociation can be represented by the chemical equation: \(\mathrm{HNO}_3 \rightarrow \mathrm{H}^{+} + \mathrm{NO}_3^{-}\). This implies that the concentration of \(\mathrm{H}^+\) will be equal to the initial concentration of \(\mathrm{HNO}_3\).

Determine the Concentration of \(\left[\mathrm{H}^{+}\right]\)

Since \(\mathrm{HNO}_3\) completely dissociates, the concentration of hydrogen ions \(\left[\mathrm{H}^{+}\right]\) in the solution is the same as the concentration of the nitric acid solution, which is \(0.344 \mathrm{M}\).

Apply the Ion Product of Water

For water at 25°C, the ion product \(K_w\) is equal to \(1.0 \times 10^{-14}\). This relation is expressed as \(\left[\mathrm{H}^{+}\right] \times \left[\mathrm{OH}^{-}\right] = 1.0 \times 10^{-14}\). We can use this product to find the concentration of hydroxide ions \(\left[\mathrm{OH}^{-}\right]\).

Calculate \(\left[\mathrm{OH}^{-}\right]\)

Using the ion product of water, \(K_w = 1.0 \times 10^{-14}\), we can find \(\left[\mathrm{OH}^{-}\right]\) by rearranging the equation: \(\left[\mathrm{OH}^{-}\right] = \frac{1.0 \times 10^{-14}}{\left[\mathrm{H}^{+}\right]}\). Substituting \(\left[\mathrm{H}^{+}\right] = 0.344\), we get \[\left[\mathrm{OH}^{-}\right] = \frac{1.0 \times 10^{-14}}{0.344} = 2.91 \times 10^{-14} \mathrm{M}.\]

Key concepts

Dissociation of Strong Acids

When it comes to strong acids like nitric acid (\(\text{HNO}_3\)), they have a fascinating chemical behavior. In solution, these acids dissociate completely. This means that every single molecule of the acid breaks apart into its ions. For \(\text{HNO}_3\), the dissociation can be written as: \(\text{HNO}_3 \rightarrow \text{H}^{+} + \text{NO}_3^{-}\)This process effectively separates the hydrogen ions (\(\text{H}^{+}\)) from the nitrate ions (\(\text{NO}_3^{-}\)). Because strong acids dissociate completely, the concentration of \(\text{H}^{+}\) ions in the solution is identical to the initial concentration of the acid itself.In practical terms, if you start with a solution that is 0.344 M in \(\text{HNO}_3\), this means your \(\text{H}^{+}\) concentration will also be 0.344 M. This straightforward calculation highlights the potency of strong acids in providing hydrogen ions, an important aspect in acid-base chemistry.

Ion Product of Water

Water is not just a passive solvent; it has its own chemical processes going on. A key factor in these processes is its constant ion product, \(K_w\). At 25°C, this is an established value of \(1.0 \times 10^{-14}\). This constant describes the equilibrium concentrations of hydrogen ions \(\left [ \text{H}^+ \right ]\) and hydroxide ions \(\left [ \text{OH}^- \right ]\) in pure water:\[\left [ \text{H}^+ \right ] \times \left [ \text{OH}^- \right ] = 1.0 \times 10^{-14}\]This equation is a powerful tool in acid-base chemistry. It remains stable regardless of the presence or absence of acids and bases in the water. By knowing the concentration of one type of ion, we can find the other by rearranging this equation.For instance, in the case of the 0.344 M nitric acid solution, we can find the \(\left [ \text{OH}^- \right ]\) by knowing \(\left [ \text{H}^+ \right ] = 0.344\) and using the formula:\[\left [ \text{OH}^- \right ] = \frac{1.0 \times 10^{-14}}{0.344}\]

Hydroxide Ion Concentration

The concentration of hydroxide ions (\(\text{OH}^-\)) is just as important as hydrogen ion concentration in understanding a solution's acidity or basicity. In our 0.344 M nitric acid solution, calculating this is simple thanks to the ion product of water:- Start with \(K_w = 1.0 \times 10^{-14}\).- Divide this value by the known \(\left [ \text{H}^+ \right ]\) concentration to find \(\left [ \text{OH}^- \right ]\): \[\left [ \text{OH}^- \right ] = \frac{1.0 \times 10^{-14}}{0.344}\] - Performing this calculation gives us \(\left [ \text{OH}^- \right ] = 2.91 \times 10^{-14} \text{ M}\).This low number reflects the strong acidic nature of the solution. When acids are present, the hydroxide ion concentration drops significantly due to the inverse relationship governed by \(K_w\).This illustrates how even in acidic solutions, we can calculate basic properties, demonstrating the interconnected balance in acid-base chemistry.