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Sodium hydrogen sulfate, represented chemically as \( ext{NaHSO}_4 \), is an important compound in chemistry. It is often encountered as a white solid, which is highly soluble in water. When it dissolves in water, it dissociates into the sodium ion \( ext{Na}^+ \) and the hydrogen sulfate ion \( ext{HSO}_4^- \).

• Sodium ion \( ( ext{Na}^+) \) is a spectator ion in solution, meaning it doesn't participate in further chemical reactions.


• The hydrogen sulfate ion \( ( ext{HSO}_4^-) \) acts as a weak acid capable of further dissociation.

Understanding the behavior of sodium hydrogen sulfate in solution is crucial for many chemical equilibrium calculations, as seen in the provided exercise. It serves as a bridge between sulfuric acid's first and second dissociation steps, providing insight into the acidity and concentration of ions within a solution.

The dissociation constant, represented by \( K_a \), is a parameter that quantifies the strength of an acid in solution. It is specifically used to describe the extent to which an acid releases protons (\( ext{H}^+ \) ions) when dissolved in water.
The second dissociation constant, \( K_{a2} \), specifically pertains to the second ionization step of an acid that dissociates in multiple steps, like sulfuric acid. For sulfuric acid, this involves the conversion of \( ext{HSO}_4^- \) into \( ext{SO}_4^{2-} \) along with the release of a proton.

• \( K_{a2} \) reflects how readily \( ext{HSO}_4^- \) ionizes in its second stage.


• A smaller \( K_{a} \) value indicates weaker ionization, hence a weaker acid in that step.

For strong acids like sulfuric acid, the first dissociation constant is typically large, indicating almost complete ionization. However, \( K_{a2} \) is significantly smaller, indicating the relative difficulty of releasing the second proton.

Equilibrium constant calculations are essential for understanding the balance between reactants and products in a chemical reaction. Specifically, \( K_a \) values allow us to determine the concentration of ions at equilibrium for dissociation reactions.
To calculate \( K_{a2} \) in this exercise, we started with the relationship between \( ext{H}^+ \) concentration and pH. By converting the given pH to \( [ ext{H}^+] \), we can input these values into the equilibrium constant expression for the second dissociation of sulfuric acid:

• The formula is \( K_{a2} = \frac{[ ext{H}^+][ ext{SO}_4^{2-}]}{[ ext{HSO}_4^-]} \).


• As \( [ ext{SO}_4^{2-}] = [ ext{H}^+] \), it simplifies the calculation.


• Substituting known values and solving gives \( K_{a2} \approx 1.12 \times 10^{-2} \).

This calculation illustrates how equilibrium constants help us quantify the composition of a solution and predict how changes in conditions will affect the chemical system.

Sulfuric acid \( ( ext{H}_2 ext{SO}_4) \) is a strong acid and a vital industrial chemical with many applications. It is characterized by its two dissociation steps, where it first donates one proton and then the second, transitioning from \( ext{H}_2 ext{SO}_4 \) to \( ext{HSO}_4^- \) and finally to \( ext{SO}_4^{2-} \).

• The first dissociation involves the release of a proton to form \( ext{HSO}_4^- \), which is almost complete in aqueous solutions.


• The second dissociation, involving \( ext{HSO}_4^- \) losing a proton to form \( ext{SO}_4^{2-} \), is partial.


• This two-step dissociation is a key reason for sulfuric acid's effectiveness in processes like manufacturing fertilizers, chemicals, and petroleum products.

Understanding sulfuric acid’s dissociation pattern is crucial, as it affects how the acid behaves in solution, including changes in pH and ion concentration, pivotal for chemical equilibrium scenarios.