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Show that \(K_{\mathrm{a}} \times K_{\mathrm{b}}=K_{\mathrm{w}}\) by determining the expressions for these two reactions and multiplying them together. $$ \mathrm{HX}(\mathrm{aq}) \rightleftarrows \mathrm{H}^{+}(\mathrm{aq})+\mathrm{X}^{-}(\mathrm{aq}) $$ \(\mathrm{X}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftarrows \mathrm{HX}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})\)

Short Answer

Expert verified
By multiplying the expressions for \(K_\mathrm{a}\) and \(K_\mathrm{b}\), the result is \(K_\mathrm{w}\). This confirms the relation \(K_\mathrm{a} \times K_\mathrm{b} = K_\mathrm{w}\).

Step by step solution

01

Write the Expression for the Acid Dissociation Constant (Ka)

The reaction for the dissociation of acid \( \mathrm{HX} \) is given by:\[\mathrm{HX} \rightleftarrows \mathrm{H}^{+} + \mathrm{X}^{-}\]The equilibrium constant expression \( K_\mathrm{a} \) for this reaction is:\[K_\mathrm{a} = \frac{[\mathrm{H}^{+}][\mathrm{X}^{-}]}{[\mathrm{HX}]}\]
02

Write the Expression for the Base Dissociation Constant (Kb)

The reaction where the base \( \mathrm{X}^{-} \) reacts with water is given by:\[\mathrm{X}^{-} + \mathrm{H}_{2}\mathrm{O} \rightleftarrows \mathrm{HX} + \mathrm{OH}^{-}\]The equilibrium constant expression \( K_\mathrm{b} \) for this base reaction is:\[K_\mathrm{b} = \frac{[\mathrm{HX}][\mathrm{OH}^{-}]}{[\mathrm{X}^{-}]}\]
03

Multiply (Ka) and (Kb)

To find \( K_\mathrm{a} \times K_\mathrm{b} \), we multiply the expressions:\[K_\mathrm{a} \times K_\mathrm{b} = \left( \frac{[\mathrm{H}^{+}][\mathrm{X}^{-}]}{[\mathrm{HX}]} \right) \times \left( \frac{[\mathrm{HX}][\mathrm{OH}^{-}]}{[\mathrm{X}^{-}]} \right)\]Upon simplifying, we cancel \([\mathrm{HX}]\) and \([\mathrm{X}^{-}]\) in numerator and denominator:\[K_\mathrm{a} \times K_\mathrm{b} = [\mathrm{H}^{+}][\mathrm{OH}^{-}]\]
04

Relate to the Ion Product of Water (Kw)

The expression \([\mathrm{H}^{+}][\mathrm{OH}^{-}]\) is known as the ion product of water, denoted as \( K_\mathrm{w} \), where:\[K_\mathrm{w} = [\mathrm{H}^{+}][\mathrm{OH}^{-}]\]Therefore, by multiplying the equilibrium expressions, we have shown that:\[K_\mathrm{a} \times K_\mathrm{b} = K_\mathrm{w}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dissociation Constant
The dissociation constant, often denoted as \(K_{\mathrm{a}}\) for acids and \(K_{\mathrm{b}}\) for bases, is a fundamental concept in chemistry that describes the extent to which an acid or base dissociates in solution. For an acid, this process involves the acid molecule donating a proton (\(\mathrm{H}^{+}\)) to the surrounding water molecules, forming its conjugate base. The dissociation constant helps us understand how an acid behaves in an aqueous solution.

For the acid \(\mathrm{HX}\), the reaction can be written as:
  • \(\mathrm{HX} \rightleftharpoons \mathrm{H}^{+} + \mathrm{X}^{-}\)
Here, \(\mathrm{HX}\) dissociates into \(\mathrm{H}^{+}\) and \(\mathrm{X}^{-}\). The equilibrium constant expression \(K_{\mathrm{a}}\) for this reaction is given by:
  • \(K_{\mathrm{a}} = \frac{[\mathrm{H}^{+}][\mathrm{X}^{-}]}{[\mathrm{HX}]}\)
This expression helps us calculate the concentration of hydrogen ions in the solution, which determines the solution's pH. \(K_{\mathrm{a}}\) is crucial for comparing the strengths of different acids. A larger \(K_{\mathrm{a}}\) value indicates a stronger acid, meaning it dissociates more in solution.
Equilibrium Constant
The equilibrium constant, denoted generally as \(K\), plays a major role in determining the direction and extent of a chemical reaction at equilibrium. When dealing with acids and bases, the equilibrium constant can specifically refer to \(K_{\mathrm{a}}\) for acids and \(K_{\mathrm{b}}\) for bases.

For the reaction involving the base \(\mathrm{X}^{-}\), it can be written as:
  • \(\mathrm{X}^{-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{HX} + \mathrm{OH}^{-}\)
Here, \(\mathrm{X}^{-}\) acts as the base, accepting a proton from water to form \(\mathrm{HX}\) and hydroxide ions (\(\mathrm{OH}^{-}\)). The equilibrium constant expression \(K_{\mathrm{b}}\) for this reaction is:
  • \(K_{\mathrm{b}} = \frac{[\mathrm{HX}][\mathrm{OH}^{-}]}{[\mathrm{X}^{-}]}\)
The value of \(K_{\mathrm{b}}\) provides insight into the base's ability to produce \(\mathrm{OH}^{-}\) ions, influencing the solution's basicity. A larger \(K_{\mathrm{b}}\) indicates a stronger base. Understanding \(K_{\mathrm{a}}\) and \(K_{\mathrm{b}}\) helps chemists predict how acids and bases will interact in different chemical environments.
Ion Product of Water
The ion product of water, denoted as \(K_{\mathrm{w}}\), is a special equilibrium constant that reflects the self-ionization of water. This process is essential for understanding acid-base equilibrium in any aqueous solution. The expression for \(K_{\mathrm{w}}\) is derived from the equilibrium concentrations of hydrogen ions (\(\mathrm{H}^{+}\)) and hydroxide ions (\(\mathrm{OH}^{-}\)) formed in the reaction.

For pure water, the self-ionization reaction can be written as:
  • \(\mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{H}^{+} + \mathrm{OH}^{-}\)
The equilibrium constant expression for this reaction, \(K_{\mathrm{w}}\), is defined as:
  • \(K_{\mathrm{w}} = [\mathrm{H}^{+}][\mathrm{OH}^{-}]\)
At 25 °C, \(K_{\mathrm{w}}\) has a value of approximately \(1.0 \times 10^{-14}\). This value is constant and applies to any aqueous solution, regardless of whether the solution is neutral, acidic, or basic. The relationship between \(K_{\mathrm{a}}\), \(K_{\mathrm{b}}\), and \(K_{\mathrm{w}}\) explains how the strengths of acids and bases are interrelated: \(K_{\mathrm{a}} \times K_{\mathrm{b}} = K_{\mathrm{w}}\). Understanding \(K_{\mathrm{w}}\) is vital for calculating the pH and pOH of a solution, thereby predicting the solution’s acidity or basicity.

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