Question

For an acid \(\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} ; K_{1}\) and for a base \(\mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{OH} ; K_{2}\) Then (1) \(K_{1} \cdot K_{2}=K_{\mathrm{w}}\) (2) \(\log K_{1}+\log K_{2}=\log K_{\mathrm{w}}\) (3) \(\mathrm{p} K_{1}+\mathrm{p} K_{2}=\mathrm{p} K_{\mathrm{w}}\) (4) all are correct

Short answer

All given statements are correct.

Step-by-step solution

Understand the given reactions

Identify the given acid-base reactions. For the acid: \(\textrm{CH}_{3} \textrm{COOH} + \textrm{H}_{2} \textrm{O} \rightleftharpoons \textrm{CH}_{3} \textrm{COO}^{-} + \textrm{H}_3 \textrm{O}^{+}\) For the base: \(\textrm{CH}_{3} \textrm{COO}^{-} + \textrm{H}_{2} \textrm{O} \rightleftharpoons \textrm{CH}_3 \textrm{COOH} + \textrm{OH}^{-}\) Identify the equilibrium constants, \(K_1\) for the acid reaction and \(K_2\) for the base reaction.

Relationship between \(K_1\) and \(K_2\)

Recognize that the product of the acid and base equilibrium constants equals the ion-product constant for water, \(\textrm{K}_w\). Therefore, \(\textrm{K}_1 \times \textrm{K}_2 = \textrm{K}_w\)

Logarithmic relationship

Recall that the logarithm of a product is the sum of the logarithms: \(\text{log}(\textrm{K}_1 \times \textrm{K}_2) = \text{log}(\textrm{K}_1) + \text{log}(\textrm{K}_2) = \text{log}(\textrm{K}_w)\)

Relationship in pK values

Remember that \(\textrm{p}K\) is the negative logarithm of the equilibrium constant: \(\text{p}K = -\text{log} K \). By this definition, \(\text{p}K_1 + \text{p}K_2 = \text{p}K_w\)

Conclusion

Summarize the relationship between the constants: 1. \(K_1 \times K_2 = K_w\) 2. \(\text{log} K_1 + \text{log} K_2 = \text{log} K_w\) 3. \(\text{p}K_1 + \text{p}K_2 = \text{p}K_w\). This means all given statements are correct.

Key concepts

acid-base equilibrium

Acid-base equilibrium refers to the point at which the rate of the forward reaction (acid dissociation) equals the rate of the reverse reaction (base association). This is a dynamic equilibrium where concentrations of reactants and products remain constant over time.
Consider the reaction for acetic acid: \(\mathrm{CH_{3}COOH} + \mathrm{H_{2}O} \rightleftharpoons \mathrm{CH_{3}COO^{-}} + \mathrm{H_{3}O^{+}}\). Here, acetic acid (\(\mathrm{CH_{3}COOH}\)) dissociates in water to form acetate ions (\(\mathrm{CH_{3}COO^{-}}\)) and hydronium ions (\(\mathrm{H_{3}O^{+}}\)).
At equilibrium, the concentrations of these ions no longer change, and the equilibrium constant \(K_{1}\) characterizes the acid strength.
Similarly, for the base reaction of acetate ions with water: \(\mathrm{CH_{3}COO^{-}} + \mathrm{H_{2}O} \rightleftharpoons \mathrm{CH_{3}COOH} + \mathrm{OH^{-}}\), the equilibrium constant \(K_{2}\) reflects the base strength.

ion-product constant

The ion-product constant for water, denoted as \(\mathrm{K_{w}}\), is a fundamental feature of water's properties at a given temperature. It represents the product of the concentrations of hydronium and hydroxide ions in water. Mathematically, \(\mathrm{K_{w} = [H_{3}O^{+}] [OH^{-}] = 1.0 \times 10^{-14}}\) at 25°C.
This constant establishes a key relationship between acid and base dissociation reactions. For any conjugate acid-base pair, the product of their equilibrium constants (\(K_{1}\) and \(K_{2}\)) always equals \(\mathrm{K_{w}}\). Let's recapture this using the given reactions:
For acetic acid: \(\mathrm{CH_{3}COOH} + \mathrm{H_{2}O} \rightleftharpoons \mathrm{CH_{3}COO^{-}} + \mathrm{H_{3}O^{+}}\) (with \(K_{1}\)); and for acetate ion: \(\mathrm{CH_{3}COO^{-}} + \mathrm{H_{2}O} \rightleftharpoons \mathrm{CH_{3}COOH} + \mathrm{OH^{-}}\) (with \(K_{2}\)).
At equilibrium: \(\mathrm{K_{1} \cdot K_{2} = K_{w}}\)

logarithm relationship in chemistry

Logarithms simplify mathematical relationships involving multiplication and exponents, which is highly useful in chemistry. One important relationship is: \(\log (ab) = \log (a) + \log (b)\).
Applying this to equilibrium constants, we get: \(\log (K_{1} \cdot K_{2}) = \log (K_{1}) + \log (K_{2})\). Because we know \(K_{1} \cdot K_{2} = K_{\mathrm{w}}\), this transforms to: \(\log (K_{\mathrm{w}}) = \log (K_{1}) + \log (K_{2})\), demonstrating that the sum of logarithms of the equilibrium constants equals the logarithm of \(K_{\mathrm{w}}\).
This logarithmic relationship is invaluable for simplifying the multiplicative link between constants in acid-base chemistry.

pK values

pK values provide a more intuitive way to express acidity and basicity. The term \(\mathrm{p}K\) is defined as the negative logarithm of an equilibrium constant: \(\mathrm{p}K = - \log K\).
This means: \(\mathrm{p}K_{1} = - \log K_{1}\) and \(\mathrm{p}K_{2} = - \log K_{2}\). Given the previous logarithm relationships, we have: \(\mathrm{p}K_{\mathrm{w}} = - \log K_{\mathrm{w}}\). So: \(\mathrm{p}K_{1} + \mathrm{p}K_{2} = - \log K_{1} - \log K_{2} = -\log (K_{1} \cdot K_{2}) = - \log K_{\mathrm{w}} = \mathrm{p}K_{\mathrm{w}}\).
This shows that the sum of the pK values for an acid and its conjugate base equals the pK value for water, creating an easy-to-understand and consistent method for comparing the strengths of acids and bases.