Phenol, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH},\) has a \(K_{a}\) of \(1.3 \times 10^{-10}.\) (a) Write out the \(K_{a}\) reaction for phenol. (b) Calculate \(K_{b}\) for phenol's conjugate base. (c) Is phenol a stronger or weaker acid than water?

Short Answer

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(a) The Ka reaction for phenol is: \(\mathrm{C_{6}H_{5}OH}\enspace\rightleftharpoons\enspace\mathrm{C_{6}H_{5}O^{-}} + \mathrm{H^{+}}\), with the Ka expression being: \(K_{a} = \frac{[\mathrm{C_{6}H_{5}O^{-}}][\mathrm{H^{+}}]}{[\mathrm{C_{6}H_{5}OH}]}\). (b) The Kb value for phenol's conjugate base (C6H5O-) is approximately \(7.69 \times 10^{-5}\). (c) Phenol is a stronger acid than water, as its Ka value (\(1.3 \times 10^{-10}\)) is larger than the Ka value of water (\(1.8 \times 10^{-16}\)).

Step by step solution

01

(a) Ka Reaction for Phenol

To write the Ka reaction for phenol, we need to dissociate it into its ions. Phenol (C6H5OH) will dissociate into a phenol ion (C6H5O-) and a proton (H+). The reaction is: \(\mathrm{C_{6}H_{5}OH}\enspace\rightleftharpoons\enspace\mathrm{C_{6}H_{5}O^{-}} + \mathrm{H^{+}}\) The Ka expression for this reaction is: \(K_{a} = \frac{[\mathrm{C_{6}H_{5}O^{-}}][\mathrm{H^{+}}]}{[\mathrm{C_{6}H_{5}OH}]}\)
02

(b) Calculation of Kb for Phenol's Conjugate Base

First, we need to find the expression for Kb from the Ka. The relationship between Ka and Kb is given by the following equation: \(K_{w} = K_{a} \times K_{b}\) Here, \(K_{w}\) is the ion product for water, which is equal to \(1.0 \times 10^{-14}\) at 25°C. Now, we can calculate the Kb value for the phenol's conjugate base (C6H5O-) by rearranging the above equation: \(K_{b} = \frac{K_{w}}{K_{a}}\) Plugging in the given values, we get: \(K_{b} = \frac{1.0 \times 10^{-14}}{1.3 \times 10^{-10}} = \frac{1.0}{1.3} \times 10^{-4} = 7.69 \times 10^{-5}\) So, the Kb value for the phenol's conjugate base is approximately \(7.69 \times 10^{-5}\).
03

(c) Comparison of Phenol's Acidity to Water

We will determine if phenol is a stronger or weaker acid than water by comparing their Ka values. The Ka value for water is approximately \(1.8 \times 10^{-16}\). Comparing the Ka values, we see that the Ka value of phenol (\(1.3 \times 10^{-10}\)) is larger than the Ka value of water (\(1.8 \times 10^{-16}\)). Since a higher Ka value indicates a stronger acid, phenol is a stronger acid than water.

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Key Concepts

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

Ka Reaction for Phenol
The acid dissociation constant (Ka) measures how completely an acid dissociates in solution. A larger Ka value indicates a stronger acid. For phenol (\(C_{6}H_{5}OH\)), the Ka reaction represents its tendency to donate a proton to the surrounding medium, revealing its acidic nature.

Phenol dissociates as follows: \[C_{6}H_{5}OH \rightleftharpoons C_{6}H_{5}O^{-} + H^{+}\]The above reaction indicates that a molecule of phenol releases a hydrogen ion (\(H^{+}\)), turning into its conjugate base (\(C_{6}H_{5}O^{-}\)). The equilibrium expression for the reaction is:\[K_{a} = \frac{[C_{6}H_{5}O^{-}][H^{+}]}{[C_{6}H_{5}OH]}\]Understanding this concept is crucial to assessing the strength of phenol as an acid compared to other substances. A compound's tendency to lose a proton, quantified by the Ka value, directly influences its reactivity and interactions in chemical processes.
Kb Calculation for Conjugate Base
When phenol loses a proton, it forms a conjugate base, phenolate (\(C_{6}H_{5}O^{-}\)). To assess the basicity of this conjugate base, we compute the Kb value, which represents the base dissociation constant. The Kb calculation requires understanding the inverse relationship between the strength of the acid (Ka) and its conjugate base (Kb). This relationship is conceptualized through the equation:\[K_{w} = K_{a} \times K_{b}\]where \(K_{w}\) is the ion product of water, a constant value of \(1.0 \times 10^{-14} \) at 25°C. To find Kb, rearrange the equation:\[K_{b} = \frac{K_{w}}{K_{a}}\]After substituting the given values for Kw and Ka, the Kb value for the conjugate base of phenol can be calculated. This step is essential for students to evaluate the basicity of various substances and comprehend their behavior in different chemical environments.
Comparing Acidity
To compare the acidity of different substances, we look at their Ka values. Phenol, with a Ka value of \(1.3 \times 10^{-10}\) is considered for comparison. Water, a well-known neutral substance, has a much lower Ka value of \(1.8 \times 10^{-16}\).

These values imply that phenol donates protons more readily than water, classifying it as a stronger acid relative to water. The concept of comparing acidity is fundamental in chemistry, as it influences solubility, reaction rates, and buffer capacities in solutions. Grasping this concept helps students predict how different acids will behave in chemical reactions and how they can alter the environment of a chemical system.

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Most popular questions from this chapter

Consider two solutions, solution \(\mathrm{A}\) and solution \(\mathrm{B} .\left[\mathrm{H}^{+}\right]\) in solution \(\mathrm{A}\) is 250 times greater than that in solution B. What is the difference in the pH values of the two solutions?

Ephedrine, a central nervous system stimulant, is used in nasal sprays as a decongestant. This compound is a weak organic base: $$\mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ON}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ONH}^{+}(a q)+\mathrm{OH}^{-}(a q)$$ A 0.035\(M\) solution of ephedrine has a pH of 11.33 . (a) What are the equilibrium concentrations of \(\mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ON}, \mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ONH}^{+},\) and \(\mathrm{OH}^{-} ?\) (b) Calculate \(K_{b}\) for ephedrine.

Carbon dioxide in the atmosphere dissolves in raindrops to produce carbonic acid \(\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right),\) causing the pH of clean, unpolluted rain to range from about 5.2 to 5.6. What are the ranges of \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) in the raindrops?

A particular sample of vinegar has a pH of \(2.90 .\) If acetic acid is the only acid that vinegar contains \(\left(K_{a}=1.8 \times 10^{-5}\right)\) calculate the concentration of acetic acid in the vinegar.

Predict the stronger acid in each pair: (a) HNO \(_{3}\) or HNO \(_{2};\) \((\mathbf{b})\mathrm{H}_{2} \mathrm{S}\) or \(\mathrm{H}_{2} \mathrm{O} ;(\mathbf{c}) \mathrm{H}_{2} \mathrm{SO}_{4}\) or \(\mathrm{H}_{2} \mathrm{SeO}_{4}(\mathbf{d}) \mathrm{CH}_{3} \mathrm{COOH}\) or \(\mathrm{CCl}_{3}\) \(\mathrm{COOH}.\)

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