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Chapter 13: Problem 55

Phenol, once known as carbolic acid, \(\mathrm{HC}_{6} \mathrm{H}_{5} \mathrm{O}\), is a weak acid. It was one of the first antiseptics used by Lister. Its \(K_{\mathrm{a}}\) is \(1.1 \times 10^{-10} .\) A solution of phenol is prepared by dissolving \(14.5 \mathrm{~g}\) of phenol in enough water to make \(892 \mathrm{~mL}\) of solution. For this solution, calculate (a) \(\mathrm{pH}\) (b) \% ionization

Short Answer

Expert verified
Question: Calculate the pH and percentage ionization of a phenol solution with given mass (14.5 g) and volume (892 mL). Answer: (a) The pH of the phenol solution is 5.36. (b) The percentage ionization of the phenol solution is approximately 0.00254%.

Step by step solution

01

Molarity Calculation

To calculate the molarity of the solution, we first need to convert the mass of phenol into moles and then use the volume of the solution. For phenol, \(\mathrm{HC}_{6} \mathrm{H}_{5} \mathrm{O}\), the molar mass is \(94.11 \mathrm{~g/mol}\). To calculate moles, we will use the following formula: moles = mass / molar mass. moles = (14.5 g) / (94.11 g/mol) = 0.154 moles Now, convert the volume of the solution into liters: 892 mL = 0.892 L. Finally, calculate the molarity using moles and volume: M = moles / volume = (0.154 moles) / (0.892 L) = 0.173 M
02

ICE Table

An ICE (Initial, Change, Equilibrium) table will help us track the concentrations of all species involved in the reaction. Define the initial concentrations as [HC6H5O] = 0.173 M, [H+] = [C6H5O-] = 0. The Ka expression for phenol is: \(K_a = \dfrac{[H+][C_{6}H_{5}O^-]}{[HC_{6}H_{5}O]}\) Let x represent the change in concentrations at equilibrium. The ICE table is as follows: Initial: 0.173 0 0 Change: -x +x +x Equilibrium: 0.173-x x x
03

Solving for x

Plug the Ka expression into the ICE table: \(1.1\times10^{-10} = \dfrac{x^2}{0.173 - x}\) Given that Ka is very small, we can make the assumption that x will be small compared to 0.173. So, we can write: \(1.1\times10^{-10} \approx \dfrac{x^2}{0.173}\) Now, solve for x: \(x^2 = (1.1\times10^{-10})(0.173)\) \(x = \sqrt{(1.1\times10^{-10})(0.173)}\) \(x = 4.4\times10^{-6}\)
04

Calculate pH

We found that [H+] = x = 4.4\(\times10^{-6}\). To calculate pH, we use the formula: pH = \(-\log[H+]\): pH = - log(4.4\(\times10^{-6}\)) ≈ 5.36
05

Calculate Percentage Ionization

To calculate the percentage ionization, we will use the following formula: % ionization = \( \dfrac{[H+]_{\textit{at}~\textit{equilibrium}}}{[HC_{6}H_{5}O]_{\textit{initial}}} \times 100\) % ionization = \(\dfrac{4.4\times10^{-6}}{0.173} \times 100\) % ionization ≈ 0.00254% (a) The pH of the solution is 5.36. (b) The percentage ionization is approximately 0.00254%.

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Key Concepts

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

Phenol Properties
Phenol, also known as hydroxybenzene, is a compound that plays a significant role in organic chemistry due to its aromatic ring and hydroxyl group. This structure lends phenol certain properties that make it both useful and interesting in chemical contexts.

  • Solubility: Phenol is moderately soluble in water, but it is more soluble in organic solvents due to its aromatic ring.
  • Acidity: Phenol is a weak acid, which means that it does not fully dissociate in water. This partial ionization is due to the stability of the phenoxide anion that forms when phenol donates a hydrogen ion (proton).
  • Antiseptic Nature: Historically, phenol has been used as an antiseptic, as referenced in the original exercise mentioning Lister. However, due to its toxic effects at higher concentrations, its use in this capacity has diminished.

Understanding these properties helps explain why phenol ionizes only slightly in water.
Acid Ionization Constant (Ka)
The acid ionization constant, or Ka, is a quantitative measure of the strength of an acid in solution. It is the equilibrium constant for the chemical reaction in which an acid donates a proton to water to form its conjugate base and the hydronium ion (H3O+).

The general expression for Ka of a weak acid HA in water is:
\[ K_a = \frac{[A^-][H_3O^+]}{[HA]} \]

For phenol, the expression becomes:
\[ K_a = \frac{[C_6H_5O^-][H^+]}{[HC_6H_5O]} \]

A smaller Ka value implies a weaker acid that ionizes less in water. In the example of phenol, with its Ka of \(1.1 \times 10^{-10}\), the small value indicates that phenol is only slightly ionized in aqueous solution, which aligns with its classification as a weak acid.
Percentage Ionization
Percentage ionization quantifies the extent to which an acid ionizes in a particular solution. It is especially relevant for weak acids like phenol, which do not fully dissociate in water. This metric is found by comparing the concentration of ionized acid to the initial concentration of the acid before ionization occurs.

The formula to calculate percentage ionization is:
\[ \text{% ionization} = \left(\frac{[\text{Ionized Acid}]}{[\text{Initial Acid}]}\right) \times 100 \% \]

In the exercise involving phenol, the concentration of ionized acid (hydrogen ions) at equilibrium (\(x\)) is used to gauge how much of the phenol has ionized. Given that only a tiny fraction ionizes, as evidenced by the calculations, the percentage ionization is extremely low, which illustrates the weak nature of phenol as an acid. The concept of percentage ionization is critical in understanding how different concentrations and strengths of acids behave in solution.

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Most popular questions from this chapter

According to the Brønsted-Lowry theory, which of the following would you expect to act as an acid? Which as a base? (a) \(\mathrm{CHO}_{2}^{-}\) (b) \(\mathrm{NH}_{4}^{+}\) (c) \(\mathrm{HSO}_{3}^{-}\)

According to the Bronsted-Lowry theory, which of the following would you expect to act as an acid? Which as a base? (a) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}{ }^{+}\) (b) \(\mathrm{HClO}\) (c) \(\mathrm{CN}^{-}\)

Write the overall chemical equation and calculate \(K\) for the complete ionization of oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\)

The \(\mathrm{pH}\) of a household ammonia cleaning solution is \(11.68\). How many grams of ammonia are needed in a 1.25-L solution to give the same pH?

Consider the process $$\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \quad \Delta H^{\circ}=55.8 \mathrm{~kJ}$$ (a) Will the \(\mathrm{pH}\) of pure water at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) be \(7.0 ?\) (b) If not, calculate the \(\mathrm{pH}\) of pure water at \(37^{\circ} \mathrm{C}\).
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