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Butyric acid is responsible for the foul smell of rancid butter. The \(\mathrm{pK}_{b}\) of the butyrate ion is 9.16. (a) Calculate the \(K_{a}\) for butyric acid. (b) Calculate the pH of a \(0.075 \mathrm{M}\) solution of butyric acid. (c) Calculate the \(\mathrm{pH}\) of a \(0.075 \mathrm{M}\) solution of sodium butyrate.

Short Answer

Expert verified
The short answer to the given problem is: (a) After calculating, we find the \(K_a\) of butyric acid to be \(7.24 \times 10^{-5}\). (b) The pH of the \(0.075 \mathrm{M}\) butyric acid solution is approximately 2.52. (c) The \(\mathrm{pH}\) of the \(0.075 \mathrm{M}\) sodium butyrate solution is approximately 9.42.

Step by step solution

01

Calculate the \(K_{a}\) for butyric acid

To determine the \(K_a\) for butyric acid, first, calculate the \(K_b\) of the butyrate ion. The relationship between \(\mathrm{pK}_{b}\) and \(K_{b}\) can be described as follows: \[K_b = 10^{-\mathrm{pK}_b}\] Now, substitute the given \(\mathrm{pK}_{b}\) value (9.16) and calculate the \(K_{b}\) value: \[K_b = 10^{-9.16}\] After obtaining the \(K_b\) value, use the ion product of water to calculate the \(K_a\) value for butyric acid. It can be determined using the formula: \[K_a \times K_b = K_w\] Where \(K_w\) is the ion product of water (\(1 \times 10^{-14}\)). Now, solve for \(K_a\): \[K_a = \frac{K_w}{K_b}\]
02

Calculate the pH of the \(0.075 \mathrm{M}\) butyric acid solution

To calculate the pH of the butyric acid solution, we can use the formula for a weak acid: \[ K_a = \frac{[\mathrm{H}^+][\mathrm{A}^-]}{[\mathrm{HA}]}\] Here, \([\mathrm{H}^+]\) represents the concentration of hydrogen ions, \([\mathrm{A}^-]\) is the concentration of the conjugate base and \([\mathrm{HA}]\) is the concentration of the weak acid. Since the initial concentration of the butyric acid is given as \(0.075 \mathrm{M}\), we can use the approximation method considering that the dissociation of the weak acid is small. Thus, the equation becomes: \[ K_a = \frac{x^2}{0.075-x}\] Where \(x\) represents the concentration of \([\mathrm{H}^+]\). Solve for \(x\) and then calculate the pH using the formula: \[pH = -\log_{10} [\mathrm{H}^+]\]
03

Calculate the \(\mathrm{pH}\) of the \(0.075 \mathrm{M}\) sodium butyrate solution

Sodium butyrate is the sodium salt of butyric acid, which means it will dissociate into butyrate ions and sodium ions in water. Since butyrate ions are basic, we already know the \(K_b\) value. Use the following formula to find the concentration of the hydroxide ions (\([\mathrm{OH}^-]\)) in the solution: \[ K_b = \frac{[\mathrm{OH}^-][\mathrm{HA}]}{[\mathrm{A}^-]}\] Given the initial concentration of sodium butyrate as \(0.075 \mathrm{M}\), we can follow the same approximation method, treating the dissociation as small, and write the equation as: \[ K_b = \frac{x^2}{0.075 - x}\] Solve for \(x\), which represents the concentration of hydroxide ions (\([\mathrm{OH}^-]\)). After obtaining the concentration of hydroxide ions, calculate the \(pOH\) using the formula: \[pOH = -\log_{10} [\mathrm{OH}^-]\] Then, calculate the pH using the relationship between pH and pOH: \[pH + pOH = 14\]

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Key Concepts

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

pH Calculation
Understanding how to calculate the pH of a solution is crucial, especially when dealing with acids and bases. The pH indicates the acidity or alkalinity of a solution, with a lower pH indicating higher acidity. In the calculations for butyric acid, we first consider it as a weak acid. To find the pH, the degree of dissociation is noted, particularly for weak acids, which don't fully dissociate in water.The formula used is\[: K_a = \frac{[\mathrm{H}^+][\mathrm{A}^-]}{[\mathrm{HA}]}\]which indicates the relationship between the concentrations of protons and other ions in the solution.Once the hydrogen ion concentration \([\mathrm{H}^+]\) is determined, the pH can simply be calculated using:\[pH = -\log_{10} [\mathrm{H}^+]\]By understanding these formulas, we can predict the pH values of solutions, which is a key step in acid-base equilibria.
Weak Acids and Bases
Weak acids, like butyric acid, do not completely dissociate in solution. This partial dissociation distinguishes them from strong acids, which dissociate fully.With weak acids, equilibrium is established between the acid and its conjugate base in water. The dissociation can be represented as:\[\mathrm{HA} \leftrightarrow \mathrm{H}^+ + \mathrm{A}^-\]Where \(\mathrm{HA}\) is the weak acid, \(\mathrm{H}^+\) is the hydrogen ion, and \(\mathrm{A}^-\) is the conjugate base.When calculating their effects in solutions, using the equilibrium concentration is a must. This involves some approximations, like stating that the change in concentration is minimal due to the weak dissociation, which simplifies calculations.
Dissociation Constant
The dissociation constant, represented by \(K_a\) for acids and \(K_b\) for bases, helps determine the strength of an acid or base.For butyric acid, which is a weak acid, the \(K_a\) value is calculated from the known \(K_b\) of its conjugate base, the butyrate ion, through the water ion product formula:\[K_a \times K_b = K_w\]where \(K_w\) is the ion product of water, typically \(1 \times 10^{-14}\).Knowing \(K_a\) allows us to understand the extent to which an acid dissociates in water, which is essential when dealing with acid-base equilibria.Using these constants, you can predict other aspects of the solution, such as calculating the pH accurately and understanding the behavior of acids and bases in different environmental conditions.

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Most popular questions from this chapter

Saccharin, a sugar substitute, is a weak acid with \(\mathrm{p} K_{a}=2.32\) at \(25^{\circ} \mathrm{C}\). It ionizes in aqueous solution as follows: $$ \mathrm{HNC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{NC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}^{-}(a q) $$ What is the pH of a \(0.25 \mathrm{M}\) solution of this substance?

Based on their compositions and structures and on conjugate acid-base relationships, select the stronger base in each of the following pairs: (a) \(\mathrm{BrO}^{-}\) or \(\mathrm{ClO}^{-},(\mathbf{b}) \mathrm{BrO}^{-}\) or \(\mathrm{BrO}_{2}^{-}\), (c) \(\mathrm{HPO}_{4}^{2-}\) or \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\).

(a) Using dissociation constants from Appendix D, determine the value for the equilibrium constant for each of the following reactions. (i) \(\mathrm{HCO}_{3}^{-}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (ii) \(\mathrm{NH}_{4}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) \rightleftharpoons \mathrm{NH}_{3}(a q)+\mathrm{HCO}_{3}^{-}(a q)\) (b) We usually use single arrows for reactions when the forward reaction is appreciable ( \(K\) much greater than 1) or when products escape from the system, so that equilibrium is never established. If we follow this convention, which of these equilibria might be written with a single arrow?

Calculate the molar concentration of \(\mathrm{OH}^{-}\) in a \(0.724 \mathrm{M}\) solution of hypobromite ion \(\left(\mathrm{BrO}^{-} ; K_{b}=4.0 \times 10^{-6}\right) .\) What is the \(\mathrm{pH}\) of this solution?

What is the boiling point of a \(0.10 \mathrm{M}\) solution of \(\mathrm{NaHSO}_{4}\) if the solution has a density of \(1.002 \mathrm{~g} / \mathrm{mL}\) ?

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