Chapter 15: Problem 14
The dissociation constant of a substituted benzoic acid at \(25^{\circ} \mathrm{C}\) is \(1.0 \times 10^{-4}\). The pH of a \(0.01 \mathrm{M}\) solution of its sodium salt is ___ .
Short Answer
Expert verified
The pH of a 0.01 M solution of the sodium salt of substituted benzoic acid is 8.
Step by step solution
01
Understand the Dissociation
The sodium salt of a substituted benzoic acid, when dissolved in water, dissociates completely to produce benzoate anions and sodium cations. Since the solution is of a salt of a weak acid with a strong base, the benzoate anion will hydrolyze in water and affect the pH of the solution.
02
Write the Hydrolysis Reaction
The hydrolysis reaction of the benzoate ion in water can be written as: \(C_6H_5COO^- + H_2O \leftrightarrow C_6H_5COOH + OH^-\).
03
Write the Expression for the Hydrolysis Constant
Using the given dissociation constant of the benzoic acid (Ka), the hydrolysis constant (Kb) for the benzoate ion can be calculated using the relation \(K_w = K_a \times K_b\), where \(K_w\) is the ion-product constant for water at \(25^\circ C\), which is \(1.0 \times 10^{-14}\).
04
Calculate the Hydrolysis Constant (Kb)
Calculate Kb using the relation: \(K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-4}}\), which yields \(K_b = 1.0 \times 10^{-10}\).
05
Write the Hydrolysis Equilibrium Expression
The equilibrium expression for hydrolysis is \(K_b = \frac{[OH^-][C_6H_5COOH]}{[C_6H_5COO^-]}\). We can assume that the concentration of the benzoate ion at equilibrium is approximately equal to the initial concentration of the salt, \(0.01 M\), since the degree of hydrolysis is small.
06
Calculate the Concentration of Hydroxide Ions
Let \(x\) be the concentration of hydroxide ions produced. The expression becomes \(K_b = \frac{x^2}{0.01 - x}\). Since Kb is much smaller than the initial concentration, we can assume \(0.01 - x \approx 0.01\). The expression simplifies to \(K_b = \frac{x^2}{0.01}\). Solving for \(x\), we get \(x = \sqrt{K_b \times 0.01}\).
07
Solve for the Hydroxide Ion Concentration
Plugging in the values we get \(x = \sqrt{1.0 \times 10^{-10} \times 0.01}\) which yields \(x = 1.0 \times 10^{-6} M\).
08
Calculate the pOH of the Solution
The pOH of the solution can be calculated as \(pOH = -\log{[OH^-]}\), which gives us \(pOH = -\log{(1.0 \times 10^{-6})} = 6\).
09
Calculate the pH of the Solution
Since pH + pOH = 14 for aqueous solutions at 25 degrees Celsius, we can find the pH: pH = 14 - pOH, which gives us pH = 14 - 6 = 8.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Dissociation Constant
Understanding the dissociation constant, often symbolized as Ka, is crucial when studying the behavior of weak acids in solution. The dissociation constant provides insight into the degree to which an acid releases protons (H+) when dissolved in water. A higher Ka value indicates a stronger acid, which dissociates more completely into its ions.
For example, a substituted benzoic acid with a dissociation constant of \(1.0 \times 10^{-4}\) suggests it is a weak acid. This weak acid only partially ionizes in water, leading to an equilibrium between the un-ionized acid and the ionized products. Knowing Ka is essential for predicting the behavior of the acid's salts in solution, specifically their hydrolysis and the resulting pH values.
For example, a substituted benzoic acid with a dissociation constant of \(1.0 \times 10^{-4}\) suggests it is a weak acid. This weak acid only partially ionizes in water, leading to an equilibrium between the un-ionized acid and the ionized products. Knowing Ka is essential for predicting the behavior of the acid's salts in solution, specifically their hydrolysis and the resulting pH values.
Hydrolysis of Salts
When a salt derived from a weak acid and a strong base, like the sodium salt of benzoic acid, is dissolved in water, it undergoes hydrolysis. This process involves the salt reacting with water, leading to the formation of the acid and hydroxide ions (OH-).
The reaction for this process can be written as \(C_6H_5COO^- + H_2O \leftrightarrow C_6H_5COOH + OH^-\). The benzoate ions (\(C_6H_5COO^-\)) accept protons from water, slightly increasing the solution's pH due to the formation of OH- ions. A proper understanding of hydrolysis is key to predicting the qualitative changes in pH when salts are dissolved in water.
The reaction for this process can be written as \(C_6H_5COO^- + H_2O \leftrightarrow C_6H_5COOH + OH^-\). The benzoate ions (\(C_6H_5COO^-\)) accept protons from water, slightly increasing the solution's pH due to the formation of OH- ions. A proper understanding of hydrolysis is key to predicting the qualitative changes in pH when salts are dissolved in water.
pOH and pH Relationship
In aqueous solutions, the pH and pOH values are interconnected by the relationship pH + pOH = 14, applicable at 25 degrees Celsius. The pH measures the concentration of hydrogen ions (H+), while pOH measures the concentration of hydroxide ions (OH-).
For instance, if we calculate the pOH of a solution to be 6, as done in the benzoate ion example, this tells us that the solution is basic since pOH values below 7 indicate basicity. To find the pH, we subtract the pOH from 14, giving us a pH higher than 7, affirming the basic nature of the solution.
For instance, if we calculate the pOH of a solution to be 6, as done in the benzoate ion example, this tells us that the solution is basic since pOH values below 7 indicate basicity. To find the pH, we subtract the pOH from 14, giving us a pH higher than 7, affirming the basic nature of the solution.
Buffer Solution Preparation
Buffer solutions are designed to resist changes in pH when small amounts of acids or bases are added. The preparation of a buffer usually involves mixing a weak acid with its conjugate base or a weak base with its conjugate acid. An example includes a mixture of acetic acid and sodium acetate.
Such solutions work through the common ion effect and the equilibrium established between the weak acid/base and its conjugate. By having both present, the addition of H+ ions can be countered by the conjugate base, while the addition of OH- ions can be neutralized by the weak acid, maintaining the pH within a narrow range. Understanding this principle is essential for creating stable environments in many chemical and biological applications.
Such solutions work through the common ion effect and the equilibrium established between the weak acid/base and its conjugate. By having both present, the addition of H+ ions can be countered by the conjugate base, while the addition of OH- ions can be neutralized by the weak acid, maintaining the pH within a narrow range. Understanding this principle is essential for creating stable environments in many chemical and biological applications.