Write equations showing how each weak base ionizes water to form OH–. Also write the corresponding expression for Kb. c. C2H5NH2
Ch.17 - Acids and Bases
Chapter 17, Problem 92
Amphetamine (C9H13N) is a weak base with a pKb of 4.2. Calculate the pH of a solution containing an amphetamine concentration of 225 mg/L.
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Convert the concentration of amphetamine from mg/L to mol/L by dividing the mass concentration by the molar mass of amphetamine (C9H13N).
Use the formula for the base dissociation constant, Kb, which is related to pKb by the equation: \( K_b = 10^{-pK_b} \).
Set up the equilibrium expression for the dissociation of amphetamine in water: \( C_9H_{13}N + H_2O \rightleftharpoons C_9H_{13}NH^+ + OH^- \).
Use the initial concentration of amphetamine and the expression for Kb to set up an ICE (Initial, Change, Equilibrium) table and solve for the concentration of OH⁻ ions at equilibrium.
Calculate the pOH of the solution using the concentration of OH⁻ ions, and then convert pOH to pH using the relation: \( pH + pOH = 14 \).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Weak Bases and pKb
A weak base is a substance that partially ionizes in solution, establishing an equilibrium between the un-ionized base and its ions. The pKb value indicates the strength of the base; a lower pKb corresponds to a stronger base. In this case, amphetamine has a pKb of 4.2, which helps determine its ability to accept protons in solution.
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pH and pOH Relationship
pH is a measure of the hydrogen ion concentration in a solution, while pOH measures the hydroxide ion concentration. The relationship between pH and pOH is given by the equation pH + pOH = 14 at 25°C. Understanding this relationship is crucial for calculating the pH of a solution containing a weak base like amphetamine.
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Concentration and Molarity
Concentration refers to the amount of solute present in a given volume of solution, often expressed in molarity (moles per liter). To calculate the pH of the amphetamine solution, it is necessary to convert the given mass (225 mg/L) into molarity by using the molar mass of amphetamine. This conversion is essential for applying the equilibrium expressions related to weak bases.
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