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Chapter 13: Problem 85

At \(25^{\circ} \mathrm{C}\), a \(0.20 \mathrm{M}\) solution of methylamine \(\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)\) is \(5.0 \%\) ionized. What is \(K_{b}\) for methylamine?

Short Answer

Expert verified
Answer: The Kb value for the methylamine solution is approximately 5.26 x 10^(-4).

Step by step solution

01

Understand the Ionization Reaction

Methylamine (CH3NH2) is a weak base. When it undergoes ionization, it reacts with water (H2O) to produce a hydroxide ion (OH-) and a methylammonium ion (CH3NH3+). The ionization equation is: CH3NH2 + H2O ↔ CH3NH3+ + OH-
02

Calculate Initial Ionized Species Concentrations

Since the solution is initially 5.0% ionized, we can calculate the concentration of CH3NH2 that has ionized: 0.05 * 0.20 M CH3NH2 = 0.01 M CH3NH2 ionized The ion concentrations will be the same for CH3NH3+ and OH- since they are produced in a 1:1 ratio: [CH3NH3+] = [OH-] = 0.01 M
03

Calculate the Remaining CH3NH2 Concentration

Subtract the ionized CH3NH2 concentration from the initial concentration to find the remaining CH3NH2 concentration: 0.20 M - 0.01 M = 0.19 M CH3NH2
04

Calculate Kb using Ion Concentrations and Base Equation

Now, we can use the calculated concentrations and the base-dissociation constant equation to find Kb: Kb = ([CH3NH3+] [OH-]) / [CH3NH2] Kb = (0.01 M * 0.01 M) / 0.19 M Kb ≈ 5.26 x 10^(-4) The Kb value for methylamine is approximately 5.26 x 10^(-4).

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