The pH of a solution prepared by dissolving 0.350 mol of solid methylamine hydrochloride (CH₃NH₃Cl)in 1.00 L of 1.10 M methylamine (CH₃NH₂)is __________. The K_b for methylamine is 4.40 × 10^-4. (Assume the final volume is 1.00 L.) A)1.66 B)2.86 C)10.28 D)11.14 E)10.61

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Quizplus · Accepted Answer

First, let's write the balanced chemical equation for the reaction between methylamine and its conjugate acid, methylamine hydrochloride: CH3NH3+ (aq) + H2O (l) ↔ CH3NH2 (aq) + H3O+ (aq) Using the given values, we can calculate the initial concentration of CH3NH2 before any reaction takes place: Initial [CH3NH2] = 1.10 M To calculate the pH of the solution, we need to determine the concentration of H3O+ ions at equilibrium. This can be done by using the equilibrium constant expression for the above equation: K_b = [CH3NH2][H3O+] / [CH3NH3+] Since the initial concentration of CH3NH3+ is equal to the amount of solid CH3NH3Cl dissolved, which is 0.350 mol/L, we can substitute in the known values: 4.40 × 10^-4 = [CH3NH2][H3O+] / (0.350 M) We can solve for [H3O+]: [H3O+] = (4.40 × 10^-4)(0.350 M) / [CH3NH2] To determine [CH3NH2], we need to use the stoichiometry of the reaction. Since 1 mole of CH3NH3+ reacts with 1 mole of H2O to produce 1 mole of CH3NH2 and 1 mole of H3O+, we can assume that the final concentration of CH3NH2 is equal to the initial concentration of CH3NH3+ that reacted: [CH3NH2] = 0.350 M - [H3O+] Substituting this expression into the equation for [H3O+], we get: [H3O+] = (4.40 × 10^-4)(0.350 M) / (0.350 M - [H3O+]) Simplifying the expression: [H3O+]2 - (4.40 × 10^-4)(0.350 M)[H3O+] - (4.40 × 10^-4)(0.350 M) = 0 This is a quadratic equation that can be solved using the quadratic formula. The positive root corresponds to the concentration of H3O+ at equilibrium: [H3O+] = 1.74 × 10^-12 M Finally, we can calculate the pH: pH = -log[H3O+] = 11.14

The PH of a Solution Prepared by Dissolving 0