Suppose that 50.0 mL of 0.25 M CH3NH2(aq) is titrated with 0.35 M HCl(aq).
(a) What is the initial pH of the 0.25 M CH3NH2(aq)?
(b) What is the pH after the addition of 15.0 mL of 0.35 M HCl(aq)?
(c) What volume of 0.35 M HCl(aq) is required to reach halfway to the stoichiometric point?
mL
(d) Calculate the pH at the half stoichiometric point.
(e) What volume of 0.35 M HCl(aq) is required to reach the stoichiometric point?
mL
(f) Calculate the pH at the stoichiometric point.
You work this th same way as the C6H5COOH problem EXCEPT this one is a base and not an acid. That makes the hydrolysis equation slight different.
To answer these questions, we need to understand the concept of titration and the behavior of the acid and base involved.
In this case, we have a titration between a weak base, CH3NH2(aq), and a strong acid, HCl(aq). The goal is to determine various pH values at different points throughout the titration process.
(a) To find the initial pH of the CH3NH2(aq) solution, we need to determine if it is a basic, acidic, or neutral solution. CH3NH2 is a weak base, so we can use the Kb expression to find the concentration of hydroxide ions (OH-) and then convert it to pH.
Kb = [CH3NH3+][OH-] / [CH3NH2]
Since the initial concentration of CH3NH2 is 0.25 M, we can assume that the concentration of CH3NH2 remains 0.25 M until the equivalence point.
We can calculate OH- concentration using Kb and then convert it to pH using the pOH-pH relationship: pH = 14 - pOH.
(b) After the addition of 15.0 mL of 0.35 M HCl(aq), we need to calculate the new concentrations of CH3NH2 and HCl in the solution. This will involve stoichiometry and the concept of limiting reagent. Once we have the new concentrations, we can calculate the pOH and then convert it to pH.
(c) To determine the volume of HCl required to reach halfway to the stoichiometric point, we need to consider the stoichiometry of the reaction between CH3NH2 and HCl. The stoichiometric point occurs when the moles of HCl added equals the moles of CH3NH2 initially present. Halfway to the stoichiometric point means half of the moles of CH3NH2 have reacted. We can use the balanced chemical equation and stoichiometry to calculate the volume.
(d) At the half stoichiometric point, we can calculate the moles and concentrations of CH3NH2 and its conjugate acid, CH3NH3+. The solution will contain a mixture of CH3NH2 and CH3NH3+. We can use the Henderson-Hasselbalch equation to calculate the pH of the solution.
(e) To find the volume of HCl required to reach the stoichiometric point, we need to consider the stoichiometry again. The stoichiometric point occurs when the moles of HCl added equal the moles of CH3NH2 initially present. We can use the balanced chemical equation and stoichiometry to calculate the volume.
(f) At the stoichiometric point, we can calculate the moles and concentrations of the products, NH4+ and Cl-. The solution will contain NH4+ and Cl- ions. We can use the concentration of NH4+ to calculate the pOH and then convert it to pH.
Overall, to answer these questions, you will need to apply concepts like stoichiometry, acid-base equilibrium, and the Henderson-Hasselbalch equation.