Equilibrium Chemistry: Calculating pH and Concentrations of Methylamine Solution

How can we calculate the pH and concentrations of CH3NH2 and CH3NH+3 in a 0.0443 M methylamine solution?

Given the Kb of CH3NH2 is 4.47 x 10^-4, what are the possible values for pH, [CH3NH2], and [CH3NH+3] in the solution?

Answer:

The correct answer is a) pH = 9.22, [CH3NH2] = 0.0443 M, [CH3NH+3] = 4.47 x 10^-4 M.

To calculate the concentrations and pH of a solution, we can use the equilibrium equation, base ionization constant (Kb), and the concept of pH and pOH. By setting up the equilibrium expression, the concentrations can be determined. The pH can then be found by first calculating the pOH and subtracting it from 14.

The pH and concentrations of CH3NH2 and CH3NH+3 in a 0.0443 M methylamine solution can be calculated using the equilibrium equation associated with a base reaction and the concept of Kb (base ionization constant). The methylamine (CH3NH2) can accept an H+ from water, forming the methylammonium ion (CH3NH+3) and hydroxide ion (OH-). We can set up the Kb expression: Kb = [CH3NH+3][OH-]/[CH3NH2].

Given the initial concentration of the methylamine and the value of Kb, we can solve for [CH3NH+3] and [OH-] using the quadratic formula. Then, the pH of the solution can be found by first finding the pOH (which is -log [OH-]) and then subtracting that value from 14 (as pH + pOH = 14).

For a 0.0443 M methylamine solution with a Kb of 4.47 x 10^-4, the correct values for pH, [CH3NH2], and [CH3NH+3] are pH = 9.22, [CH3NH2] = 0.0443 M, and [CH3NH+3] = 4.47 x 10^-4 M.

Understanding equilibrium chemistry and the calculations involved in determining pH and concentrations of solutions can provide valuable insights into chemical reactions and their behavior in solution.